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NCERT Exemplar Class 12 Maths Solutions Chapter 3 Matrices
NCERT Exemplar Class 12 Maths Solutions Chapter 3, Matrix is one of the most interesting chapters to study. Matrices are much faster and more efficient than the usual direct-solving method. In our daily lives, we see tables, spreadsheets, seating charts, and cinema bookings arranged in rows and columns resembling matrices. So, what is a matrix? A matrix is a rectangular arrangement of numbers, symbols, or expressions in rows and columns . The numbers inside a matrix are called its elements . It is usually written as A=[aij], where aij denotes the element in the i -th row and j -th column. Matrices are used in various fields like computer graphics, economics, and engineering to solve systems of equations, data representation, and transformations. They are a key concept in linear algebra.
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- The Sub-Topics That Are Covered Under The Class 12 Maths NCERT Exemplar Solutions Chapter 3 Matrices Are:
- NCERT Exemplar Class 12 Maths Solutions
- Importance of NCERT Exemplar Class 12 Maths Solutions Chapter 3
- NCERT Exemplar Class 12 Maths Solutions Chapter
NCERT Exemplar Class 12 Math chapter 3 solutions cover various matrix-related topics like the types, the operations on two or matrices, invertible matrices, etc. It is a highly scoring chapter of NCERT Class 12 Maths Solutions that a student can utilize to gain higher scores in their exams.
| Class 12 Maths Chapter 3 exemplar solutions Exercise: 3.3 Page number: 52-64 Total questions: 101 |
Question:1
If a matrix has 28 elements, what are the possible orders it can have? What if it has 13 elements?
Answer:
In mathematics, a matrix is a rectangular array that includes numbers, expressions, symbols, and equations which are placed in an arrangement of rows and columns. The number of rows and columns that are arranged in the matrix is called the order or dimension of the matrix. By rule, the rows are listed first and then the columns.
It is given that the matrix has 28 elements.
So, according to the rule of the matrix,
If the given matrix has $m n$ elements, then the dimension of the order can be given by $m * n$, where $m$ and $n$ are natural numbers.
So, if a matrix has 28 elements, which is $m n=28$, then the following possible orders can be found:
$\because m n=28$
Take m and n to be any number, so that, when they are multiplied, we get 28.
So, let $\mathrm{m}=1$ and $\mathrm{n}=28$.
Then, $m \times n=1 \times 28(=28)$
$\Rightarrow 1 \times 28$ is a possible order of the matrix with 28 elements
Take $\mathrm{m}=2$ and $\mathrm{n}=14$.
Then, $m \times n=2 \times 14(=28)$
$\Rightarrow 2 \times 14$ is a possible order of the matrix with 28 elements.
Take $\mathrm{m}=4$ and $\mathrm{n}=7$.
Then, $m \times n=4 \times 7(=28)$
$\Rightarrow 4 \times 7$ is a possible order of the matrix with 28 elements.
Take $\mathrm{m}=7$ and $\mathrm{n}=4$.
Then, $m \times n=7 \times 4(=28)$
$\Rightarrow 7 \times 4$ is a possible order of the matrix with 28 elements.
Take $\mathrm{m}=14$ and $\mathrm{n}=2$.
Then, $m \times n=14 \times 2(=28)$
$\Rightarrow 14 \times 2$ is a possible order of the matrix having 28 elements.
Take $\mathrm{m}=28$ and $\mathrm{n}=1$.
Then, $m \times n=28 \times 1(=28)$
$\Rightarrow 28 \times 1$ is a possible order of the matrix with 28 elements.
The following are the possible orders that a matrix having 28 elements can have:
$1 \times 28,2 \times 14,4 \times 7,7 \times 4,14 \times 2$ and $28 \times 1$
If the given matrix consisted of 13 elements, then its possible order can be found out in a similar way as given above:
Here, $m n=13$.
Take $m$ and $n$ to be any numbers so that when multiplied we get 13.
Take $\mathrm{m}=1$ and $\mathrm{n}=13$.
Then, $m \times n=1 \times 13(=13)$
$\Rightarrow 1 \times 13$ is a possible order of the matrix with 13 elements.
Take $\mathrm{m}=13$ and $\mathrm{n}=1$.
Then, $m \times n=13 \times 1(=13)$
$\Rightarrow 13 \times 1$ is a possible order of the matrix with 13 elements.
Thus, the possible orders of the matrix consisting of 13 elements are as follows:
$1 \times 13$ and $13 \times 1$
Question:2
Answer:
i) We have, $\mathrm{A}=\left[\begin{array}{ccc}\mathrm{a} & 1 & x \\ 2 & \sqrt{3} & x^2-y \\ 0 & 5 & \frac{-2}{5}\end{array}\right]$ the order of matrix A is $3 \times 3$
ii) We have, $\mathrm{A}=\left[\begin{array}{ccc}\mathrm{a} & 1 & x \\ 2 & \sqrt{3} & x^2-y \\ 0 & 5 & \frac{-2}{5}\end{array}\right]$ the number of elements are $3 \times 3=9$
iii) We have, $\mathrm{A}=\left[\begin{array}{ccc}\mathrm{a} & 1 & x \\ 2 & \sqrt{3} & x^2-y \\ 0 & 5 & \frac{-2}{5}\end{array}\right]$
Since, $\mathrm{a}_{\mathrm{ij}}$ is the element lying in the $\mathrm{i}^{\text {th }}$ row an $\mathrm{j}^{\text {th }}$ column We have $a_{23}=x^2-y, a_{31}=0, a_{12}=1$.
Question:3
Answer:
i) Let $A=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right]_{2 \times 2}$
Given that $\mathrm{a}_{\mathrm{ij}}=\frac{(\mathrm{i}-2 \mathrm{j})^2}{2}$
$\begin{aligned} & \mathrm{a}_{11}=\frac{(1-2 \times 1)^2}{2}=\frac{1}{2} \\ & \mathrm{a}_{12}=\frac{(1-2 \times 2)^2}{2}=\frac{9}{2} \\ & \mathrm{a}_{21}=\frac{(2-2 \times 1)^2}{2}=0 \\ & \mathrm{a}_{22}=\frac{(2-2 \times 2)^2}{2}=2\end{aligned}$
Hence, the matrix $\mathrm{A}=\left[\begin{array}{cc}\frac{1}{2} & \frac{9}{2} \\ 0 & 2\end{array}\right]$
ii) Let $A=\left[\begin{array}{ll}a_{13} & a_{12} \\ a_{21} & a_{22}\end{array}\right]_{2 \times 2}$
Given that ${ }^` \mathrm{a}_{\mathrm{ij}}=|-2 \mathrm{i}+3 \mathrm{j}|$
$\begin{aligned} & a_{11}=|-2 \times 1+3 \times 1|=1 \\ & a_{12}=|-2 \times 1+3 \times 2|=4 \\ & a_{21}=|-2 \times 2+3 \times 1|=-1 \\ & a_{22}=|-2 \times 2+3 \times 2|=2\end{aligned}$
Hence, the matrix $A=\left[\begin{array}{cc}1 & 4 \\ -1 & 2\end{array}\right]$
Question:4
Construct a 3 × 2 matrix whose elements are given by $a_{ij} = e^{ix}\sin jx$
Answer:
$\operatorname{Let} A=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32}\end{array}\right]_{3 \times 2}$ Given that $\mathrm{a}_{\mathrm{ij}}=\mathrm{e}^{\mathrm{i} . \mathrm{x}} \sin \mathrm{j} \mathrm{x}$
$\begin{aligned} & a_{11}=e^x \sin x \\ & a_{12}=e^x \sin 2 x \\ & a_{21}=e^{2 x} \sin x \\ & a_{22}=e^{2 x} \sin 2 x \\ & a_{31}=e^{3 x} \sin x \\ & a_{32}=e^{3 x} \sin 2 x\end{aligned}$
Hence, the matrix $\mathrm{A}=\left[\begin{array}{cc}\mathrm{e}^x \sin x & \mathrm{e}^x \sin 2 x \\ \mathrm{e}^{2 x} \sin x & \mathrm{e}^{2 x} \sin 2 x \\ \mathrm{e}^{3 x} s \sin x & \mathrm{e}^{3 x} \sin 2 x\end{array}\right]$
Question:5
Answer:
Given that $\mathrm{A}=\mathrm{B}$
$\Rightarrow\left[\begin{array}{cc}\mathrm{a}+4 & 3 \mathrm{~b} \\ 8 & -6\end{array}\right]=\left[\begin{array}{cc}2 \mathrm{a}+2 & \mathrm{~b}^2+2 \\ 8 & \mathrm{~b}^2-5 \mathrm{~b}\end{array}\right]$
Equating the corresponding elements, we get
$\begin{aligned} & a+4=2 a+2 \\ & 3 b=b^2+2 \\ & b^2-5 b=-6 \\ & \Rightarrow 2 a-a=2 \\ & b^2-3 b+2=0 \\ & b^2-5 b+6=0 \\ & \therefore a=2 \\ & \therefore b^2-3 b+2=0\end{aligned}$
$\begin{aligned} & \Rightarrow b^2-2 b-b+2=0 \\ & \Rightarrow b(b-2)-1(b-2)=0 \\ & \Rightarrow(b-1)(b-2)=0 \\ & \therefore b=1,2 \\ & \therefore b^2-5 b+6=0 \\ & b^2-3 b-2 b+6=0 \\ & \Rightarrow b(b-3)-2(b-3)=0 \\ & \Rightarrow(b-2)(b-3)=0 \\ & \Rightarrow b=2,3\end{aligned}$
But here 2 is common.
Hence, the value of $a=2$ and $b=2$.
Question:6
If possible, find the sum of the matrices A and B, where
Answer:
We have, $\mathrm{A}=\left[\begin{array}{cc}\sqrt{3} & 1 \\ 2 & 3\end{array}\right]_{2 \times 2}$, and $\mathrm{B}=\left[\begin{array}{lll}x & y & z \\ a & \mathrm{~b} & 6\end{array}\right]_{2 \times 3}$
Here, A and B are of different orders.
Two matrices A and B are confirmable for addition only if the order of both matrices A and B is the same.
Hence, the sum of matrices A and B is not possible.
Question:7
If $\begin{array}{l} X=\left[\begin{array}{lll} 3 & 1 & -1 \\ 5 & -2 & -3 \end{array}\right] \\ Y=\left[\begin{array}{lll} 2 & 1 & -1 \\ 7 & 2 & 4 \end{array}\right] \end{array}$ find
(i) X + Y
(ii) 2X - 3Y
(iii) A matrix Z such that X + Y + Z is a zero matrix.
Answer:
Given that $\mathrm{X}=\left[\begin{array}{ccc}3 & 1 & -1 \\ 5 & -2 & -3\end{array}\right]$ and $\mathrm{Y}=\left[\begin{array}{ccc}2 & 1 & -1 \\ 7 & 2 & 4\end{array}\right]$
i) $\begin{aligned} & \mathrm{X}+\mathrm{Y}=\left[\begin{array}{ccc}3 & 1 & -1 \\ 5 & -2 & -3\end{array}\right]+\left[\begin{array}{ccc}2 & 1 & -1 \\ 7 & 2 & 4\end{array}\right] \\ & =\left[\begin{array}{ccc}3+2 & 1+1 & -1-1 \\ 5+7 & -2+2 & -3+4\end{array}\right] \\ & =\left[\begin{array}{ccc}5 & 2 & -2 \\ 12 & 0 & 1\end{array}\right]\end{aligned}$
ii) $\begin{aligned} & 2 X-3 Y=2\left[\begin{array}{ccc}3 & 1 & -1 \\ 5 & -2 & -3\end{array}\right]-3\left[\begin{array}{ccc}2 & 1 & -1 \\ 7 & 2 & 4\end{array}\right] \\ & =\left[\begin{array}{ccc}2 \times 3 & 2 \times 1 & -2 \times 1 \\ 2 \times 5 & -2 \times 2 & -2 \times 3\end{array}\right]-\left[\begin{array}{ccc}3 \times 2 & 1 \times 3 & -1 \times 3 \\ 3 \times 7 & 3 \times 2 & 3 \times 4\end{array}\right] \\ & =\left[\begin{array}{ccc}6 & 2 & -2 \\ 10 & -4 & -6\end{array}\right]-\left[\begin{array}{ccc}6 & 3 & -3 \\ 21 & 6 & 12\end{array}\right] \\ & =\left[\begin{array}{ccc}6-6 & 2-3 & -2+3 \\ 10-21 & -4-6 & -6-12\end{array}\right] \\ & =\left[\begin{array}{ccc}0 & -1 & 1 \\ -11 & -10 & -18\end{array}\right]\end{aligned}$
iii) $\begin{aligned} & \mathrm{X}+\mathrm{Y}+\mathrm{Z}=0 \\ & \Rightarrow\left[\begin{array}{ccc}3 & 1 & -1 \\ 5 & -2 & -3\end{array}\right]+\left[\begin{array}{ccc}2 & 1 & -1 \\ 7 & 2 & 4\end{array}\right]+\left[\begin{array}{lll}\mathrm{a} & \mathrm{b} & \mathrm{c} \\ \mathrm{d} & \mathrm{e} & \mathrm{f}\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]\end{aligned}$
Where $Z=\left[\begin{array}{lll}a & b & c \\ d & e & f\end{array}\right]$
$\begin{aligned} & \Rightarrow\left[\begin{array}{ccc}3+2+\mathrm{a} & 1+1+\mathrm{b} & -1-1+\mathrm{c} \\ 5+7+\mathrm{d} & -2+2+\mathrm{e} & -3+4+\mathrm{f}\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right] \\ & \Rightarrow\left[\begin{array}{ccc}5+\mathrm{a} & 2+\mathrm{b} & -2+\mathrm{c} \\ 12+\mathrm{d} & \mathrm{e} & 1+\mathrm{f}\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]\end{aligned}$
Equating the corresponding elements, we get
$\begin{aligned} & 5+\mathrm{a}=0 \\ & \Rightarrow \mathrm{a}=-5,2+\mathrm{b}=0 \\ & \Rightarrow \mathrm{~b}=-2-2+\mathrm{c}=0 \\ & \Rightarrow \mathrm{c}=2 \\ & 12+\mathrm{d}=0 \\ & \Rightarrow \mathrm{~d}=-12, \mathrm{e}=0,1+\mathrm{f}=0 \\ & \Rightarrow \mathrm{f}=-1\end{aligned}$
Hence, the matrix $Z=\left[\begin{array}{ccc}-5 & -2 & 2 \\ -12 & 0 & -1\end{array}\right]$
Question:8
Answer:
The given equation can be written as
$\begin{aligned} & {\left[\begin{array}{cc}2 x^2 & 2 x \\ 3 x & x^2\end{array}\right]+\left[\begin{array}{cc}16 & 10 x \\ 8 & 8 x\end{array}\right]=\left[\begin{array}{cc}\left(2 x^2+16\right) & 18 \\ 20 & 12 x\end{array}\right]} \\ & \Rightarrow\left[\begin{array}{cc}2 x^2+16 & 12 x \\ 3 x+8 & x^2+8 x\end{array}\right]=\left[\begin{array}{cc}2 x^2+16 & 48 \\ 20 & 12 x\end{array}\right]\end{aligned}$
Equating the corresponding elements we get
$\begin{aligned} & 12 \mathrm{x}=48 \\ & 3 \mathrm{x}+8=20 \\ & \mathrm{x}^2+8 \mathrm{x}=12 \mathrm{x} \\ & \therefore \mathrm{x}=\frac{48}{12}=4 \\ & 3 \mathrm{x}=20-8=12 \\ & \Rightarrow \mathrm{x}^2=12 \mathrm{x}-8 \mathrm{x}=4 \mathrm{x} \\ & \Rightarrow \mathrm{x}^2-4 \mathrm{x}=0 \\ & \mathrm{x}=0, \mathrm{x}=4 \\ & \therefore \mathrm{x}=4\end{aligned}$
Hence, the non-zero value of $x$ is 4.
Question:9
Answer:
Given that $A=\left[\begin{array}{ll}0 & 1 \\ 1 & 1\end{array}\right]$ and $B=\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]$
$\begin{aligned} & \mathrm{A}+\mathrm{B}=\left[\begin{array}{ll}0 & 1 \\ 1 & 1\end{array}\right]+\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right] \\ & \Rightarrow \mathrm{A}+\mathrm{B}=\left[\begin{array}{ll}0+0 & 1-1 \\ 1+1 & 1+0\end{array}\right] \\ & \Rightarrow \mathrm{A}+\mathrm{B}=\left[\begin{array}{ll}0 & 0 \\ 2 & 1\end{array}\right] \\ & \mathrm{A}-\mathrm{B}=\left[\begin{array}{ll}0 & 1 \\ 1 & 1\end{array}\right]-\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right] \\ & \Rightarrow \mathrm{A}-\mathrm{B}=\left[\begin{array}{ll}0-0 & 1+1 \\ 1-1 & 1-0\end{array}\right] \\ & \Rightarrow \mathrm{A}-\mathrm{B}=\left[\begin{array}{ll}0 & 2 \\ 0 & 1\end{array}\right]\end{aligned}$
$\begin{aligned} & \therefore(A+B) \cdot(A-B)=\left[\begin{array}{ll}0 & 0 \\ 2 & 1\end{array}\right],\left[\begin{array}{ll}0 & 2 \\ 0 & 1\end{array}\right] \\ & =\left[\begin{array}{ll}0+0 & 0+0 \\ 0+0 & 4+1\end{array}\right] \\ & =\left[\begin{array}{ll}0 & 0 \\ 0 & 5\end{array}\right]\end{aligned}$
Now, R.H.S. $=\mathrm{A}^2-\mathrm{B}^2$
$\begin{aligned} & =\mathrm{A} \cdot \mathrm{A}-\mathrm{B} \cdot \mathrm{B} \\ & =\left[\begin{array}{ll}0 & 1 \\ 1 & 1\end{array}\right]\left[\begin{array}{ll}0 & 1 \\ 1 & 1\end{array}\right]-\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right] \\ & =\left[\begin{array}{ll}0+1 & 0+1 \\ 0+1 & 1+1\end{array}\right]-\left[\begin{array}{cc}0-1 & 0+0 \\ 0+0 & -1+0\end{array}\right]\end{aligned}$
$\begin{aligned} & =\left[\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right]-\left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right] \\ & =\left[\begin{array}{ll}1 & +1 \\ 1 & 1-0 \\ 1 & 2+1\end{array}\right] \\ & =\left[\begin{array}{ll}2 & 1 \\ 1 & 3\end{array}\right]\end{aligned}$
Hence, $\left[\begin{array}{ll}0 & 0 \\ 0 & 5\end{array}\right] \neq\left[\begin{array}{cc}2 & 10 \\ 1 & 3\end{array}\right]$
Hence, $(A+B) \cdot(A-B) \neq A^2-B^2$
Question:10
Answer:
Given that $\left[\begin{array}{lll}1 & x & 1\end{array}\right]\left[\begin{array}{ccc}1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2\end{array}\right]\left[\begin{array}{l}1 \\ 2 \\ x\end{array}\right]=0$
$\begin{aligned} & \Rightarrow\left[\begin{array}{lll}1+2 x+15 & 3+5 x+3 & 2+x+2\end{array}\right]\left[\begin{array}{l}1 \\ 2 \\ x\end{array}\right]=0 \\ & \Rightarrow\left[\begin{array}{lll}2 x+16 & 5 x+6 & x+4\end{array}\right]\left[\begin{array}{l}1 \\ 2 \\ x\end{array}\right]=0 \\ & \Rightarrow\left[\begin{array}{ll}2 \mathrm{x}+16+10 \mathrm{x}+12+\mathrm{x}^2+4 \mathrm{x}\end{array}\right]=0 \\ & \Rightarrow \mathrm{x}^2+16 \mathrm{x}+28=0 \\ & \Rightarrow \mathrm{x}^2+14 \mathrm{x}+2 \mathrm{x}+28=0 \\ & \Rightarrow \mathrm{x}(\mathrm{x}+14)+2(\mathrm{x}+14)=0 \\ & \Rightarrow(\mathrm{x}+2)(\mathrm{x}+14)=0 \\ & \mathrm{x}+2=0 \quad \text { or } \quad \mathrm{x}+14=0 \\ & \therefore \mathrm{x}=-2 \quad \text { or } \quad \mathrm{x}=-14\end{aligned}$
Hence, the values of $x$ are -2 and -14.
Question:11
Answer:
Given that $\mathrm{A}=\left[\begin{array}{cc}5 & 3 \\ -1 & -2\end{array}\right]$
$\begin{aligned} & \mathrm{A}^2=\mathrm{A} \cdot \mathrm{A} \\ & =\left[\begin{array}{cc}5 & 3 \\ -1 & -2\end{array}\right]\left[\begin{array}{cc}5 & 3 \\ -1 & -2\end{array}\right] \\ & =\left[\begin{array}{cc}25-3 & 15-6 \\ -5+2 & -3+4\end{array}\right] \\ & =\left[\begin{array}{cc}22 & 9 \\ -3 & 1\end{array}\right] \\ & \mathrm{A}^2-3 \mathrm{~A}-7 \mathrm{I}=\mathrm{O}\end{aligned}$
L.H.S. $\left[\begin{array}{cc}2 & 9 \\ -3 & 1\end{array}\right]-3\left[\begin{array}{cc}5 & 3 \\ -1 & -2\end{array}\right]-7\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]$
$\begin{aligned} & \Rightarrow\left[\begin{array}{cc}22 & 9 \\ -3 & 1\end{array}\right]-\left[\begin{array}{cc}15 & 9 \\ -3 & -6\end{array}\right]-\left[\begin{array}{cc}7 & 0 \\ 0 & 7\end{array}\right] \\ & \Rightarrow\left[\begin{array}{cc}22-15-7 & 9-9-0 \\ -3+3-0 & 1+6-7\end{array}\right] \\ & \Rightarrow\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right] \text { R.H.S. }\end{aligned}$
We are given $\mathrm{A}^2-3 \mathrm{~A}-7 \mathrm{I}=0$
$\begin{aligned} & \Rightarrow \mathrm{A}^{-1}\left[\mathrm{~A}^2-3 \mathrm{~A}-7 \mathrm{I}\right]=\mathrm{A}^{-1} \mathrm{O} \ldots .\left[\text { Pre-multiplying both sides by } \mathrm{A}^{-1}\right] \\ & \Rightarrow \mathrm{A}^{-1} \mathrm{~A} \cdot \mathrm{~A}-3 \mathrm{~A}^{-1} \cdot \mathrm{~A}-7 \mathrm{~A}^{-1} \mathrm{I}=\mathrm{O} \ldots . .\left[\mathrm{A}^{-1} \mathrm{O}=\mathrm{O}\right] \\ & \Rightarrow \mathrm{I} \cdot \mathrm{A}-3 \mathrm{I}-7 \mathrm{~A}-1 \mathrm{I}=\mathrm{O} \\ & \Rightarrow \mathrm{A}-3 \mathrm{I}-7 \mathrm{~A}^{-1}=\mathrm{O} \\ & \Rightarrow-7 \mathrm{~A}^{-1}=3 \mathrm{I}-\mathrm{A} \\ & \Rightarrow \mathrm{A}^{-1}=\frac{1}{-7}[3 \mathrm{I}-\mathrm{A}]\end{aligned}$
$\begin{aligned} & \Rightarrow \mathrm{A}^{-1}=\frac{1}{-7}\left[3\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)-\left(\begin{array}{cc}5 & 3 \\ -1 & -2\end{array}\right)\right] \\ & =\frac{1}{-7}\left[3\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)-\left(\begin{array}{cc}5 & 3 \\ -1 & -2\end{array}\right)\right] \\ & =1(-7)\left[\begin{array}{cc}3-5 & 0-3 \\ 0+1 & 3+2\end{array}\right] \\ & =\frac{1}{-7}\left[\begin{array}{cc}-2 & -3 \\ 1 & 5\end{array}\right]\end{aligned}$
Hence, $\mathrm{A}^{-1}=-\frac{1}{7}\left[\begin{array}{cc}-2 & -3 \\ 1 & 5\end{array}\right]$
Question:12
Answer:
We have $\left[\begin{array}{ll}2 & 1 \\ 3 & 2\end{array}\right] \mathrm{A}\left[\begin{array}{cc}-3 & 2 \\ 5 & -3\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$ or $\mathrm{PAQ}=\mathrm{I}$,
Where $\mathrm{P}=\left[\begin{array}{ll}2 & 1 \\ 3 & 2\end{array}\right]$ and $\mathrm{Q}=\left[\begin{array}{cc}-3 & 2 \\ 5 & -3\end{array}\right]$
$\begin{aligned} & \therefore \mathrm{P}^{-1} \mathrm{PAQ}=\mathrm{P}^{-1} \mathrm{I} \\ & \Rightarrow \mathrm{IQA}=\mathrm{P}^{-1} \\ & \Rightarrow \mathrm{AQ}=\mathrm{P}^{-1} \\ & \Rightarrow \mathrm{AQQ}^{-1}=\mathrm{P}^{-1} \mathrm{Q}^{-1} \\ & \Rightarrow \mathrm{AI}=\mathrm{P}^{-1} \mathrm{Q}^{-1} \\ & \Rightarrow \mathrm{~A}=\mathrm{P}^{-1} \mathrm{Q}^{-1}\end{aligned}$
Now adj. $\mathrm{P}=\left[\begin{array}{cc}2 & -1 \\ -3 & 2\end{array}\right]$ and $|\mathrm{P}|=1$
$\therefore \mathrm{P}^{-1}=\left[\begin{array}{cc}2 & -1 \\ -3 & 2\end{array}\right]$
$\begin{aligned} & \therefore \mathrm{Q}^{-1}=\left[\begin{array}{ll}3 & 2 \\ 5 & 3\end{array}\right] \\ & \Rightarrow \mathrm{A}=\mathrm{P}^{-1} \mathrm{Q}^{-1} \\ & =\left[\begin{array}{cc}2 & -1 \\ -3 & 2\end{array}\right]\left[\begin{array}{ll}3 & 2 \\ 5 & 3\end{array}\right] \\ & =\left[\begin{array}{cc}6-5 & 4-3 \\ -9+10 & -6+6\end{array}\right] \\ & =\left[\begin{array}{ll}1 & 1 \\ 1 & 0\end{array}\right]\end{aligned}$
Question:13
Answer:
We have, $\left[\begin{array}{l}4 \\ 1 \\ 3\end{array}\right] \mathrm{A}=\left[\begin{array}{lll}-4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3\end{array}\right]$
Let $\mathrm{A}=\left[\begin{array}{lll}x & y & z\end{array}\right]$
$\begin{aligned} & \therefore\left[\begin{array}{l}4 \\ 1 \\ 3\end{array}\right]\left[\begin{array}{lll}x & y & z\end{array}\right]=\left[\begin{array}{lll}-4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3\end{array}\right] \\ & \Rightarrow\left[\begin{array}{ccc}4 x & 4 y & 4 z \\ x & y & z \\ 3 x & 3 y & 3 z\end{array}\right]\left[\begin{array}{lll}-4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3\end{array}\right]\end{aligned}$
Comparing elements of both sides
$\begin{aligned} & 4 x=-4 \\ & \Rightarrow x=-1 \\ & 4 y=8 \\ & y=2\end{aligned}$
And 4z = 4
⇒ z = 1
$\therefore \mathrm{A}=\left[\begin{array}{lll}-1 & 2 & 1\end{array}\right]$
Question:14
Answer:
Here, $B=\left[\begin{array}{lll}2 & 1 & 2 \\ 1 & 2 & 4\end{array}\right]_{2 \times 3}$ and $A=\left[\begin{array}{cc}3 & -4 \\ 1 & 1 \\ 2 & 0\end{array}\right]_{3 \times 2}$
$\begin{aligned} & \therefore \mathrm{BA}=\left[\begin{array}{ll}6+1+4 & -8+1+0 \\ 3+2+8 & -4+2+0\end{array}\right]_{2 \times 2} \\ & \Rightarrow \mathrm{BA}=\left[\begin{array}{ll}11 & -7 \\ 13 & -2\end{array}\right]\end{aligned}$
L.H.S: $(\mathrm{BA})^2=(\mathrm{BA}) \cdot(\mathrm{BA})$
$\begin{aligned} & =\left[\begin{array}{ll}11 & -7 \\ 13 & -2\end{array}\right]\left[\begin{array}{ll}11 & -7 \\ 13 & -2\end{array}\right] \\ & \Rightarrow\left[\begin{array}{cc}121-91 & -77+14 \\ 143-26 & -91+4\end{array}\right] \\ & \Rightarrow\left[\begin{array}{cc}30 & -63 \\ 117 & -87\end{array}\right]\end{aligned}$
$\begin{aligned} & \text { R.H.S: B }{ }^2=\mathrm{B} \cdot \mathrm{B} \\ & =\left[\begin{array}{lll}2 & 1 & 2 \\ 1 & 2 & 4\end{array}\right]_{2 \times 3} \cdot\left[\begin{array}{lll}2 & 1 & 2 \\ 1 & 2 & 4\end{array}\right]_{2 \times 3}\end{aligned}$
Here, the number of columns of the first
i.e., 3 is not equal to the number of rows of the second matrix i.e., 2.
So, $\mathrm{B}^2$ is not possible.
Similarly, $\mathrm{A}^2$ is also not possible.
Hence, $(\mathrm{BA})^2 \cdot \mathrm{~B}^2 \mathrm{~A}^2$
Question:15
Answer:
$\begin{aligned} & \mathrm{BA}=\left[\begin{array}{ll}4 & 1 \\ 2 & 3 \\ 1 & 2\end{array}\right]_{3 \times 2}\left[\begin{array}{lll}2 & 1 & 2 \\ 1 & 2 & 4\end{array}\right]_{2 \times 3} \\ & \mathrm{BA}=\left[\begin{array}{lll}8+1 & 4+2 & 8+4 \\ 4+3 & 2+6 & 4+12 \\ 2+2 & 1+4 & 2+8\end{array}\right]_{3 \times 3} \\ & =\left[\begin{array}{lll}9 & 6 & 12 \\ 7 & 8 & 16 \\ 4 & 5 & 10\end{array}\right]_{3 \times 3}\end{aligned}$
Now $A B=\left[\begin{array}{lll}2 & 1 & 2 \\ 1 & 2 & 4\end{array}\right]_{2 \times 3}\left[\begin{array}{ll}4 & 1 \\ 2 & 3 \\ 1 & 2\end{array}\right]_{3 \times 2}$
$\begin{aligned} & =\left[\begin{array}{cc}8+2+2 & 2+3+4 \\ 4+4+4 & 1+6+8\end{array}\right]_{2 \times 2} \\ & =\left[\begin{array}{cc}12 & 9 \\ 12 & 15\end{array}\right]_{2 \times 2}\end{aligned}$
Hence, $\mathrm{BA}=\left[\begin{array}{lll}9 & 6 & 12 \\ 7 & 8 & 16 \\ 4 & 5 & 10\end{array}\right]$ and $\mathrm{AB}=\left[\begin{array}{cc}12 & 9 \\ 12 & 15\end{array}\right]$.
Question:16
An example is shown for A ≠ O, B ≠ O, AB = O.
Answer:
Let $A=\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right]$ and $B=\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]$
$\begin{aligned} & \mathrm{AB}=\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right] \\ & \Rightarrow \mathrm{AB}=\left[\begin{array}{cc}1-1 & 1-1 \\ -1+1 & -1+1\end{array}\right] \\ & =\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]=\mathrm{O}\end{aligned}$
Hence, $A=\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right]$ and $B=\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]$
Question:17
Answer:
$\begin{aligned} & \text { Here, } \mathrm{A}=\left[\begin{array}{lll}2 & 4 & 0 \\ 3 & 9 & 6\end{array}\right], \mathrm{B}=\left[\begin{array}{ll}1 & 4 \\ 2 & 8 \\ 1 & 3\end{array}\right] \\ & \mathrm{AB}=\left[\begin{array}{lll}2 & 4 & 0 \\ 3 & 9 & 6\end{array}\right]\left[\begin{array}{ll}1 & 4 \\ 2 & 8 \\ 1 & 3\end{array}\right]\end{aligned}$
$\begin{aligned} & =\left[\begin{array}{cc}2+8+0 & 8+32+0 \\ 3+18+6 & 12+72+18\end{array}\right] \\ & =\left[\begin{array}{cc}10 & 40 \\ 27 & 102\end{array}\right]\end{aligned}$
$\begin{aligned} & \text { L.H.S. }(A B)^{\prime}=\left[\begin{array}{cc}10 & 27 \\ 40 & 102\end{array}\right] \\ & \text { Now } B=\left[\begin{array}{ll}1 & 4 \\ 2 & 8 \\ 1 & 3\end{array}\right]\end{aligned}$
$\begin{aligned} & \Rightarrow \mathrm{B}^{\prime}=\left[\begin{array}{lll}1 & 2 & 1 \\ 4 & 8 & 3\end{array}\right] \\ & \mathrm{A}=\left[\begin{array}{lll}2 & 4 & 0 \\ 3 & 9 & 6\end{array}\right] \\ & \Rightarrow \mathrm{A}^{\prime}=\left[\begin{array}{cc}2 & 3 \\ 4 & 90 \\ 0 & 6\end{array}\right]\end{aligned}$
R.H.S. $\mathrm{B}^{\prime} \mathrm{A}^{\prime}=\left[\begin{array}{lll}1 & 2 & 1 \\ 4 & 8 & 3\end{array}\right]\left[\begin{array}{ll}2 & 3 \\ 4 & 9 \\ 0 & 6\end{array}\right]$
$\begin{aligned} & =\left[\begin{array}{cc}2+8+0 & 3+18+6 \\ 8+32+0 & 1272+18\end{array}\right] \\ & =\left[\begin{array}{cc}10 & 27 \\ 40 & 102\end{array}\right] \\ & =\text { L.H.S. }\end{aligned}$
Hence, L.H.S. = R.H.S.
Question:18
Answer:
Given that: $\mathrm{x}=x\left[\begin{array}{l}2 \\ 1\end{array}\right]+y\left[\begin{array}{l}3 \\ 5\end{array}\right]+\left[\begin{array}{c}-8 \\ -11\end{array}\right]=\mathrm{O}$
L.H.S. $x\left[\begin{array}{l}2 \\ 1\end{array}\right]+y\left[\begin{array}{l}3 \\ 5\end{array}\right]+\left[\begin{array}{c}-8 \\ -11\end{array}\right]=\mathrm{O}$
$\begin{aligned} & \Rightarrow\left[\begin{array}{c}2 x \\ x\end{array}\right]+\left[\begin{array}{c}3 y \\ 5 y\end{array}\right]+\left[\begin{array}{c}-8 \\ -11\end{array}\right]=\mathrm{O} \\ & \Rightarrow\left[\begin{array}{c}2 x+3 y-8 \\ x+5 y-11\end{array}\right]=\left[\begin{array}{l}0 \\ 0\end{array}\right]\end{aligned}$
Comparing the corresponding elements of both sides, we get,
$\begin{aligned} & 2 x+3 y-8=0 \\ & \Rightarrow 2 x+3 y=8 \\ & x+5 y-11=0 \\ & \Rightarrow x+5 y=11\end{aligned}$
Multiplying equation (1) by 1 and equation (2) by 2, and then subtracting, we get,
$\begin{aligned} 2 x+3 y & =8 \\ 2 x+10 y & =22 \\ (-)-(-) & -(-) \\ -7 y & =-14\end{aligned}$
$\therefore \mathrm{y}=2$
$\begin{aligned} & x+5 \times 2=11 \\ & \Rightarrow x+10=11 \\ & x=11-10=1\end{aligned}$
Hence, the values of x and y are 1 and 2, respectively.
Question:19
Answer:
We have the given matrix equations,
$2 \mathrm{X}+3 \mathrm{Y}=\left[\begin{array}{ll} 2 & 3 \\ 4 & 0 \end{array}\right], 3 \mathrm{X}+2 \mathrm{Y}=\left[\begin{array}{cc} -2 & 2 \\ 1 & -5 \end{array}\right]$
By subtracting equation (i) from (ii), we get
$\begin{array}{l} (3 X+2 Y)-(2 X+3 Y)=\left[\begin{array}{cc} -2 & 2 \\ 1 & -5 \end{array}\right]-\left[\begin{array}{cc} 2 & 3 \\ 4 & 0 \end{array}\right] \\ \Rightarrow 3 X+2 Y-2 X-3 Y=\left[\begin{array}{cc} -2-2 & 2-3 \\ 1-4 & -5-0 \end{array}\right] \\ \Rightarrow 3 X-2 X+2 Y-3 Y=\left[\begin{array}{cc} -4 & -1 \\ -3 & -5 \end{array}\right] \\ \Rightarrow X-Y=\left[\begin{array}{ll} -4 & -1 \\ -3 & -5 \end{array}\right] \end{array}$
By adding equations (i) and (ii), we get
$\begin{aligned} &(3 X+2 Y)+(2 X+3 Y)=\left[\begin{array}{cc} -2 & 2 \\ 1 & -5 \end{array}\right]+\left[\begin{array}{ll} 2 & 3 \\ 4 & 0 \end{array}\right]\\ &\Rightarrow 3 \mathrm{X}+2 \mathrm{Y}+2 \mathrm{X}+3 \mathrm{Y}=\left[\begin{array}{cc} -2+2 & 2+3 \\ 1+4 & -5+0 \end{array}\right]\\ &\Rightarrow 3 \mathrm{X}+2 \mathrm{X}+2 \mathrm{Y}+3 \mathrm{Y}=\left[\begin{array}{cc} 0 & 5 \\ 5 & -5 \end{array}\right]\\ &\Rightarrow 5 X+5 Y=\left[\begin{array}{cc} 0 & 5 \\ 5 & -5 \end{array}\right]\\ &\Rightarrow 5(X+Y)=\left[\begin{array}{cc} 0 & 5 \\ 5 & -5 \end{array}\right]\\ &\Rightarrow X+Y=\frac{1}{5}\left[\begin{array}{cc} 0 & 5 \\ 5 & -5 \end{array}\right]\\ &\Rightarrow X+Y=\left[\begin{array}{ll} \frac{1}{5} \times 0 & \frac{1}{5} \times 5 \\ \frac{1}{5} \times 5 & \frac{1}{5} \times-5 \end{array}\right]\\ &\Rightarrow X+Y=\left[\begin{array}{cc} 0 & 1 \\ 1 & -1 \end{array}\right] \end{aligned}$
By adding equations (iii) and (iv), we get
$\\ (\mathrm{X}-\mathrm{Y})+(\mathrm{X}+\mathrm{Y})=\left[\begin{array}{ll} -4 & -1 \\ -3 & -5 \end{array}\right]+\left[\begin{array}{cc} 0 & 1 \\ 1 & -1 \end{array}\right] \\ \Rightarrow \mathrm{X}-\mathrm{Y}+\mathrm{X}+\mathrm{Y}=\left[\begin{array}{ll} -4+0 & -1+1 \\ -3+1 & -5-1 \end{array}\right] \\ \Rightarrow \mathrm{X}+\mathrm{X}-\mathrm{Y}+\mathrm{Y}=\left[\begin{array}{ll} -4 & 0 \\ -2 & -6 \end{array}\right] \\ \Rightarrow 2 \mathrm{X}=\left[\begin{array}{lc} -4 & 0 \\ -2 & -6 \end{array}\right] \\ \Rightarrow \mathrm{X}=\frac{1}{2}\left[\begin{array}{lc} -4 & 0 \\ -2 & -6 \end{array}\right] \\ \Rightarrow \mathrm{X}=\left[\begin{array}{ll} \frac{1}{2} \times-4 & \frac{1}{2} \times 0 \\ \frac{1}{2} \times-2 & \frac{1}{2} \times-6 \end{array}\right] \\ \Rightarrow \mathrm{X}=\left[\begin{array}{lc} -2 & 0 \\ -1 & -3 \end{array}\right]$
Substituting the matrix A in equation (iv), we get
$\begin{array}{l} {\left[\begin{array}{lc} -2 & 0 \\ -1 & -3 \end{array}\right]+\mathrm{Y}=\left[\begin{array}{cc} 0 & 1 \\ 1 & -1 \end{array}\right]} \\ \Rightarrow \mathrm{Y}=\left[\begin{array}{cc} 0 & 1 \\ 1 & -1 \end{array}\right]-\left[\begin{array}{cc} -2 & 0 \\ -1 & -3 \end{array}\right] \\ \Rightarrow \mathrm{Y}=\left[\begin{array}{cc} 0-(-2) & 1-0 \\ 1-(-1) & -1-(-3) \end{array}\right] \\ \Rightarrow \mathrm{Y}=\left[\begin{array}{cc} 2 & 1 \\ 1+1 & -1+3 \end{array}\right] \\ \Rightarrow \mathrm{Y}=\left[\begin{array}{cc} 2 & 1 \\ 2 & 2 \end{array}\right] \\ \mathrm{X}=\left[\begin{array}{lc} -2 & 0 \\ -1 & -3 \end{array}\right]_{\text {and }} \mathrm{Y}=\left[\begin{array}{ll} 2 & 1 \\ 2 & 2 \end{array}\right] \end{array}$
Question:20
If $A = [3\: \: 5], B = [7\: \: 3]$, then find a non-zero matrix C such that AC = BC.
Answer:
We have, $\mathrm{A}=\left[\begin{array}{ll}3 & 5\end{array}\right]_{1 \times 2}$ and $\mathrm{B}=\left[\begin{array}{ll}7 & 3\end{array}\right]_{1 \times 2}$
For $\mathrm{AC}=\mathrm{BC}$
We have order of $\mathrm{C}=2 \times \mathrm{n}$
For $\mathrm{n}=1$
Let $\mathrm{C}=\left[\begin{array}{l}x \\ y\end{array}\right]$
$\therefore \mathrm{AC}=\left[\begin{array}{ll}3 & 5\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=[(3 x+5 y]$
And $\mathrm{BC}=\left[\begin{array}{ll}7 & 3\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=[3 \mathrm{x}+5 \mathrm{y}]$
For $\mathrm{AC}=\mathrm{BC}$,
$\begin{aligned} & {[3 \mathrm{x}+5 \mathrm{y}]=[7 \mathrm{x}+3 \mathrm{y}]} \\ & \Rightarrow 3 \mathrm{x}+5 \mathrm{y}=7 \mathrm{x}+3 \mathrm{y} \\ & \Rightarrow 4 \mathrm{x}=2 \mathrm{y} \\ & \Rightarrow \mathrm{x}=\frac{1}{2} y \\ & \Rightarrow \mathrm{y}=2 \mathrm{x} \\ & \therefore \mathrm{C}=\left[\begin{array}{c}x \\ 2 x\end{array}\right]\end{aligned}$
We see that on taking C of order $2 \times 1,2 \times 2,2 \times 3, \ldots$, we get
$\mathrm{C}=\left[\begin{array}{c}x \\ 2 x\end{array}\right],\left[\begin{array}{cc}x & x \\ 2 x & 2 x\end{array}\right],\left[\begin{array}{ccc}x & x & x \\ 2 x & 2 x & 2 x\end{array}\right] \ldots$
In general,
$\mathrm{C}=\left[\begin{array}{c}\mathrm{k} \\ 2 \mathrm{k}\end{array}\right],\left[\begin{array}{cc}\mathrm{k} & \mathrm{k} \\ 2 \mathrm{k} & 2 \mathrm{k}\end{array}\right]$ etc $\ldots$
Where k is any real number.
Question:21
Given an example of matrices A, B, and C such that AB = AC, where A is an on-zero matrix, but B ≠ C.
Answer:
Let $\mathrm{A}=\left[\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right], \mathrm{B}=\left[\begin{array}{ll}1 & 1 \\ 1 & 0\end{array}\right]$ and $\mathrm{C}=\left[\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right] \quad \ldots \ldots .[\because \mathrm{B} \neq \mathrm{C}]$
$\therefore \mathrm{AB}=\left[\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 1 & 0\end{array}\right]=\left[\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right] \ldots . .(\mathrm{i})$
And $A C=\left[\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right]=\left[\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right]$
We have $\mathrm{AB}=\mathrm{AC}$ but $\mathrm{B} \neq \mathrm{C}$
Question:22
Answer:
i) We have, $\mathrm{A}=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]$, $\mathrm{B}=\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right]$ and $\mathrm{C}=\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]$
$\begin{aligned} & \mathrm{AB}=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right] \\ & =\left[\begin{array}{cc}2+6 & 3-8 \\ -4+3 & -6-4\end{array}\right] \\ & =\left[\begin{array}{cc}8 & -5 \\ -1 & -10\end{array}\right]\end{aligned}$
And $(\mathrm{AB}) \mathrm{C}=\left[\begin{array}{cc}8 & -5 \\ -1 & -10\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]$
$\begin{aligned} & =\left[\begin{array}{cc}8+5 & 0 \\ -1+10 & 0\end{array}\right] \\ & =\left[\begin{array}{cc}13 & 0 \\ 9 & 0\end{array}\right] \ldots . .(\mathrm{i})\end{aligned}$
Again, $(\mathrm{BC})=\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]$
$\begin{aligned} & =\left[\begin{array}{ll}2-3 & 0 \\ 3+4 & 0\end{array}\right] \\ & =\left[\begin{array}{cc}-1 & 0 \\ 7 & 0\end{array}\right]\end{aligned}$
And $A(B C)=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}-1 & 0 \\ 7 & 0\end{array}\right]$
$\begin{aligned} & =\left[\begin{array}{cc}-1+14 & 0 \\ 2+7 & 0\end{array}\right] \\ & =\left[\begin{array}{cc}13 & 0 \\ 9 & 0\end{array}\right]\end{aligned}$
From (i) and (ii), we get
$\therefore(\mathrm{AB}) \mathrm{C}=\mathrm{A}(\mathrm{BC})$
ii) We have, $\mathrm{A}=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right], \mathrm{B}=\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right]$ and $\mathrm{C}=\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]$
$\begin{aligned} & \mathrm{B}+\mathrm{C}=\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right]+\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right] \\ & =\left[\begin{array}{cc}3 & 3 \\ 2 & -4\end{array}\right]\end{aligned}$
$\Rightarrow \mathrm{A} \cdot(\mathrm{B}+\mathrm{C})=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}3 & 3 \\ 2 & -4\end{array}\right]$
$=\left[\begin{array}{cc}3+4 & 3-8 \\ -6+2 & -6-4\end{array}\right]$
$=\left[\begin{array}{cc}7 & -5 \\ -4 & -10\end{array}\right] \ldots .$. (iii)
$\mathrm{AB}=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right]$
$\begin{aligned} & =\left[\begin{array}{cc}2+6 & 3-8 \\ -4+3 & -6-4\end{array}\right] \\ & =\left[\begin{array}{cc}8 & -5 \\ -1 & -10\end{array}\right]\end{aligned}$
And $A C=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]$
$\begin{aligned} & =\left[\begin{array}{cc}1-2 & 0 \\ -2-1 & 0\end{array}\right] \\ & =\left[\begin{array}{ll}-1 & 0 \\ -3 & 0\end{array}\right]\end{aligned}$
$\therefore \mathrm{AB}+\mathrm{AC}=\left[\begin{array}{cc}8 & -5 \\ -1 & -10\end{array}\right]+\left[\begin{array}{cc}-1 & 0 \\ -3 & 0\end{array}\right]$
$=\left[\begin{array}{cc}7 & -5 \\ -4 & -10\end{array}\right]$
$\mathrm{A} \cdot(\mathrm{B}+\mathrm{C})=\mathrm{A} \cdot \mathrm{B}+\mathrm{A} \cdot \mathrm{C}$
Question:23
Answer:
Given that,
$\begin{aligned} & \mathrm{P}=\left[\begin{array}{lll}x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z\end{array}\right] \text { and } \mathrm{Q}=\left[\begin{array}{lll}\mathrm{a} & 0 & 0 \\ 0 & \mathrm{~b} & 0 \\ 0 & 0 & \mathrm{c}\end{array}\right] \\ & \mathrm{PQ}=\left[\begin{array}{lll}x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z\end{array}\right]\left[\begin{array}{lll}\mathrm{a} & 0 & 0 \\ 0 & \mathrm{~b} & 0 \\ 0 & 0 & \mathrm{c}\end{array}\right]\end{aligned}$
$\begin{aligned} \mathrm{PQ} & =\left[\begin{array}{ccc}x \mathrm{a}+0+0 & 0+0+0 & 0+0+0 \\ 0+0+0 & 0+y \mathrm{~b}+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+z \mathrm{c}\end{array}\right] \\ \mathrm{PQ} & =\left[\begin{array}{ccc}x \mathrm{a} & 0 & 0 \\ 0 & y \mathrm{~b} & 0 \\ 0 & 0 & z \mathrm{c}\end{array}\right]\end{aligned}$
Now $\mathrm{QP}=\left[\begin{array}{lll}\mathrm{a} & 0 & 0 \\ 0 & \mathrm{~b} & 0 \\ 0 & 0 & \mathrm{c}\end{array}\right]\left[\begin{array}{lll}x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z\end{array}\right]$
$\begin{aligned} & \mathrm{QP}=\left[\begin{array}{ccc}x \mathrm{a}+0+0 & 0+0+0 & 0+0+0 \\ 0+0+0 & 0+y \mathrm{~b}+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+z \mathrm{c}\end{array}\right] \\ & \mathrm{QP}=\left[\begin{array}{ccc}x \mathrm{a} & 0 & 0 \\ 0 & y \mathrm{~b} & 0 \\ 0 & 0 & z \mathrm{c}\end{array}\right]\end{aligned}$
Hence, $\mathrm{PQ}=\mathrm{QP}$.
Question:24
Answer:
We have, $\left[\begin{array}{lll}2 & 1 & 3\end{array}\right]_{1 \times 3}\left[\begin{array}{ccc}-1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1\end{array}\right]_{3 \times 3}\left[\begin{array}{c}1 \\ 0 \\ -1\end{array}\right]_{3 \times 1}=\mathrm{A}$
$\therefore\left[\begin{array}{lll}2 & 1 & 3\end{array}\right]_{1 \times 3}\left[\begin{array}{c}-1+0+1 \\ -1+0+0 \\ 0+0-1\end{array}\right]_{3 \times 1}=\mathrm{A}$
$\begin{aligned} & \Rightarrow\left[\begin{array}{lll}2 & 1 & 3\end{array}\right]_{1 \times 3}\left[\begin{array}{c}0 \\ -1 \\ -1\end{array}\right]_{3 \times 1}{ }^`=\mathrm{A} \\ & \Rightarrow[0-1-3]=\mathrm{A} \\ & \Rightarrow \mathrm{A}=[-4]\end{aligned}$
Question:25
Answer:
We have $\mathrm{A}=\left[\begin{array}{ll}2 & 1\end{array}\right], \mathrm{B}=\left[\begin{array}{lll}5 & 3 & 4 \\ 8 & 7 & 6\end{array}\right]$ and $\mathrm{C}=\left[\begin{array}{ccc}-1 & 2 & 1 \\ 1 & 0 & 2\end{array}\right]$
$\begin{aligned} & \therefore \mathrm{A}(\mathrm{B}+\mathrm{C})=\left[\begin{array}{ll}2 & 1\end{array}\right]\left[\begin{array}{lll}5-1 & 3+2 & 4+1 \\ 8+1 & 7+0 & 6+2\end{array}\right] \\ & =\left[\begin{array}{ll}2 & 1\end{array}\right]\left[\begin{array}{lll}4 & 5 & 5 \\ 9 & 7 & 8\end{array}\right] \\ & =\left[\begin{array}{lll}8+9 & 10+7 & 10+8\end{array}\right] \\ & =\left[\begin{array}{lll}17 & 17 & 18\end{array}\right] \ldots \ldots .(\mathrm{i})\end{aligned}$
Now $A B=\left[\begin{array}{ll}2 & 1\end{array}\right]\left[\begin{array}{lll}5 & 3 & 4 \\ 8 & 7 & 6\end{array}\right]$
$\begin{aligned} & =\left[\begin{array}{lll}10+8 & 6+7 & 8+6\end{array}\right] \\ & =\left[\begin{array}{lll}18 & 3 & 14\end{array}\right]\end{aligned}$
And $A C=\left[\begin{array}{ll}2 & 1\end{array}\right]\left[\begin{array}{ccc}-1 & 2 & 1 \\ 1 & 0 & 2\end{array}\right]$
$\begin{aligned} & =\left[\begin{array}{lll}-2+1 & 4+0 & 2+2\end{array}\right] \\ & {\left[\begin{array}{lll}-1 & 4 & 4\end{array}\right]} \\ & \therefore \mathrm{AB}+\mathrm{AC}=\left[\begin{array}{lll}18 & 13 & 14\end{array}\right]+\left[\begin{array}{lll}-1 & 4 & 4\end{array}\right] \\ & =\left[\begin{array}{lll}17 & 17 & 18\end{array}\right] \ldots . .(\mathrm{ii)}\end{aligned}$
From equations (i) and (ii)
$\mathrm{A}(\mathrm{B}+\mathrm{C})=(\mathrm{AB}+\mathrm{AC})$
Question:26
Answer:
We have, $\mathrm{A}=\left[\begin{array}{ccc}1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]$
$\begin{aligned} & \therefore \mathrm{A}^2=\mathrm{A} \cdot \mathrm{A} \\ & =\left[\begin{array}{ccc}1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]\left[\begin{array}{ccc}1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right] \\ & =\left[\begin{array}{ccc}1+0+0 & 0+0-1 & -1+0-1 \\ 2+2+0 & 0+1+3 & -2+3+3 \\ 0+2+0 & 0+1+1 & 0+3+1\end{array}\right] \\ & =\left[\begin{array}{ccc}1 & -1 & -2 \\ 4 & 4 & 4 \\ 2 & 2 & 4\end{array}\right]\end{aligned}$
$\therefore \mathrm{A}^2+\mathrm{A}=\left[\begin{array}{ccc}1 & -1 & -2 \\ 4 & 4 & 4 \\ 2 & 2 & 4\end{array}\right]+\left[\begin{array}{ccc}1 & 0 & -4 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]$
$=\left[\begin{array}{ccc}2 & -1 & -3 \\ 6 & 5 & 7 \\ 2 & 3 & 5\end{array}\right] \ldots \ldots .(\mathrm{i})$
Now, $A+I=\left[\begin{array}{ccc}1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]+\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$=\left[\begin{array}{ccc}2 & 0 & -1 \\ 2 & 2 & 3 \\ 0 & 1 & 2\end{array}\right]$
So, $A(A+I)=\left[\begin{array}{ccc}1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]\left[\begin{array}{ccc}2 & 0 & -1 \\ 2 & 2 & 3 \\ 0 & 1 & 2\end{array}\right]$
$=\left[\begin{array}{ccc}2+0+0 & 0+0-1 & -1+0-2 \\ 4+2+0 & 0+2+3 & -2+3+6 \\ 0+2+0 & 0+2+1 & 0+3+2\end{array}\right]$
$=\left[\begin{array}{ccc}2 & -1 & -3 \\ 6 & 5 & 7 \\ 2 & 3 & 5\end{array}\right]$
From (i) and (ii)
We get $\mathrm{A}^2+\mathrm{A}=\mathrm{A}(\mathrm{A}+\mathrm{I})$
Question:27
Answer:
i) Given that: $\mathrm{A}=\left[\begin{array}{ccc}0 & -1 & 2 \\ 4 & 3 & -4\end{array}\right], \mathrm{B}=\left[\begin{array}{ll}4 & 0 \\ 1 & 3 \\ 2 & 6\end{array}\right]$
$\begin{aligned} & \mathrm{A}^{\prime}=\left[\begin{array}{ccc}0 & -1 & 2 \\ 4 & 3 & -4\end{array}\right]_{2 \times 3}^{\prime} \\ & =\left[\begin{array}{cc}0 & 4 \\ -1 & 3 \\ 2 & -4\end{array}\right]_{3 \times 2} \\ & \left(\mathrm{~A}^{\prime}\right)^{\prime}=\left[\begin{array}{cc}0 & 4 \\ -1 & 3 \\ 2 & -4\end{array}\right]_{3 \times 2}^{\prime} \\ & =\left[\begin{array}{ccc}0 & -1 & 2 \\ 4 & 3 & -4\end{array}\right]_{2 \times 3} \\ & =\mathrm{A}\end{aligned}$
ii) Given that: $\mathrm{A}=\left[\begin{array}{ccc}0 & -1 & 2 \\ 4 & 3 & -4\end{array}\right], \mathrm{B}=\left[\begin{array}{ll}4 & 0 \\ 1 & 3 \\ 2 & 6\end{array}\right]$
L.H.S. $A B=\left[\begin{array}{ccc}0 & -1 & 2 \\ 4 & 3 & -4\end{array}\right]_{2 \times 3}\left[\begin{array}{ll}4 & 0 \\ 1 & 3 \\ 2 & 6\end{array}\right]_{3 \times 2}$
$\begin{aligned} & =\left[\begin{array}{cc}0-1+4 & 0-3+12 \\ 16+3-8 & 0+9-24\end{array}\right]_{2 \times 2} \\ & =\left[\begin{array}{cc}3 & 9 \\ 11 & -15\end{array}\right]_{2 \times 2}\end{aligned}$
$\begin{aligned} & (\mathrm{AB})^{\prime}=\left[\begin{array}{cc}3 & 11 \\ 9 & -15\end{array}\right]_{2 \times 2} \\ & \text { R.H.S. } \mathrm{B}^{\prime}=\left[\begin{array}{ll}4 & 0 \\ 1 & 3 \\ 2 & 6\end{array}\right]^{\prime}\end{aligned}$
$\begin{aligned} & =\left[\begin{array}{lll}4 & 1 & 2 \\ 0 & 3 & 6\end{array}\right] \\ & \mathrm{A}^{\prime}=\left[\begin{array}{ccc}0 & -1 & 2 \\ 4 & 3 & -4\end{array}\right]^{\prime} \\ & =\left[\begin{array}{cc}0 & 4 \\ -1 & 3 \\ 2 & -4\end{array}\right]\end{aligned}$
$\mathrm{B}^{\prime} \mathrm{A}^{\prime}=\left[\begin{array}{lll}4 & 1 & 2 \\ 0 & 3 & 6\end{array}\right]_{2 \times 3}\left[\begin{array}{cc}0 & 4 \\ -1 & 3 \\ 2 & -4\end{array}\right]_{3 \times 2}$
$\begin{aligned} & =\left[\begin{array}{cc}0-1+4 & 16+3-8 \\ 0-3+12 & 0+9-24\end{array}\right]_{2 \times 2} \\ & =\left[\begin{array}{cc}3 & 11 \\ 9 & -15\end{array}\right]_{2 \times 2}\end{aligned}$
L.H.S. = R.H.S.
Hence, $(\mathrm{AB})^{\prime}=\mathrm{B}^{\prime} \mathrm{A}^{\prime}$ is verified.
iii) $\begin{aligned} & \text { Given that: } \mathrm{A}=\left[\begin{array}{ccc}0 & -1 & 2 \\ 4 & 3 & -4\end{array}\right], \mathrm{B}=\left[\begin{array}{ll}4 & 0 \\ 1 & 3 \\ 2 & 6\end{array}\right] \\ & \text { L.H.S. kA }=\mathrm{k}\left[\begin{array}{ccc}0 & -1 & 2 \\ 4 & 3 & -4\end{array}\right]\end{aligned}$
$\begin{aligned} & =\left[\begin{array}{ccc}0 & -\mathrm{k} & 2 \mathrm{k} \\ 4 \mathrm{k} & 3 \mathrm{k} & -4 \mathrm{k}\end{array}\right] \\ & (\mathrm{kA})^{\prime}=\left[\begin{array}{cc}0 & 4 \mathrm{k} \\ -\mathrm{k} & 3 \mathrm{k} \\ 2 \mathrm{k} & -\mathrm{k}\end{array}\right]\end{aligned}$
R.H.S. $\mathrm{kA}^{\prime}=\mathrm{k}\left[\begin{array}{ccc}0 & -1 & 2 \\ 4 & 3 & -4\end{array}\right]^{\prime}$
$\begin{aligned} & =\mathrm{k}\left[\begin{array}{cc}0 & 4 \\ -1 & 3 \\ 2 & -4\end{array}\right] \\ & =\left[\begin{array}{cc}0 & 4 \mathrm{k} \\ -\mathrm{k} & 3 \mathrm{k} \\ 2 \mathrm{k} & -4 \mathrm{k}\end{array}\right]\end{aligned}$
Hence, L.H.S. = R.H.S.
$(\mathrm{kA})^{\prime}=\left(\mathrm{kA}^{\prime}\right)$ is verified.
Question:28
Answer:
i) Given that: $\mathrm{A}=\left[\begin{array}{ll}1 & 2 \\ 4 & 1 \\ 5 & 6\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{ll}1 & 2 \\ 6 & 4 \\ 7 & 3\end{array}\right]$
L.H.S. $(2 \mathrm{~A}+\mathrm{B})^{\prime}=\left[2\left(\begin{array}{ll}1 & 2 \\ 4 & 1 \\ 5 & 6\end{array}\right)+\left(\begin{array}{ll}1 & 2 \\ 6 & 4 \\ 7 & 3\end{array}\right)\right]^{\prime}$
$\begin{aligned} & =\left[\left(\begin{array}{cc}2 & 4 \\ 8 & 2 \\ 10 & 12\end{array}\right)+\left(\begin{array}{ll}1 & 2 \\ 6 & 4 \\ 7 & 3\end{array}\right)\right]^{\prime} \\ & =\left[\begin{array}{cc}2+1 & 4+2 \\ 8+6 & 2+4 \\ 10+7 & 12+3\end{array}\right]^{\prime} \\ & =\left[\begin{array}{cc}3 & 6 \\ 14 & 6 \\ 17 & 15\end{array}\right]^{\prime} \\ & =\left[\begin{array}{ccc}3 & 14 & 17 \\ 6 & 6 & 15\end{array}\right]\end{aligned}$
R.H.S. $2 A^{\prime}+\mathrm{B}^{\prime}=2\left[\begin{array}{ll}1 & 2 \\ 4 & 1 \\ 5 & 6\end{array}\right]^{\prime}+\left[\begin{array}{ll}1 & 2 \\ 6 & 4 \\ 7 & 3\end{array}\right]^{\prime}$
$\begin{aligned} & =2\left[\begin{array}{lll}1 & 4 & 5 \\ 2 & 1 & 6\end{array}\right]+\left[\begin{array}{lll}1 & 6 & 7 \\ 2 & 4 & 3\end{array}\right] \\ & =\left[\begin{array}{lll}2 & 8 & 10 \\ 4 & 2 & 12\end{array}\right]+\left[\begin{array}{lll}1 & 6 & 7 \\ 2 & 4 & 3\end{array}\right] \\ & =\left[\begin{array}{ccc}2+1 & 8+6 & 10+7 \\ 4+2 & 2+4 & 12+3\end{array}\right] \\ & =\left[\begin{array}{ccc}3 & 14 & 17 \\ 6 & 6 & 15\end{array}\right]\end{aligned}$
Hence, L.H.S. = R.H.S.
$(2 \mathrm{~A}+\mathrm{B})^{\prime}=2 \mathrm{~A}^{\prime}+\mathrm{B}^{\prime}$ is verified.
ii) Given that: $\mathrm{A}=\left[\begin{array}{ll}1 & 2 \\ 4 & 1 \\ 5 & 6\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{ll}1 & 2 \\ 6 & 4 \\ 7 & 3\end{array}\right]$
L.H.S. $(\mathrm{A}-\mathrm{B})^{\prime}=\left[\left(\begin{array}{ll}1 & 2 \\ 4 & 1 \\ 5 & 6\end{array}\right)-\left(\begin{array}{ll}1 & 2 \\ 6 & 4 \\ 7 & 3\end{array}\right)\right]^{\prime}$
$\begin{aligned} & =\left[\begin{array}{cc}1-1 & 2-2 \\ 4-6 & 1-4 \\ 5-7 & 6-3\end{array}\right]^{\prime} \\ & =\left[\begin{array}{cc}0 & 0 \\ -2 & -3 \\ -2 & 3\end{array}\right]^{\prime} \\ & =\left[\begin{array}{ccc}0 & -2 & -2 \\ 0 & -3 & 3\end{array}\right]\end{aligned}$
R.H.S. $\mathrm{A}^{\prime}-\mathrm{B}^{\prime}=\left[\begin{array}{ll}1 & 2 \\ 4 & 1 \\ 5 & 6\end{array}\right]^{\prime}-\left[\begin{array}{ll}1 & 2 \\ 6 & 4 \\ 7 & 3\end{array}\right]^{\prime}$
$\begin{aligned} & =\left[\begin{array}{lll}1 & 4 & 5 \\ 2 & 1 & 6\end{array}\right]-\left[\begin{array}{lll}1 & 6 & 7 \\ 2 & 4 & 3\end{array}\right] \\ & =\left[\begin{array}{ccc}1-1 & 4-6 & 5-7 \\ 2-2 & 1-4 & 6-3\end{array}\right] \\ & =\left[\begin{array}{ccc}0 & -2 & -2 \\ 0 & -3 & 3\end{array}\right]\end{aligned}$
Hence, L.H.S. = R.H.S. $(\mathrm{A}-\mathrm{B})^{\prime}=\mathrm{A}^{\prime}-\mathrm{B}^{\prime}$ is verified.
Question:29
Show that A’A and AA’ are both symmetric matrices for any matrix A.
Answer:
We know that,
In linear algebra, a symmetric matrix is a square matrix that is equal to its transpose. Formally, because equal matrices have equal dimensions, only square matrices can be symmetric.
We know that the transposition of AB is given by
(AB)’ = B’A’
Using this result, and by taking the transpose of A’A, we have,
Transpose of A’A = (A’A)T = (A’A)’
Using, transpose of A’A = (A’A)’
⇒ (A’A)’ = A’(A’)’
And also,
(A’)’ = A
So,
(A’A)’ = A’A
Since (A’A)’ = A’A
This means that A’A is the symmetric matrix for any matrix A.
Now, take the transpose of AA’.
Transpose of AA’ = (AA’)’
⇒ (AA’)’ = (A’)’A’ [ (AB)’ = B’A’]
⇒ (AA’)’ = AA’ [(A’)’ = A]
Since (AA’)’ = AA’
This means, AA’ is the symmetric matrix for any matrix A.
Thus, A’A and AA’ are symmetric matrices for any matrix A.
Question:30 Let A and B be square matrices of the order 3 × 3. Is $(AB)^2 = A^2B^2$? Give reasons.
Answer:
As A and B are square matrices of order $3 \times 3$.
We have, $(\mathrm{AB})^2=\mathrm{AB} \cdot \mathrm{AB}$
$\begin{aligned} & =\mathrm{A}(\mathrm{BA}) \mathrm{B} \\ & =\mathrm{A}(\mathrm{AB}) \mathrm{B} \ldots \ldots .[\mathrm{If} \mathrm{AB}=\mathrm{BA}] \\ & =\mathrm{AABB} \\ & =\mathrm{A}^2 \mathrm{~B}^2\end{aligned}$
Thus, $(\mathrm{AB})^2=\mathrm{A}^2 \mathrm{~B}^2$ is true only if $\mathrm{AB}=\mathrm{BA}$.
Question:31
Show that if A and B are square matrices such that AB = BA, then $(A + B)^2 = A^2 + 2AB + B^2$.
Answer:
Given that $A$ and $B$ are square matrices such that $A B=B A$.
So, $(\mathrm{A}+\mathrm{B})^2=(\mathrm{A}+\mathrm{B}) \cdot(\mathrm{A}+\mathrm{B})$
$\begin{aligned} & =A^2+A B+B A+B^2 \\ & =A^2+A B+A B+B^2 \ldots \ldots .[\text { Since, } A B=B A] \\ & =A^2+2 A B+B^2\end{aligned}$
Question:32.1
Answer:
Given,
$A=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right], B=\left[\begin{array}{ll} 4 & 0 \\ 1 & 5 \end{array}\right], C=\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right]$
$\begin{aligned} &\text { LHS }=A+(B+C)=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]+\left(\left[\begin{array}{cc} 4 & 0 \\ 1 & 5 \end{array}\right]+\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right]\right)\\ &\Rightarrow \mathrm{LHS}=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]+\left(\left[\begin{array}{ll} 4+2 & 0+0 \\ 1+1 & 5-2 \end{array}\right]\right) \end{aligned}$
$\begin{aligned} &\Rightarrow \mathrm{LHS}=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]+\left[\begin{array}{ll} 6 & 0 \\ 2 & 3 \end{array}\right]=\left[\begin{array}{ll} 7 & 2 \\ 1 & 6 \end{array}\right]\\ &R H S=(A+B)+C=\left(\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]+\left[\begin{array}{cc} 4 & 0 \\ 1 & 5 \end{array}\right]\right)+\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right]\\ &\Rightarrow \mathrm{RHS}=\left(\left[\begin{array}{cc} 1+4 & 2+0 \\ -1+1 & 3+5 \end{array}\right]\right)+\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right]\\ &\Rightarrow \mathrm{RHS}=\left[\begin{array}{ll} 5 & 2 \\ 0 & 8 \end{array}\right]+\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right]=\left[\begin{array}{ll} 7 & 2 \\ 1 & 6 \end{array}\right]\\ &\text { Clearly LHS }=R H S=\left[\begin{array}{ll} 7 & 2 \\ 1 & 6 \end{array}\right]\\ &\text { Hence, we have }\\ &A+(B+C)=(A+B)+C \text { ...proved } \end{aligned}$
Question:32.2
Answer:
We have to prove that: A(BC) = (AB)C
$\begin{array}{l} \text { LHS = } \mathrm{A}(\mathrm{BC})=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]\left(\left[\begin{array}{cc} 4 & 0 \\ 1 & 5 \end{array}\right]\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right]\right) \\ \Rightarrow \mathrm{LHS}=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]\left(\begin{array}{cc} 4 \times 2+0 \times 1 & 4 \times 0+0 \times(-2) \\ 1 \times 2+5 \times 1 & 1 \times 0+5 \times(-2)]) \end{array}\right. \\ \Rightarrow \mathrm{LHS}=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]\left[\begin{array}{cc} 8 & 0 \\ 7 & -10 \end{array}\right] \\ LHS= {\left[\begin{array}{cc} 22 & -20 \\ 13 & -30 \end{array}\right]} \end{array}$
$\begin{aligned} &\mathrm{RHS}=(\mathrm{AB}) \mathrm{C}=\left(\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]\left[\begin{array}{cc} 4 & 0 \\ 1 & 5 \end{array}\right]\right)\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right]\\ &\text { By matrix multiplication as done for LHS }\\ &\Rightarrow \mathrm{RHS}=\left[\begin{array}{cc} 6 & 10 \\ -1 & 15 \end{array}\right]\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right]\\ &\Rightarrow \mathrm{RHS}=\left[\begin{array}{ll} 22 & -20 \\ 13 & -30 \end{array}\right]\\ &\text { Evidently, LHS = RHS }=\left[\begin{array}{rr} 22 & -20 \\ 13 & -30 \end{array}\right]\\ &\therefore \mathrm{A}(\mathrm{BC})=(\mathrm{AB}) \mathrm{C} \ldots \text { .proved } \end{aligned}$
Question:32.3
Answer:
To prove: (a + b)B = aB + bB
Given, a = 4 and b = -2
$\\LHS =(4+(-2)) B=2\left[\begin{array}{ll}4 & 0 \\ 1 & 5\end{array}\right]=\left[\begin{array}{cc}8 & 0 \\ 2 & 10\end{array}\right]$ \\$\mathrm{RHS}=\mathrm{aB}+\mathrm{bB}=4\left[\begin{array}{ll}4 & 0 \\ 1 & 5\end{array}\right]-2\left[\begin{array}{ll}4 & 0 \\ 1 & 5\end{array}\right]$ \\$\Rightarrow \mathrm{RHS}=\left[\begin{array}{cc}16 & 0 \\ 4 & 20\end{array}\right]-\left[\begin{array}{cc}8 & 0 \\ 2 & 10\end{array}\right]=\left[\begin{array}{cc}8 & 0 \\ 2 & 10\end{array}\right]$
It is clear that, $\mathrm{LHS}=\mathrm{RHS}=\left[\begin{array}{cc}8 & 0 \\ 2 & 10\end{array}\right]$
Hence, we have,
(a + b)B = aB + bB …proved
Question:32.4
Show that:
a(C - A) = aC -aA
Answer:
We have,
$\begin{aligned} A & =\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right] \\ B & =\left[\begin{array}{cc}4 & 0 \\ 1 & 5\end{array}\right] \\ C & =\left[\begin{array}{cc}2 & 0 \\ 1 & -2\end{array}\right]\end{aligned}$
And $a=4, b=-2$
And $a(C-A)=4(C-A)$
$=\left[\begin{array}{cc}4 & -8 \\ 8 & -20\end{array}\right]$
Also, $\mathrm{aC}-\mathrm{aA}=4 \mathrm{C}-4 \mathrm{~A}$
$\begin{aligned} & =\left[\begin{array}{cc}8 & 0 \\ 4 & -8\end{array}\right]-\left[\begin{array}{cc}4 & 8 \\ -4 & 12\end{array}\right] \\ & =\left[\begin{array}{cc}4 & -8 \\ 8 & -20\end{array}\right] \\ & =\mathrm{a}(\mathrm{C}-\mathrm{A})\end{aligned}$
Clearly $L H S=\mathrm{RHS}=\left[\begin{array}{cc}4 & -8 \\ 8 & -20\end{array}\right]$
Hence proved.
Question:32.5
Answer:
To prove: $(AT)^{}T = A$
In the transpose of a matrix, the rows of the matrix become the columns.
$\\\text { LHS }=\left(A^{T}\right)^{T}\\=\left(\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]^{\mathrm{T}}\right)^{\mathrm{T}}=\left[\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right]^{\mathrm{T}}=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]=\mathrm{A}=\mathrm{RHS}$
Hence, proved.
Question:32.6
Answer:
a) To prove: $(bA)^T = bA^T$
As, LHS = $(bA)^T = (-2A)^T=(-2\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right])^T$
$(bA)^T = (-2A)^T=\left[\begin{array}{cc} -2 & -4 \\ 2 & -6 \end{array}\right]^T=\left[\begin{array}{cc} -2 & 2 \\ -4 & -6 \end{array}\right]$
$\begin{aligned} &\text { Similarly, }\\ &\mathrm{RHS}=-2\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]^{\mathrm{T}}=-2\left[\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right]=\left[\begin{array}{cc} -2 & 2 \\ -4 & -6 \end{array}\right]\\ &\text { Hence proved }L H S=R H S=\left[\begin{array}{cc} -2 &2 \\ -4 & -6 \end{array}\right]\\ &\text { Then, }(b A)^{T}=b A^{\top} \ldots \text { proved } \end{aligned}$
Question:32.7
Answer:
To prove: $(AB)^T = B^T A^T$
By multiplying the matrices and taking the transpose, we get,
$\begin{aligned} & \therefore \text { LHS }=\left(\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right]\left[\begin{array}{ll}4 & 0 \\ 1 & 5\end{array}\right]\right)^{\mathrm{T}} \\ & \Rightarrow \text { LHS }=\left[\begin{array}{cl}1 \times 4+2 \times 1 & 1 \times 0+2 \times 5 \\ -1 \times 4+3 \times 1 & -1 \times 0+3 \times 5\end{array}\right]^{\mathrm{T}}=\left[\begin{array}{cc}6 & 10 \\ -1 & 15\end{array}\right]^{\mathrm{T}} \\ & \therefore \text { LHS }=\left[\begin{array}{cc}6 & -1 \\ 10 & 15\end{array}\right]\end{aligned}$
As $\mathrm{RHS}=\mathrm{B}^{\top} \mathrm{A}^{\top}$
By taking the transpose of matrices and then multiplying, we get,
$\begin{aligned} & \text { RHS }=\left[\begin{array}{ll}4 & 0 \\ 1 & 5\end{array}\right]^{\mathrm{T}}\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right]^{\mathrm{T}}=\left[\begin{array}{ll}4 & 1 \\ 0 & 5\end{array}\right]\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right] \\ & \Rightarrow \text { RHS }=\left[\begin{array}{l}4 \times 1+1 \times(2) \\ 4 \times(-1)+1 \times 3 \\ 0 \times 1+5 \times(2) \\ 0 \times(-1)+5 \times 3\end{array}\right]=\left[\begin{array}{cc}6 & -1 \\ 10 & 15\end{array}\right]\end{aligned}$
We have, LHS = RHS = $\left[\begin{array}{cc}6 & -1 \\ 10 & 15\end{array}\right]$
Hence $(A B)^{\top}=B^{\top} A^{\top}$... proved
Question:32.8
Answer:
We have,
$\begin{aligned} & A=\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right] \\ & B=\left[\begin{array}{ll}4 & 0 \\ 1 & 5\end{array}\right] \\ & C=\left[\begin{array}{cc}2 & 0 \\ 1 & -2\end{array}\right] a\end{aligned}$
And $a=4, b=-2$
$\begin{aligned} & (\mathrm{A}-\mathrm{B})=\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right]-\left[\begin{array}{ll}4 & 0 \\ 1 & 5\end{array}\right] \\ & =\left[\begin{array}{cc}1-4 & 2-0 \\ -1-1 & 3-5\end{array}\right] \\ & =\left[\begin{array}{cc}-3 & 2 \\ -2 & -2\end{array}\right]\end{aligned}$
$\begin{aligned} & \therefore(\mathrm{A}-\mathrm{B}) \mathrm{C}=\left[\begin{array}{cc}-3 & 2 \\ -2 & -2\end{array}\right]\left[\begin{array}{cc}2 & 0 \\ 1 & -2\end{array}\right] \\ & =\left[\begin{array}{cc}-4 & -4 \\ -6 & 4\end{array}\right]\end{aligned}$
Now, $A C=\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right]\left[\begin{array}{cc}2 & 0 \\ 1 & -2\end{array}\right]$
$=\left[\begin{array}{ll}4 & -4 \\ 1 & -6\end{array}\right]$
And $B C=\left[\begin{array}{ll}4 & 0 \\ 1 & 5\end{array}\right]\left[\begin{array}{cc}2 & 0 \\ 1 & -2\end{array}\right]$
$=\left[\begin{array}{cc}8 & 0 \\ 7 & -10\end{array}\right]$
$\begin{aligned} & \therefore \mathrm{AC}-\mathrm{BC}=\left[\begin{array}{ll}4-8 & -4-0 \\ 1-7 & -6+10\end{array}\right] \\ & =\left[\begin{array}{cc}-4 & -4 \\ -6 & 4\end{array}\right] \\ & =(\mathrm{A}-\mathrm{B}) \mathrm{C}\end{aligned}$
Hence proved.
Question:32.9
Show that:
$(A - B)^T = A^T - B^T$
Answer:
To Prove: $(A - B)^T = A^T - B^T$
$(A-B)=\left[\begin{array}{cc} 1-4 & 2-0 \\ -1-1 & 3-5 \end{array}\right]=\left[\begin{array}{cc} -3 & 2 \\ -2 & -2 \end{array}\right]$
$(A-B)^T=\left[\begin{array}{cc} -3 & -2 \\ 2 & -2 \end{array}\right]$
$\\A^{T}-B^{T}=\left[\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right]-\left[\begin{array}{cc} 4 & 1 \\ 0 & 5 \end{array}\right]\\=\left[\begin{array}{cc} -3 & -2 \\ 2 & -2 \end{array}\right]\\=(A-B)^{T}$
Hence $(A-B)^{T}=A^T-B^T$.
Question:33
Answer:
As $A=\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right],$
$A^2=\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$
According to the rule of matrix multiplication:
$\\$$ \mathrm{A}^{2}=\left[\begin{array}{cc} \cos \theta \times \cos \theta+\sin \theta \times(-\sin \theta) & \cos \theta \times \sin \theta+\sin \theta \times \cos \theta \\ -\cos \theta \times \sin \theta+(-\sin \theta \times \cos \theta) & \cos \theta \times \cos \theta+\sin \theta \times(-\sin \theta) \end{array}\right] $$ \\ \Rightarrow \mathrm{A}^{2}=\left[\begin{array}{cc}\cos ^{2} \theta-\sin ^{2} \theta & 2 \sin \theta \cos \theta \\ -2 \sin \theta \cos \theta & \cos ^{2} \theta-\sin ^{2} \theta\end{array}\right]$
We know that:
$\\2 \sin \theta \cos \theta=\sin 2 \theta$ and $\cos ^{2} \theta-\sin ^{2} \theta=\cos 2 \theta \\\therefore A^{2}=\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{array}\right]_{\ldots}$
Hence, proved.
Question:34
Answer:
$\begin{aligned} &\text { As, LHS }=(A+B)^{2}=\left(\left[\begin{array}{cc} 0 & -x \\ x & 0 \end{array}\right]+\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]\right)^{2}\\ &\Rightarrow \mathrm{LHS}=\left[\begin{array}{cc} 0 & 1-\mathrm{x} \\ 1+\mathrm{x} & 0 \end{array}\right]^{2}=\left[\begin{array}{cc} 0 & 1-\mathrm{x} \\ 1+\mathrm{x} & 0 \end{array}\right]\left[\begin{array}{cc} 0 & 1-\mathrm{x} \\ 1+\mathrm{x} & 0 \end{array}\right]\\ &\text { By the rule of matrix multiplication we can write LHS as - }\\ &\text { LHS }=\left[\begin{array}{cc} 0+(1-x)(1+x) & 0 \\ 0 & (1+x)(1-x) \end{array}\right]\\ &\Rightarrow \mathrm{LHS}=\left[\begin{array}{cc} 1-\mathrm{x}^{2} & 0 \\ 0 & 1-\mathrm{x}^{2} \end{array}\right]\\ &\text { Given } x^{2}=-1\\ &\therefore \mathrm{LHS}=\left[\begin{array}{cc} 1-(-1) & 0 \\ 0 & 1-(-1) \end{array}\right]=\left[\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right]\\ \end{aligned}$
$\begin{aligned} &\mathrm{RHS}=\mathrm{A}^{2}+\mathrm{B}^{2}=\left[\begin{array}{cc} 0 & -\mathrm{x} \\ \mathrm{x} & 0 \end{array}\right]^{2}+\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]^{2}\\ &\Rightarrow \mathrm{RHS}=\left[\begin{array}{cc} 0 & -\mathrm{x} \\ \mathrm{x} & 0 \end{array}\right]\left[\begin{array}{cc} 0 & -\mathrm{x} \\ \mathrm{x} & 0 \end{array}\right]+\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right] \end{aligned}$
By the rule of matrix multiplication, we can write-
$\mathrm{RHS}=\left[\begin{array}{cc}-\mathrm{x}^{2} & 0 \\ 0 & -\mathrm{x}^{2}\end{array}\right]+\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}1-\mathrm{x}^{2} & 0 \\ 0 & 1-\mathrm{x}^{2}\end{array}\right]$
Given $x^{2}=-1$
$\therefore \mathrm{RHS}=\left[\begin{array}{cc}1-(-1) & 0 \\ 0 & 1-(-1)\end{array}\right]=\left[\begin{array}{ll}2 & 0 \\ 0 & 2\end{array}\right]$
We have, $\mathrm{RHS}=\mathrm{LHS}=\left[\begin{array}{ll}2 & 0 \\ 0 & 2\end{array}\right]$
Hence, $(A+B)^{2}=A^{2}+B^{2}$. -proved
Question:35
Answer:
We need to prove that
$\begin{array}{l} A^{2}=I=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ \because A=\left[\begin{array}{ccc} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{array}\right] \\ \therefore A^{2}=\left[\begin{array}{ccc} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{array}\right]\left[\begin{array}{ccc} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{array}\right] \end{array}$
According to the rule of matrix multiplication, we have-
$\begin{aligned} &A^{2}=\left[\begin{array}{ccc} 0 \times 0+1 \times 4+(-1) \times 3 & 0 \times 1+1 \times(-3)+(-1) \times(-3) & 0 \times(-1)+1 \times 4+(-1) \times 4 \\ 4 \times 0+(-3) \times 4+4 \times 3 & 4 \times 1+(-3) \times(-3)+4 \times(-3) & 4 \times(-1)+(-3) \times 4+4 \times 4 \\ 3 \times 0+(-3) \times 4+4 \times 3 & 3 \times 1+(-3) \times(-3)+4 \times(-3) & 3 \times(-1)+(-3) \times 4+4 \times 4 \end{array}\right]\\ &\Rightarrow \mathrm{A}^{2}=\left[\begin{array}{ccc} 4-3 & -3+3 & 4-4 \\ -12+12 & 4+9-12 & -4-12+16 \\ -12+12 & 3+9-12 & -3+16-12 \end{array}\right]\\ &\therefore \mathrm{A}^{2}=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=I\\ &\text { Hence Proved } \end{aligned}$
Question:36
Prove by Mathematical Induction that $(A')^n = (A^n)'$, where n ∈ N for any square matrix A.
Answer:
Let $\mathrm{P}(\mathrm{n}):\left(\mathrm{A}^{\prime}\right)^{\mathrm{n}}=\left(\mathrm{A}^{\mathrm{n}}\right)^{\prime}$
$\therefore \mathrm{P}(1):\left(\mathrm{A}^{\prime}\right)=(\mathrm{A})^{\prime}$
$\Rightarrow \mathrm{A}^{\prime}=\mathrm{A}^{\prime}$
$\Rightarrow \mathrm{P}(1)$ is true.
Now, let $\mathrm{P}(\mathrm{k})=\left(\mathrm{A}^{\prime}\right)^{\mathrm{k}}=\left(\mathrm{A}^{\mathrm{k}}\right)^{\prime}$
Where $k \in N$
And $\mathrm{P}(\mathrm{k}+1):\left(\mathrm{A}^{\prime}\right)^{\mathrm{k}+1}=\left(\mathrm{A}^{\prime}\right)^{\mathrm{k}} \mathrm{A}^{\prime}$
$=\left(\mathrm{A}^{\prime}\right)^{\mathrm{k}^{\prime}} \mathrm{A}^{\prime}$
$=\left(\mathrm{AA}^{\mathrm{k}}\right)^{\prime} \quad . . . . .\left(\mathrm{As}(\mathrm{AB})^{\prime}=\mathrm{B}^{\prime} \mathrm{A}^{\prime}\right)$
$=\left(\mathrm{A}^{\mathrm{k}+1}\right)^{\prime}$
Thus $\mathrm{P}(1)$ is true and whenever $\mathrm{P}(\mathrm{k})$ is true $\mathrm{P}(\mathrm{k}+1)$ is true, So, $\mathrm{P}(\mathrm{n})$ is true for all $\mathrm{n} \in \mathrm{N}$
Question:37.1
Let $A=\begin{bmatrix} 1 &3 \\-5 & 7 \end{bmatrix}$
To apply elementary row transformations, we can say that:
A = IA where I is the identity matrix
We proceed with solving the problem in such a way that LHS becomes I and the transformations in I give us a new matrix such that.
I = XA
And this X is called the inverse of A = A-1.
So we get:
$\begin{aligned} &\left[\begin{array}{cc} 1 & 3 \\ -5 & 7 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \mathrm{A}\\ &\text { By Applying } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}+5 \mathrm{R}_{1}\\ &\Rightarrow\left[\begin{array}{cc} 1 & 3 \\ 0 & 22 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 5 & 1 \end{array}\right] \mathrm{A}\\ &\text { By Applying } \mathrm{R}_{2} \rightarrow(1 / 22) \mathrm{R}_{2}\\ &\Rightarrow\left[\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ \frac{5}{22} & \frac{1}{22} \end{array}\right] \mathrm{A}\\ &\text { By Applying } \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-3 \mathrm{R}_{2}\\ &\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} \frac{7}{22} & -\frac{3}{22} \\ \frac{5}{22} & \frac{1}{22} \end{array}\right] \mathrm{A}\\ &\text { As we have an Identity matrix in LHS. }\\ \end{aligned}$
$\begin{aligned} &\therefore A^{-1}=\left[\begin{array}{ll} \frac{7}{22} & -\frac{3}{22} \\ \frac{5}{22} & \frac{1}{22} \end{array}\right] \end{aligned}$
Question:37.2
Answer:
Let $B=\begin{bmatrix} 1 &-3 \\-2 & 6 \end{bmatrix}$
To apply elementary row transformations, we write:
B = IB where I is the identity matrix
We proceed with solving the problem in such a way that LHS becomes I and the transformations in I give us a new matrix such that.
I = XB
And this X is called inverse of $B = B^{-1}$
So we get,
$\begin{array}{l} {\left[\begin{array}{cc} 1 & -3 \\ -2 & 6 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \mathrm{B}} \end{array}$
By Applying R2→ R2 + 2R1
$\Rightarrow\left[\begin{array}{cc} 1 & -3 \\ 0 & 0 \end{array}\right]=\left[\begin{array}{cc} 1 & 0 \\ 2 & 1 \end{array}\right] \mathrm{A}$
We have all zeroes in one of the rows of the matrix in the LHS.
So by any means, we can't make an identity matrix in LHS.
∴ The inverse of B does not exist.
$B^{-1}$ does not exist.
Question:38
Answer:
Given that: $\left[\begin{array}{cc}x y & 4 \\ z+6 & x+y\end{array}\right]=\left[\begin{array}{cc}8 & w \\ 0 & 6\end{array}\right]$
Equating the corresponding elements,
$\begin{aligned} & x y=8 \\ & w=4 \\ & z+6=0 \\ & \Rightarrow z=-6, x+y=6\end{aligned}$
Now, solving $x+y=6$ $\qquad$
And $x y=8$ $\qquad$ (ii)
From equation (i), $y=6-x$ $\qquad$
Putting the value of $y$ in equation (ii), we get,
$\begin{aligned} & x(6-x)=8 \\ & \Rightarrow 6 x-x^2=8 \\ & \Rightarrow x^2-6 x+8=0 \\ & \Rightarrow x^2-4 x-2 x+8=0 \\ & \Rightarrow x(x-4)-2(x-4)=0 \\ & \Rightarrow(x-4)(x-2)=0 \\ & \therefore x=4,2\end{aligned}$
From equation (iii)
$y=2,4$
Hence, $x=4$ or $2, y=2$ or $4, z=-6$ and $w=4$.
Question:39
Answer:
Given that:
3A + 5B + 2C = O = null matrix
We have to determine the value of C,
$\begin{array}{l} \text { As, } 3\left[\begin{array}{lr} 1 & 5 \\ 7 & 12 \end{array}\right]+5\left[\begin{array}{ll} 9 & 1 \\ 7 & 8 \end{array}\right]+2 C=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \\ \Rightarrow\left[\begin{array}{ll} 3 & 15 \\ 21 & 36 \end{array}\right]+\left[\begin{array}{ll} 45 & 5 \\ 35 & 40 \end{array}\right]+2 C=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \\ \Rightarrow 2 C+\left[\begin{array}{ll} 48 & 20 \\ 56 & 76 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \\ \therefore 2 C=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]-\left[\begin{array}{ll} 48 & 20 \\ 56 & 76 \end{array}\right] \\ \Rightarrow 2 C=\left[\begin{array}{ll} -48 & -20 \\ -56 & -76 \end{array}\right] \\ \therefore C=\frac{1}{2}\left[\begin{array}{ll} -48 & -20 \\ -56 & -76 \end{array}\right]=\left[\begin{array}{ll} -24 & -10 \\ -28 & -38 \end{array}\right] \end{array}$
Question:40
If $A=\begin{bmatrix} 3 &-5 \\-4 & 2 \end{bmatrix}$ then find $A^2 - 5A - 14I$. Hence, obtain $A^3$.
Answer:
Given that: $\mathrm{A}=\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]$
$\mathrm{A}^2=\mathrm{A} \cdot \mathrm{A}$
$\begin{aligned} & =\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right] \\ & =\left[\begin{array}{cc}9+20 & -15-10 \\ -12-8 & 20+4\end{array}\right] \\ & =\left[\begin{array}{cc}29 & -25 \\ -20 & 24\end{array}\right]\end{aligned}$
$\therefore \mathrm{A}^2-5 \mathrm{~A}-14 \mathrm{I}=\left[\begin{array}{cc}29 & -25 \\ -20 & -24\end{array}\right]-5\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]-14\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$\begin{aligned} & =\left[\begin{array}{cc}29 & -25 \\ -20 & 24\end{array}\right]-\left[\begin{array}{cc}15 & -25 \\ -20 & 10\end{array}\right]-\left[\begin{array}{cc}14 & 0 \\ 0 & 14\end{array}\right] \\ & =\left[\begin{array}{cc}29 & -25 \\ -20 & 24\end{array}\right]-\left[\begin{array}{cc}29 & -25 \\ -20 & 24\end{array}\right] \\ & =\left[\begin{array}{cc}29-29 & -25+25 \\ -20+20 & 24-24\end{array}\right] \\ & =\left[\begin{array}{cc}0 & 0 \\ 0 & 0\end{array}\right]\end{aligned}$
Hence, $\mathrm{A}^2-5 \mathrm{~A}-14 \mathrm{I}=0$
Now, multiplying both sides by A, we get,
$A^2 \cdot A-5 A \cdot A-14 I A=0 A$
$\begin{aligned} & \Rightarrow A^3-5 A^2-14 A=0 \\ & \Rightarrow A^3=5 A^2+14 A \\ & \Rightarrow A^3=5\left[\begin{array}{cc}29 & -25 \\ -20 & 24\end{array}\right]+14\left[\begin{array}{cc}3 & -5 \\ -4 & -2\end{array}\right]\end{aligned}$
$\begin{aligned} & =\left[\begin{array}{cc}145 & -125 \\ -100 & 120\end{array}\right]+\left[\begin{array}{cc}42 & -70 \\ -56 & 28\end{array}\right] \\ & =\left[\begin{array}{cc}145+42 & -125-70 \\ -100-56 & 120+28\end{array}\right] \\ & =\left[\begin{array}{cc}187 & -195 \\ -156 & 148\end{array}\right]\end{aligned}$
Hence, $A^3=\left[\begin{array}{cc}187 & -195 \\ -156 & 148\end{array}\right]$
Question:41
Answer:
Given that: $3\left[\begin{array}{ll}\mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{d}\end{array}\right]=\left[\begin{array}{cc}\mathrm{a} & 6 \\ -1 & 2 \mathrm{~d}\end{array}\right]+\left[\begin{array}{cc}4 & \mathrm{a}+\mathrm{b} \\ \mathrm{c}+\mathrm{d} & 3\end{array}\right]$
$\left[\begin{array}{ll}3 \mathrm{a} & 3 \mathrm{~b} \\ 3 \mathrm{c} & 3 \mathrm{~d}\end{array}\right]=\left[\begin{array}{cc}\mathrm{a}+4 & 6+\mathrm{a}+\mathrm{b} \\ -1+\mathrm{c}+\mathrm{d} & 2 \mathrm{~d}+3\end{array}\right]$
Equating the corresponding elements, we get,
$\begin{aligned} & 3 \mathrm{a}=\mathrm{a}+4 \\ & \Rightarrow 3 \mathrm{a}-\mathrm{a}=4 \\ & \Rightarrow 2 \mathrm{a}=4 \\ & \Rightarrow \mathrm{a}=2 \\ & 3 \mathrm{~b}=6+\mathrm{a}+\mathrm{b} \\ & \Rightarrow 3 \mathrm{~b}-\mathrm{b}-\mathrm{a}=6 \\ & \Rightarrow 2 \mathrm{~b}-\mathrm{a}=6 \\ & \Rightarrow 2 \mathrm{~b}-2=6 \\ & \Rightarrow 2 \mathrm{~b}=8 \\ & \Rightarrow \mathrm{~b}=4\end{aligned}$
$\begin{aligned} & 3 \mathrm{c}=-1+\mathrm{c}+\mathrm{d} \\ & \Rightarrow 3 \mathrm{c}-\mathrm{c}-\mathrm{d}=-1 \\ & \Rightarrow 2 \mathrm{c}-\mathrm{d}=-1\end{aligned}$
And $3 \mathrm{~d}=2 \mathrm{~d}+3$
$\begin{aligned} & \Rightarrow 3 d-2 d=3 \\ & \Rightarrow d=3\end{aligned}$
Now $2 \mathrm{c}-\mathrm{d}=-1$
$\begin{aligned} & \Rightarrow 2 \mathrm{c}-3=-1 \\ & \Rightarrow 2 \mathrm{c}=3-1 \\ & \Rightarrow 2 \mathrm{c}=2 \\ & \therefore \mathrm{c}=1 \\ & \therefore \mathrm{a}=2, \mathrm{~b}=4, \mathrm{c}=1 \text { and } \mathrm{d}=3 .\end{aligned}$
Question:42
Answer:
Order of matrix $\left[\begin{array}{cc}2 & -1 \\ 1 & 0 \\ -3 & 4\end{array}\right]$ is $3 \times 2$ and the matrix
$\left[\begin{array}{ccc}-1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15\end{array}\right]$ is $3 \times 3$
$\therefore$ Order of matrix A must be $2 \times 3$
Let $\mathrm{A}=\left[\begin{array}{lll}\mathrm{a} & \mathrm{b} & \mathrm{c} \\ \mathrm{d} & \mathrm{e} & \mathrm{f}\end{array}\right]_{2 \times 3}$
So, $\left[\begin{array}{cc}2 & -1 \\ 1 & 0 \\ -3 & 4\end{array}\right]\left[\begin{array}{lll}\mathrm{a} & \mathrm{b} & \mathrm{c} \\ \mathrm{d} & \mathrm{e} & \mathrm{f}\end{array}\right]=\left[\begin{array}{ccc}-1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15\end{array}\right]$
$\left[\begin{array}{ccc}2 \mathrm{a}-\mathrm{d} & 2 \mathrm{~b}-\mathrm{e} & 2 \mathrm{c}-\mathrm{f} \\ \mathrm{a}+0 & \mathrm{~b}+0 & \mathrm{c}+0 \\ -3 \mathrm{a}+4 \mathrm{~d} & -3 \mathrm{~b}+4 \mathrm{e} & -3 \mathrm{c}+4 \mathrm{f}\end{array}\right]=\left[\begin{array}{ccc}-1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 5\end{array}\right]$
Equating the corresponding elements, we get,
$\begin{aligned} & 2 a-d=-1 \text { and } a=1 \\ & \Rightarrow 2 \times 1-d=-1 \\ & \Rightarrow d=2+1 \\ & \Rightarrow d=3\end{aligned}$
$\begin{aligned} & 2 \mathrm{~b}-\mathrm{e}=-8 \text { and } \mathrm{b}=-2 \\ & \Rightarrow 2(-2)-\mathrm{e} \\ & \Rightarrow-8 \\ & \Rightarrow-4-\mathrm{e}=-8 \\ & \Rightarrow \mathrm{e}=4 \\ & 2 \mathrm{c}-\mathrm{f}=-10 \text { and } \mathrm{c}=-5 \\ & \Rightarrow 2(-5)-\mathrm{f}=-10 \\ & \Rightarrow-10-\mathrm{f}=-10 \\ & \Rightarrow \mathrm{f}=0 \\ & \mathrm{a}=1, \mathrm{~b}=-2, \mathrm{c}=-5, \mathrm{~d}=3, \mathrm{e}=4 \text { and } \mathrm{f}=0\end{aligned}$
Hence, $A=\left[\begin{array}{ccc}1 & -2 & -5 \\ 3 & 4 & 0\end{array}\right]$.
Question:43
If $A=\begin{bmatrix} 1 &2 \\4 &1 \end{bmatrix}$ find $A^2 + 2A + 7I$
Answer:
We are given the following matrix A such that,
$A=\begin{bmatrix} 1 &2 \\4 &1 \end{bmatrix}$
$\begin{array}{l} \because \mathrm{A}^{2}=\mathrm{A} . \mathrm{A} \\ \Rightarrow \mathrm{A}^{2}=\left[\begin{array}{ll} 1 & 2 \\ 4 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 2 \\ 4 & 1 \end{array}\right] \end{array}$
According to the rule of matrix multiplication, we get
$\begin{aligned} &A^{2}=\left[\begin{array}{ll} 1 \times 1+2 \times 4 & 1 \times 2+2 \times 1 \\ 4 \times 1+1 \times 4 & 4 \times 2+1 \times 1 \end{array}\right]\\ &\Rightarrow A^{2}=\left[\begin{array}{ll} 9 & 4 \\ 8 & 9 \end{array}\right]\\ &\therefore \mathrm{A}^{2}+2 \mathrm{~A}+71=\left[\begin{array}{ll} 9 & 4 \\ 8 & 9 \end{array}\right]+2\left[\begin{array}{ll} 1 & 2 \\ 4 & 1 \end{array}\right]+7\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\\ &\Rightarrow \mathrm{A}^{2}+2 \mathrm{~A}+7 \mathrm{I}=\left[\begin{array}{ll} 9 & 4 \\ 8 & 9 \end{array}\right]+\left[\begin{array}{ll} 2 & 4 \\ 8 & 2 \end{array}\right]+\left[\begin{array}{ll} 7 & 0 \\ 0 & 7 \end{array}\right]\\ &\Rightarrow A^{2}+2 A+7 I=\left[\begin{array}{ll} 9+2+7 & 4+4+0 \\ 8+8+0 & 9+2+7 \end{array}\right]\\ &\Rightarrow \mathrm{A}^{2}+2 \mathrm{~A}+7I=\left[\begin{array}{cc} 18 & 8 \\ 16 & 18 \end{array}\right] \ldots \mathrm{ans} \end{aligned}$
Question:44
Answer:
Here, $\mathrm{A}=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$
Given that: $\mathrm{A}^{-1}=\mathrm{A}^{\prime}$
Pre-multiplying both sides by A
$\mathrm{AA}^{-1}=\mathrm{AA}^{\prime}$
$\begin{aligned} & \Rightarrow \mathrm{I}=\mathrm{AA}^{\prime} \quad \ldots \ldots .\left[\because \mathrm{AA}^{-1}=\mathrm{I}\right] \\ & \Rightarrow\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right] \\ & \Rightarrow\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}\cos ^2 \alpha+\sin ^2 \alpha & -\sin \alpha \cos \alpha+\sin \alpha \cos \alpha \\ -\sin \alpha \cos \alpha+\cos \alpha \sin \alpha & \sin ^2 \alpha+\cos ^2 \alpha\end{array}\right] \\ & \Rightarrow\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\end{aligned}$
Hence, it is true for all values of a.
Question:45
Answer:
A matrix is said to be skew-symmetric if A = -A’
Let, A = $\begin{bmatrix} 0 &a &3 \\2 & b & -1\\c &1 &0 \end{bmatrix}$
As A is a skew-symmetric matrix.
∴ A = -A’
$\begin{array}{l} \left[\begin{array}{ccc} 0 & a & 3 \\ 2 & b & -1 \\ c & 1 & 0 \end{array}\right]=-\left[\begin{array}{ccc} 0 & a & 3 \\ 2 & b & -1 \\ c & 1 & 0 \end{array}\right]^{T} \\\\ \left[\begin{array}{ccc} 0 & a & 3 \\ 2 & b & -1 \\ c & 1 & 0 \end{array}\right]=-\left[\begin{array}{ccc} 0 & 2 & c \\ a & b & 1 \\ 3 & -1 & 0 \end{array}\right] \\\\ {\left[\begin{array}{ccc} 0 & a & 3 \\ 2 & b & -1 \\ c & 1 & 0 \end{array}\right]=\left[\begin{array}{ccc} 0 & -2 & -c \\ -a & -b & -1 \\ -3 & 1 & 0 \end{array}\right]} \end{array}$
Equating the respective elements of both matrices, as both matrices are equal to each other, we have,
a = -2 ; c = -3 ; b = -b ⇒ 2b = 0 ⇒ b = 0
Thus, we get,
a = -2 , b = 0 and c = -3
Question:46
Answer:
We have, $\mathrm{P}(\mathrm{x})=\left[\begin{array}{cc}\cos x & \operatorname{six} \\ -\sin x & \cos x\end{array}\right]$
$\therefore \mathrm{P}(\mathrm{y})=\left[\begin{array}{cc}\cos y & \sin y \\ -\sin y & \cos y\end{array}\right]$
Now,
$\mathrm{P}(\mathrm{x}) \cdot \mathrm{P}(\mathrm{y})=\left[\begin{array}{cc}\cos x & \sin x \\ -\sin x & \cos x\end{array}\right]\left[\begin{array}{cc}\cos y & \sin y \\ -\sin y & \cos y\end{array}\right]$
$\begin{aligned} & =\left[\begin{array}{cc}\cos x \cdot \cos y-\sin x \cdot \sin y & \cos x \cdot \sin y+\sin x \cdot \cos y \\ -\sin x \cdot \cos y-\cos x \cdot \sin y & -\sin x \cdot \sin y+\cos x \cdot \cos y\end{array}\right] \\ & =\left[\begin{array}{cc}\cos (x+y) & \sin (x+y) \\ -\sin (x+y) & \cos (x+y)\end{array}\right]\end{aligned}$
$=P(x+y) \ldots \ldots(i)$
Also,
$\mathrm{P}(\mathrm{y}) \cdot \mathrm{P}(\mathrm{x})=\left[\begin{array}{cc}\cos y & \sin y \\ -\sin y & \cos y\end{array}\right]\left[\begin{array}{cc}\cos x & \sin x \\ -\sin x & \cos x\end{array}\right]$
$=\left[\begin{array}{cc}\cos y \cdot \cos x-\sin y \cdot \sin x & \cos y \cdot \sin x+\sin y \cdot \cos x \\ -\sin y \cdot \cos x-\sin x \cdot \cos y & -\sin y \cdot \sin x+\cos y \cdot \cos x\end{array}\right]$
$=\left[\begin{array}{cc}\cos (x+y) & \sin (x+y) \\ -\sin (x+y) & \cos (x+y)\end{array}\right]$
$\mathrm{P}(\mathrm{x}) \cdot(\mathrm{y})=\mathrm{P}(\mathrm{x}+\mathrm{y})=\mathrm{P}(\mathrm{y}) \cdot \mathrm{P}(\mathrm{x})$
Question:47
If A is square matrix such that $A^2 = A$, show that $(I + A)^3 = 7A + I$.
Answer:
We know that,
A. I = I. A
So, A and I are commutative.
Thus, we can expand $(\mathrm{I}+\mathrm{A})^3$ like real numbers expansion.
So, $(I+A)^3=I^3+3 I^2 A+3 \mathrm{IA}^2+A^3$
$\begin{aligned} & =\mathrm{I}+3 \mathrm{IA}+3 \mathrm{~A}^2+\mathrm{AA}^2 \ldots . .\left(\mathrm{As} \mathrm{I}^{\mathrm{n}}=\mathrm{I}, \mathrm{n} \in \mathrm{N}\right) \\ & =\mathrm{I}+3 \mathrm{~A}+3 \mathrm{~A}+\mathrm{AA} \\ & =\mathrm{I}+3 \mathrm{~A}+3 \mathrm{~A}+\mathrm{A}^2 \\ & =\mathrm{I}+3 \mathrm{~A}+3 \mathrm{~A}+\mathrm{A} \\ & =\mathrm{I}+7 \mathrm{~A}\end{aligned}$
Hence proved,
$(I + A)^{3} = I + 7A$
Question:48
Answer:
Given that B is a skew-symmetric matrix
$\therefore \mathrm{B}^{\prime}=-\mathrm{B}$
Let $\mathrm{P}=\mathrm{A}^{\prime} \mathrm{BA}$
$\begin{aligned} & =\mathrm{A}^{\prime} \mathrm{B}^{\prime}\left(\mathrm{A}^{\prime}\right)^{\prime} \ldots \ldots .\left[(\mathrm{AB})^{\prime}=\mathrm{B}^{\prime} \mathrm{A}^{\prime}\right] \\ & =\mathrm{A}^{\prime}(-\mathrm{B}) \mathrm{A} \\ & =-\mathrm{A}^{\prime} \mathrm{BA} \\ & =-\mathrm{P}\end{aligned}$
So $\mathrm{P}^{\prime}=-\mathrm{P}$
Thus, we say that C = A’ BA is a skew-symmetric matrix.
Question:49
If AB = BA for any two square matrices, prove by mathematical induction that $(AB)^n = A^n B^n$.
Answer:
Let $\mathrm{P}(\mathrm{n}):(\mathrm{AB})^{\mathrm{n}}=\mathrm{A}^{\mathrm{n}} \mathrm{B}^{\mathrm{n}}$
So, $\mathrm{P}(1)$ : $(\mathrm{AB})^1=\mathrm{A}^1 \mathrm{~B}^1$
$\Rightarrow \mathrm{AB}=\mathrm{AB}$
So, $\mathrm{P}(1)$ is true.
Let $\mathrm{P}(\mathrm{n})$ is true for some $\mathrm{k} \in \mathrm{N}$
So, $P(k):(A B)^k=A^k B^k, k \in N$
Now $(A B)^{k+1}=(A B)^{\mathrm{k}}(\mathrm{AB}) \quad \ldots .($ Using $(\mathrm{i}))$
$\begin{aligned} & =A^k B^k(A B) \\ & =A^k B^{k-1}(B A) B \\ & =A^k B^{k-1}(A B) B \quad \ldots .(\text { As given } A B=B A) \\ & =A^k B^{k-1} A B^2 \\ & =A^k B^{k-2}(B A) B^2 \\ & =A^k B^{k-2} A B B^2 \\ & =A^k B^{k-2} A B^3\end{aligned}$
$=\mathrm{A}^{\mathrm{k}+1} \mathrm{~B}^{\mathrm{k}+1}$
Thus $\mathrm{P}(1)$ is true and whenever $\mathrm{P}(\mathrm{k})$ is true $\mathrm{P}(\mathrm{k}+1)$ is true.
So, $\mathrm{P}(\mathrm{n})$ is true for all $\mathrm{n} \in \mathrm{N}$.
Question:50
Answer:
Matrix A is such that $\mathrm{A}^{\prime}=\mathrm{A}^{-1}$
$\Rightarrow \mathrm{AA}^{\prime}=\mathrm{I}$
$\Rightarrow\left[\begin{array}{ccc}0 & 2 y & z \\ x & y & -z \\ x & -y & z\end{array}\right]\left[\begin{array}{ccc}0 & x & x \\ 2 y & y & -y \\ z & -z & z\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$\Rightarrow\left[\begin{array}{ccc}4 y^2+z^2 & 2 y^2-z^2 & -2 y^2+z^2 \\ 2 y^2-z^2 & x^2+y^2+z^2 & x^2-y^2-z^2 \\ -2^2+z^2 & x^2-y^2+z^2 & x^2+y^2+z^2\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$\begin{aligned} & \Rightarrow 4 y^2+z^2=1 \\ & 2 y^2-z^2=0 \\ & x^2+y^2+z^2=1 \\ & x^2-y^2-z^2=0\end{aligned}$
$\begin{aligned} & \Rightarrow y^2=\frac{1}{6}, z^2=\frac{1}{3}, x^2=\frac{1}{2} \\ & \Rightarrow \mathrm{x}= \pm \frac{1}{\sqrt{2}} \\ & \Rightarrow \mathrm{y}= \pm \frac{1}{\sqrt{6}}\end{aligned}$
And $z= \pm \frac{1}{\sqrt{3}}$
Question:51.1
Answer:
Let A = $\begin{bmatrix} 2 &-1 & 3\\-5 &3 &1\\-3 &2 &3 \end{bmatrix}$
To apply elementary row transformations, we write:
A = IA where I is the identity matrix
We proceed with solving the problem in such a way that LHS becomes I and the transformations in I give us a new matrix such that.t
I = XA
And this X is called inverse of $A = A^{-1}$
Note: Never apply row and column transformations simultaneously over a matrix.
So we get,
$\begin{aligned} &\left[\begin{array}{ccc} 2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \mathrm{A}\\ &\text { Applying } R_{2} \rightarrow R_{2}+R_{1}\\ &\Rightarrow\left[\begin{array}{ccc} 2 & -1 & 3 \\ -3 & 2 & 4 \\ -3 & 2 & 3 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \mathrm{A}\\ &\text { Applying } \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{2}\\ &\Rightarrow\left[\begin{array}{ccc} 2 & -1 & 3 \\ -3 & 2 & 4 \\ 0 & 0 & -1 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 1 & 1 & 0 \\ -1 & -1 & 1 \end{array}\right] \mathrm{A}\\ &\text { Applying } R_{1} \rightarrow R_{1}+R_{2}\\ &\Rightarrow\left[\begin{array}{ccc} -1 & 1 & 7 \\ -3 & 2 & 4 \\ 0 & 0 & -1 \end{array}\right]=\left[\begin{array}{ccc} 2 & 1 & 0 \\ 1 & 1 & 0 \\ -1 & -1 & 1 \end{array}\right] \mathrm{A} \end{aligned}$
Applying R 2 → R 2 - 3R 1
$\begin{aligned} &\left[\begin{array}{ccc} -1 & 1 & 7 \\ 0 & -1 & -17 \\ 0 & 0 & -1 \end{array}\right]=\left[\begin{array}{ccc} 2 & 1 & 0 \\ -5 & -2 & 0 \\ -1 & -1 & 1 \end{array}\right] \mathrm{A}\\ &\text { Applying } \mathrm{R}_{3} \rightarrow(-1) \mathrm{R}_{3}\\ &\Rightarrow\left[\begin{array}{ccc} -1 & 1 & 7 \\ 0 & -1 & -17 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 2 & 1 & 0 \\ -5 & -2 & 0 \\ 1 & 1 & -1 \end{array}\right] \mathrm{A}\\ \end{aligned}$
$\text { Applying } R_{1} \rightarrow R_{1}+R_{2}$
$\left[\begin{array}{ccc} -1 & 0 & -10 \\ 0 & -1 & -17 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} -3 & -1 & 0 \\ -5 & -2 & 0 \\ 1 & 1 & -1 \end{array}\right] A$
$\\\text { Applying } \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+10 \mathrm{R}_{3} \text { and } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}+17 \mathrm{R}_{3}\\ \Rightarrow\left[\begin{array}{ccc} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 7 & 9 & -10 \\ 12 & 15 & -17 \\ 1 & 1 & -1 \end{array}\right] \mathrm{A}\\ \text { Applying } \mathrm{R}_{1} \rightarrow(-1) \mathrm{R}_{1} \text { and } \mathrm{R}_{2} \rightarrow(-1) \mathrm{R}_{2}\\$
$\Rightarrow\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} -7 & -9 & 10 \\ -12 & -15 & 17 \\ 1 & 1 & -1 \end{array}\right] A$
$\text { As we have an Identity Matrix in LHS, }\\ \\\therefore \mathrm{A}^{-1}=\left[\begin{array}{ccc} -7 & -9 & 10 \\ -12 & -15 & 17 \\ 1 & 1 & -1 \end{array}\right]$
Question:51.2
Answer:
Let A = $\begin{bmatrix} 2 &3 &-3 \\-1 &-2 &2 \\1 &1 &-1 \end{bmatrix}$
To apply elementary row transformations, we write:
A = IA where I is the identity matrix
We proceed with solving the problem in such a way that LHS becomes I and the transformations in I give us a new matrix such that.t
I = XA
And this X is called inverse of $A = A^{-1}$
Note: Never apply row and column transformations simultaneously over a matrix.
So we get:
$\begin{array}{l} {\left[\begin{array}{ccc} 2 & 3 & -3 \\ -1 & -2 & 2 \\ 1 & 1 & -1 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \mathrm{A}} \\\\ \text { Applying } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}+\mathrm{R}_{3} \\ {\left[\begin{array}{ccc} 2 & 3 & -3 \\ 0 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}\right] \mathrm{A}}\\ \\ \text { Applying } \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-2 \mathrm{R}_{3} \\ {\left[\begin{array}{ccc} 0 & 1 & -1 \\ 0 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & -2 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}\right] \mathrm{A}} \\\\ \text { Applying } \mathrm{R}_{2} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2} \\ {\left[\begin{array}{ccc} 0 & 1 & -1 \\ 0 & 0 & 0 \\ 1 & 1 & -1 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & -2 \\ -1 & 1 & -1 \\ 0 & 0 & 1 \end{array}\right] \mathrm{A}} \end{array}$
The second row of LHS contains all zeros, so we aren’t going to get any matrix in LHS.
∴ The inverse of A does not exist.
Hence, A-1 does not exist.
Question:51.3
Answer:
Let A = $\begin{bmatrix} 2 &0 &-1 \\5 &1 &0 \\0 &1 &3 \end{bmatrix}$
To apply elementary row transformations, we write:
A = IA where I is the identity matrix
We proceed with solving our problem in such a way that LHS becomes I and the transformations in I give us a new matrix such that:
I = XA
And this X is called inverse of $A = A^{-1}$
Note: Never apply row and column transformations simultaneously over a matrix.
So we get,
$\begin{aligned} &\left[\begin{array}{ccc} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \mathrm{A}\\ &\text { Applying } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-(5 / 2) \mathrm{R}_{1}\\ &\Rightarrow\left[\begin{array}{ccc} 2 & 0 & -1 \\ 0 & 1 & \frac{5}{2} \\ 0 & 1 & 3 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ -\frac{5}{2} & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A\\ &\text { Applying } \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{2}\\ &\left[\begin{array}{ccc} 2 & 0 & -1 \\ 0 & 1 & \frac{5}{2} \\ 0 & 0 & \frac{1}{2} \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ -\frac{5}{2} & 1 & 0 \\ \frac{5}{2} & -1 & 1 \end{array}\right] \mathrm{A}\\ \end{aligned}$
$\begin{aligned} &\text { Applying } R_{2} \rightarrow R_{2}-5 R_{3}\\ &\left[\begin{array}{ccc} 2 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & \frac{1}{2} \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ -15 & 6 & -5 \\ \frac{5}{2} & -1 & 1 \end{array}\right]\\ &\text { Applying } \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+2 \mathrm{R}_{3}\\ &\Rightarrow\left[\begin{array}{ccc} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & \frac{1}{2} \end{array}\right]=\left[\begin{array}{ccc} 6 & -2 & 2 \\ -15 & 6 & -5 \\ \frac{5}{2} & -1 & 1 \end{array}\right]\\ &\text { Applying } \mathrm{R}_{1} \rightarrow(1 / 2) \mathrm{R}_{1} \text { and } \mathrm{R}_{3} \rightarrow 2 \mathrm{R}_{3}\\ &\Rightarrow\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ -5 & -2 & 2 \end{array}\right] \mathrm{A}\\ \end{aligned}$
$\begin{aligned} &\text { As we have Identity matrix in LHS, we get, }\\ &\therefore \mathrm{A}^{-1}=\left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ -5 & -2 & 2 \end{array}\right] \end{aligned}$
Question:52
Answer:
If A is any matrix, then it can be written as the sum of a symmetric and skew-symmetric matrix.
Symmetric matrix is given by 1/2(A + A’)
Skew symmetric is given by 1/2(A - A’)
And A = 1/2(A + A’) + 1/2(A - A’)
Here, A = $\begin{bmatrix} 2 &3 & 1\\1 &-1 &2 \\4 &1 &2 \end{bmatrix}$
The symmetric matrix is given by –
$\Rightarrow \frac{1}{2}\left ( A+A' \right )=\frac{1}{2}\left (\left[\begin{array}{ccc} 2 & 3 & 1 \\ 1 & -1 & 2 \\ 4 & 1 & 2 \end{array}\right]+\left[\begin{array}{ccc} 2 & 3 & 1 \\ 1 & -1 & 2 \\ 4 & 1 & 2 \end{array}\right]^{\mathrm{T}} \right )$
$\begin{array}{l} \Rightarrow 1 / 2\left(A+A^{\prime}\right)={\frac{1}{2}}\left(\left[\begin{array}{ccc} 2 & 3 & 1 \\ 1 & -1 & 2 \\ 4 & 1 & 2 \end{array}\right]+\left[\begin{array}{ccc} 2 & 1 & 4 \\ 3 & -1 & 1 \\ 1 & 2 & 2 \end{array}\right]\right) \\ \end{array}$
$\Rightarrow 1 / 2\left(A+A^{\prime}\right)={\frac{1}{2}}\left(\left[\begin{array}{ccc} 2+2 & 3+1 & 1+4 \\ 1+3 & -1-1 & 2+1 \\ 4+1 & 1+2 & 2+2 \end{array}\right]\right)$
$\Rightarrow 1 / 2\left(A+A^{\prime}\right)=\left[\begin{array}{ccc} 4 & 4 & 5 \\ 4 & -2 & 3 \\ 5 & 3 & 4 \end{array}\right]=\left[\begin{array}{ccc} 2 & 2 & \frac{5}{2} \\ 2 & -1 & \frac{3}{2} \\ \frac{5}{2} & \frac{3}{2} & 2 \end{array}\right]$
Skew Symmetric matrix is given by –
$\frac{1}{2}\left[\begin{array}{ccc} 2 & 3 & 1 \\ 1 & -1 & 2 \\ 4 & 1 & 2 \end{array}\right]-\frac{1}{2}\left[\begin{array}{ccc} 2 & 3 & 1 \\ 1 & -1 & 2 \\ 4 & 1 & 2 \end{array}\right]^{\mathrm{T}}$
$\Rightarrow \frac{1}{2}\left ( A-A' \right )=\frac{1}{2}\left (\left[\begin{array}{ccc} 2 & 3 & 1 \\ 1 & -1 & 2 \\ 4 & 1 & 2 \end{array}\right]-\left[\begin{array}{ccc} 2 & 3 & 1 \\ 1 & -1 & 2 \\ 4 & 1 & 2 \end{array}\right]^{\mathrm{T}} \right )$
$\begin{array}{l} \Rightarrow 1 / 2\left(A-A^{\prime}\right)=\frac{1}{2}\left(\left[\begin{array}{ccc} 2 & 3 & 1 \\ 1 & -1 & 2 \\ 4 & 1 & 2 \end{array}\right]-\left[\begin{array}{ccc} 2 & 1 & 4 \\ 3 & -1 & 1 \\ 1 & 2 & 2 \end{array}\right]\right) \\ \end{array}$
$\Rightarrow 1 / 2\left(A-A^{\prime}\right)=\frac{1}{2}\left(\left[\begin{array}{ccc} 2-2 & 3-1 & 1-4 \\ 1-3 & -1+1 & 2-1 \\ 4-1 & 1-2 & 2-2 \end{array}\right]\right)$
$\frac{1}{2}\left[\begin{array}{ccc} 0 & 2 & -3 \\ -2 & 0 & 1 \\ 3 & -1 & 0 \end{array}\right]=\left[\begin{array}{ccc} 0 & 1 & \frac{-3}{2} \\ -1 & 0 & \frac{1}{2} \\ \frac{3}{2} & \frac{-1}{2} & 0 \end{array}\right]$
$\therefore A=\left[\begin{array}{ccc} 2 & 2 & \frac{5}{2} \\ 2 & -1 & \frac{3}{2} \\ \frac{5}{2} & \frac{3}{2} & 2 \end{array}\right]+\left[\begin{array}{ccc} 0 & 1 & \frac{-3}{2} \\ -1 & 0 & \frac{1}{2} \\ \frac{3}{2} & \frac{-1}{2} & 0 \end{array}\right]$
Question:53
The matrix $P=\begin{bmatrix} 0 &0 &4 \\0 &4 &0 \\4 &0 &0 \end{bmatrix}$ is a
A. square matrix
B. diagonal matrix
C. unit matrix
D. none
Answer:
As P has an equal number of rows and columns and thus it matches the definition of a square matrix.
The given matrix does not satisfy the definition of unit and diagonal matrices.
Hence, we can say that,
∴ Option (A) is the only correct answer.
Question:54
The total number of possible matrices of order 3 × 3 with each entry 2 or 0 is
A. 9
B. 27
C. 81
D. 512
Answer:
D)
As the above matrix has a total of 3×3 = 9 elements, then
As each element can take 2 values (0 or 2)
∴ By simply counting principles, we can say that the total number of possible matrices = total number of ways in which 9 elements can take possible values = $2^9$ = 512
It matches with option D.
Hence, we can say that,
∴ Option (D) is the only correct answer.
Question:55
If $\left[\begin{array}{cc} 2 x+y & 4 x \\ 5 x-7 & 4 x \end{array}\right]=\left[\begin{array}{cc} 7 & 7 y-13 \\ y & x+6 \end{array}\right]$ then the value of x + y is
A. x = 3, y = 1
B. x = 2, y = 3
C. x = 2, y = 4
D. x = 3, y = 3
Answer:
Given that: $\left[\begin{array}{ll}2 x+y & 4 x \\ 5 x-7 & 4 x\end{array}\right]=\left[\begin{array}{cc}7 & 7 y-13 \\ y & x+6\end{array}\right]$
Equating the corresponding elements, we get,
$2 x+y=7 \ldots \ldots .(i)$
And $4 x=x+6$
From equations (ii)
$\begin{aligned} & 4 x-x=6 \\ & 3 x=6 \\ & \therefore x=2\end{aligned}$
From equations (i)
$\begin{aligned} & 2 \times 2+y=7 \\ & 4+y=7 \\ & \therefore y=7-4=3\end{aligned}$
$\therefore$ Option(B) is the correct answer.
Question:56
If $A=\frac{1}{\pi}\left[\begin{array}{cc} \sin ^{-1}(\mathrm{x} \pi) & \tan ^{-1} \frac{\mathrm{x}}{\pi} \\ \sin ^{-1} \frac{\mathrm{x}}{\pi} & \cot ^{-1}(\pi \mathrm{x}) \end{array}\right], \mathrm{B}=\frac{1}{\pi}\left[\begin{array}{cc} -\cos ^{-1}(\mathrm{x} \pi) & \tan ^{-1} \frac{\mathrm{x}}{\pi} \\ \sin ^{-1} \frac{\mathrm{x}}{\pi} & -\tan ^{-1}(\pi \mathrm{x}) \end{array}\right]$ then A - B is equal to
A. I
B. O
C. 2I
D. $\frac{1}{2}I$
Answer:
Given that: $\mathrm{A}=\frac{1}{\pi}\left[\begin{array}{ll}\sin ^{-1}(x \pi) & \tan ^{-1}\left(\frac{x}{\pi}\right) \\ \sin ^{-1}\left(\frac{x}{\pi}\right) & \cot ^{-1}(\pi x)\end{array}\right]$
And $\mathrm{B}=\frac{1}{\pi}\left[\begin{array}{cc}-\cos ^{-1}\left(\frac{x}{\pi}\right) & \tan ^{-1}\left(\frac{x}{\pi}\right) \\ \sin ^{-1}\left(\frac{x}{\pi}\right) & -\tan ^{-1}(\pi x)\end{array}\right]$
$\mathrm{A}-\mathrm{B}=\frac{1}{\pi}\left[\begin{array}{cc}\sin ^{-1}(x \pi) & \tan ^{-1}\left(\frac{x}{\pi}\right) \\ \sin ^{-1}\left(\frac{x}{\pi}\right) & \cot ^{-1}(\pi x)\end{array}\right]-\frac{1}{\pi}\left[\begin{array}{cc}-\cos ^{-1}(x \pi) & \tan ^{-1}\left(\frac{x}{\pi}\right) \\ \sin ^{-1}\left(\frac{x}{\pi}\right) & -\tan ^{-1}(\pi x)\end{array}\right]$
$=\frac{1}{\pi}\left[\begin{array}{cc}\sin ^{-1}(x \pi)+\cos ^{-1}(x \pi) & \tan ^{-1}\left(\frac{x}{\pi}\right)-\tan ^{-1}\left(\frac{x}{\pi}\right) \\ \sin ^{-1}\left(\frac{x}{\pi}\right)-\sin ^{-1}\left(\frac{x}{\pi}\right) & \cot ^{-1}(\pi x)+\tan ^{-1}(\pi x)\end{array}\right]$
$=\frac{1}{\pi}\left[\begin{array}{ll}\frac{\pi}{2} & 0 \\ 0 & \frac{\pi}{2}\end{array}\right] \cdots \cdots\left[\begin{array}{l}\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2} \\ \tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\end{array}\right]$
$=\frac{1}{\pi} \times \frac{\pi}{2}\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$=\frac{1}{2} \mathrm{I}$
∴ Option (D) is the only correct answer.
Question:57
If A and B are two matrices of the order 3 × m and 3 × n, respectively, and m = n, then the order of matrix (5A - 2B) is
A. m × 3
B. 3 × 3
C. m × n
D. 3 × n
Answer:
As the order of A is 3 × m and the order of B is 3 × n
As m = n. So, the order of A and B is the same = 3 × m
∴ Subtraction can be carried out.
And (5A - 3B) also has the same order.
Hence, option D is correct.
Question:58
If $A= \begin{bmatrix} 0 &1 \\1 &0 \end{bmatrix}$ then $A^2$ is equal to
A. $\begin{bmatrix} 0 &1 \\1 &0 \end{bmatrix}$
B.$\begin{bmatrix} 1&0 \\1 &0 \end{bmatrix}$
C.$\begin{bmatrix} 0&1 \\0 &1 \end{bmatrix}$
D.$\begin{bmatrix} 1&0 \\0 &1 \end{bmatrix}$
Answer:
$\begin{aligned} &\text { Let } A=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]\\ &\therefore \mathrm{A}^{2}=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]\\ &\text { By the rule of matrix multiplication, we have, }\\ &\Rightarrow \mathrm{A}^{2}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \text { which matches with option (D) } \end{aligned}$
Hence, we can say that,
∴ Option (D) is the correct answer.
Question:59
If matrix $A=[a_{ij}]_{2\times 2}$, where aij = 1 if i ≠ j
aij = 0 if i = j, then $A^2$ is equal to
A. I
B. A
C. 0
D. None of these
Answer:
We are given that,
$a_{11} = 0 , a_{12} = 1 , a_{21} = 1 $ and a_{22} = 0$
$\begin{array}{l} \therefore A=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right] \\ \therefore A^{2}=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right] \end{array}$
According to the rule of matrix multiplication:
$\begin{array}{l} \therefore A^2=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ \end{array}$ which matches with option (A)
Hence, we can say that,
∴ Option (A) is the correct answer.
Question:60
The matrix $\begin{bmatrix} 1 &0 &0 \\0 &2 &0 \\0 &0 &4 \end{bmatrix}$ is a
A. Identity matrix
B. symmetric matrix
C. skew-symmetric matrix
D. none of these
Answer:
$\begin{aligned} &\text { Let } A=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{array}\right]\\ &\text { Then, }\\ &A^{\prime}=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{array}\right]^{T}=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{array}\right]=A \end{aligned}$
As, $A^T = A$
∴ It is a symmetric matrix.
Hence, we can say that,
∴ Option(B) is the correct answer.
Question:61
The matrix $\begin{bmatrix} 0 &-5 &8 \\5 &0 &12 \\-8 &-12 &0 \end{bmatrix}$ is a
A. diagonal matrix
B. symmetric matrix
C. skew-symmetric matrix
D. scalar matrix
Answer:
Let A = $\begin{bmatrix} 0 &-5 &8 \\5 &0 &12 \\-8 &-12 &0 \end{bmatrix}$
$\mathrm{A}^{\prime}=\left[\begin{array}{ccc} 0 & -5 & 8 \\ 5 & 0 & 12 \\ -8 & -12 & 0 \end{array}\right]^{\mathrm{T}}=\left[\begin{array}{ccc} 0 & 5 & -8 \\ -5 & 0 & -12 \\ 8 & 12 & 0 \end{array}\right]=-\mathrm{A}$
As $A^T = -A$
∴ It is a skew-symmetric matrix.
Hence, we can say that,
∴ Option(C) is the correct answer.
Question:62
If A is a matrix of order m × n and B is a matrix such that AB’ and B’A are both defined, then the other of matrix B is
A. m × m
B. n × n
C. n × m
D. m × n
Answer:
As AB’ is defined. So, B’ must have n rows.
∴ B has n columns.
And, B’A is also defined. As A’ has order n × m
∴ B’A to exist, B must have m rows.
∴ m × n is the order of B.
Hence, we can say that,
Option (D) is the correct answer.
Question:63
If A and B are matrices of the same order, then (AB’ - BA’) is a
A. skew-symmetric matrix
B. null matrix
C. symmetric matrix
D. unit matrix
Answer:
Let C = (AB’ - BA’)
C’ = (AB’ - BA’)’
$\\ \Rightarrow$ C’ = (AB’)’ - (BA’)’
$\\ \Rightarrow$ C’ = (B’)’ A’ - (A’)’ B’
$\\ \Rightarrow$ C’ = BA’ - AB’
$\\ \Rightarrow$ C’ = -C
∴ C is a skew-symmetric matrix.
Option (A) matches with our deduction.
Hence, we can say that,
∴ Option (A) is the correct.
Question:64
If A is a square matrix such that $A^2 = I$, then $(A - I)^3 + (A + I)^3 - 7A$ is equal to
A. A
B. I - A
C. I + A
D. 3A
Answer:
As, $(A - I)^3 + (A + I)^3 - 7A$
Use $a^3 + b^3 = (a + b)(a^2 + ab + b^2)$
Also, $A^2 = I$
$(A - I)^3 + (A + I)^3 - 7A$
$\begin{array}{l} =A^{3}-3 A^{2}+3 A-I^{3}+A^{3}+3 A^{2}+3 A+I^{3}-7 A \\ =2 A^{3}+6 A-7 A \\ =2 A^{2} \cdot A+6 A-7 A \\ =2 I \cdot A+6 A-7 A \\ =2 A+6 A-7 A=8 A-7 A=A \end{array}$
∴ then $(A - I)^3 + (A + I)^3 - 7A= A$Ourr answer is similar to option (A)
Hence, we can say that,
∴ Option (A) is the correct answer.
Question:65
For any two matrices A and B, we have
A. AB = BA
B. AB ≠ BA
C. AB = O
D. None of the above
Answer:
For any two matrices:
Not always are options A, B, and C true.
Hence, we can say that,
∴ Option (D) is the only suitable answer
Question:66
On using elementary column operations C2→ C2 — 2C1 in the following matrix equation
$\left[\begin{array}{cc} 1 & -3 \\ 2 & 4 \end{array}\right]=\left[\begin{array}{cc} 1 & -1 \\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 3 & 1 \\ 2 & 4 \end{array}\right]$ we have:
A.$\left[\begin{array}{cc}1 & -5 \\ 0 & 4\end{array}\right]=\left[\begin{array}{cc}1 & -1 \\ -2 & 2\end{array}\right]\left[\begin{array}{cc}3 & -5 \\ 2 & 0\end{array}\right]$$
B. $\left[\begin{array}{cc}1 & -5 \\ 0 & 4\end{array}\right]=\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\left[\begin{array}{cc}3 & -5 \\ -0 & 5\end{array}\right]$$
C.$\left[\begin{array}{cc}1 & -5 \\ 2 & 0\end{array}\right]=\left[\begin{array}{cc}1 & -3 \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}3 & 1 \\ -2 & 4\end{array}\right]$$
D. $\left[\begin{array}{cc}1 & -5 \\ 2 & 0\end{array}\right]=\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}3 & -5 \\ 2 & 0\end{array}\right]$$
Answer:
$\left[\begin{array}{cc} 1 & -3 \\ 2 & 4 \end{array}\right]=\left[\begin{array}{cc} 1 & -1 \\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 3 & 1 \\ 2 & 4 \end{array}\right]$
For column transformation, we operate on the post matrix.
As,
$\left[\begin{array}{cc} 1 & -3 \\ 2 & 4 \end{array}\right]=\left[\begin{array}{cc} 1 & -1 \\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 3 & 1 \\ 2 & 4 \end{array}\right]$
By Applying C 2 → C 2 — 2C 1 ,
$\left[\begin{array}{cc}1 & -5 \\ 2 & 0\end{array}\right]=\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}3 & -5 \\ 2 & 0\end{array}\right]$$Itt matches with option (D).
Hence, we can say that,
∴ Option (D) is the correct answer.
Question:67
On using elementary row operation R1→ R1 — 3R2 in the following matrix equation:
$\left[\begin{array}{ll} 4 & 2 \\ 3 & 3 \end{array}\right]=\left[\begin{array}{ll} 1 & 2 \\ 0 & 3 \end{array}\right]\left[\begin{array}{ll} 2 & 0 \\ 1 & 1 \end{array}\right]$
$\begin{array}{l} \text { A. }\left[\begin{array}{cc} -5 & -7 \\ 3 & 3 \end{array}\right]=\left[\begin{array}{cc} 1 & -7 \\ 0 & 3 \end{array}\right]\left[\begin{array}{cc} 2 & 0 \\ 1 & 1 \end{array}\right] \\ \\B.{\left[\begin{array}{cc} -5 & -7 \\ 3 & 3 \end{array}\right]=\left[\begin{array}{cc} 1 & 2 \\ 0 & 3 \end{array}\right]\left[\begin{array}{cc} -1 & -3 \\ 1 & 1 \end{array}\right]} \\ \\C.{\left[\begin{array}{cc} -5 & -7 \\ 3 & 3 \end{array}\right]=\left[\begin{array}{cc} 1 & 2 \\ 1 & -7 \end{array}\right]\left[\begin{array}{cc} 2 & 0 \\ 1 & 1 \end{array}\right]} \\\\ D.{\left[\begin{array}{cc} 4 & 2 \\ -5 & -7 \end{array}\right]=\left[\begin{array}{cc} 1 & 2 \\ -3 & -3 \end{array}\right]\left[\begin{array}{cc} 2 & 0 \\ 1 & 1 \end{array}\right]} \end{array}$
Answer:
Elementary row transformation is applied to the first matrix of the RHS.
$\left[\begin{array}{ll} 4 & 2 \\ 3 & 3 \end{array}\right]=\left[\begin{array}{ll} 1 & 2 \\ 0 & 3 \end{array}\right]\left[\begin{array}{ll} 2 & 0 \\ 1 & 1 \end{array}\right]$
By Applying R 1 → R 1 — 3R 2 we get -
$\left[\begin{array}{ll} 4 & 2 \\ 3 & 3 \end{array}\right]=\left[\begin{array}{ll} 1 & 2 \\ 0 & 3 \end{array}\right]\left[\begin{array}{ll} 2 & 0 \\ 1 & 1 \end{array}\right]$
$\left[\begin{array}{cc} -5 & -7 \\ 3 & 3 \end{array}\right]=\left[\begin{array}{cc} 1 & -7 \\ 0 & 3 \end{array}\right]\left[\begin{array}{cc} 2 & 0 \\ 1 & 1 \end{array}\right] \\$
It matches with option (A)
Hence, we can say that,
∴ Option (A) is the correct answer.
Question:68
Fill in the blanks in each of the following:
______ matrix is both symmetric and skew-symmetric matrix.
Answer:
A Zero matrix
∴ Let A be the symmetric and skew-symmetric matrix.
⇒ A’=A (Symmetric)
⇒ A’=-A (Skew-Symmetric)
Considering the above two equations,
⇒ A=-A
⇒ 2A=0
⇒ A=0 (A Zero Matrix)
Hence zero matrix is both a symmetric and skew-symmetric matrix.
Question:69
Fill in the blanks in each of the following:
The sum of two skew-symmetric matrices is always _______ matrix.
Answer:
A skew-symmetric matrix
Let A and B be any two matrices
$\therefore$ For skew-symmetric matrices
$\mathrm{A}=-\mathrm{A}^{\prime} \quad \ldots . . .(\mathrm{i})$
And B = - $\mathrm{B}^{\prime}$
Adding (i) and (ii), we get
$\begin{aligned} & \mathrm{A}+\mathrm{B}=-\mathrm{A}^{\prime}-\mathrm{B}^{\prime} \\ & \Rightarrow \mathrm{A}+\mathrm{B}=-\left(\mathrm{A}^{\prime}+\mathrm{B}^{\prime}\right)\end{aligned}$
So $\mathrm{A}+\mathrm{B}$ is skew-symmetric matrix.
Question:70
Fill in the blanks in each of the following:
The negative of a matrix is obtained by multiplying it by ________.
Answer:
The negative of a matrix is obtained by multiplying it by -1.
For example:
$\begin{aligned} \text { Let}&A=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]\\ \\ \end{aligned}$
$\text { So }\left[\begin{array}{ll} -1 & -2 \\ -3 & -4 \end{array}\right]=-1\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]\\ =-A$
Question:71
Fill in the blanks in each of the following:
The product of any matrix by the scalar _____ is the null matrix.
Answer:
The null matrix is the one in which all elements are zero.
If we want to make A = $\begin{bmatrix} 1 &2 \\3 &4 \end{bmatrix}$ a null matrix we need to multiply it by 0.
0A = $0\begin{bmatrix} 1 &2 \\3 &4 \end{bmatrix}$
$\begin{bmatrix} 0 &0 \\0 &0 \end{bmatrix}$
Hence, we can say that,
The product of any matrix by the scalar 0 is the null matrix.
Question:72
Fill in the blanks in each of the following:
A matrix that is not a square matrix is called a _____ matrix.
Answer:
Rectangular Matrix
As we know, a square matrix is one in which there is the same number of rows and columns.
Eg: A = $\begin{bmatrix} 1 &2 \\3 &4 \end{bmatrix}$
Here there are 2 rows and 2 columns.
The matrix that is not square is called a rectangular matrix as it does not have the same number of rows and columns.
Eg $\begin{bmatrix} 1 &2 &3 \\4 &5 &6 \end{bmatrix}$
Here number of rows is 2 and columns are 3.
Question:73
Fill in the blanks in each of the following:
Matrix multiplication is _____ over addition.
Answer:
Distributive
⇒ Matrix multiplication is distributive over addition.
i.e A(B+C)=AB+AC
and (A+B)C=AC+BC
Question:74
Fill in the blanks in each of the following:
If A is a symmetric matrix, then $A^3$ is a ______ matrix.
Answer:
Given A is a symmetric matrix
$\therefore \mathrm{A}^{\prime}=-\mathrm{A}$
$\operatorname{Now}\left(\mathrm{A}^3\right)^{\prime}=\left(\mathrm{A}^{\prime}\right)^3 \quad \ldots \ldots .\left[\because\left(\mathrm{A}^{\prime}\right)^{\mathrm{n}}=\left(\mathrm{A}^{\mathrm{n}}\right)^{\prime}\right]$
$=\mathrm{A}^3$
Question:75
Fill in the blanks in each of the following:
If A is a skew-symmetric matrix, then $A^2$ is a _________.
Answer:
Given A is a skew-symmetric matrix.
$\begin{aligned} & \therefore \mathrm{A}^{\prime}=-\mathrm{A} \\ & \therefore\left(\mathrm{A}^2\right)^{\prime}=\left(\mathrm{A}^{\prime}\right)^2 \\ & =(-\mathrm{A})^2 \\ & =\mathrm{A}^2\end{aligned}$
So, $\mathrm{A}^2$ is a symmetric matrix.
Question:76
Fill in the blanks in each of the following:
If A and B are square matrices of the same order, then
(i) (AB)’ = ________.
(ii) (kA)’ = ________. (k is any scalar)
(iii) [k (A - B)]’ = ________.
Answer:
(i) (AB)’ = ________.
(AB)’ = B’A’
Let A be the matrix of order m× n and B be of n× p.
A’ is of order n× m and B’ is of order p× n.
Hence, we get, B’ A’ is of order p× m.
So, AB is of order m× p.
And (AB)’ is of order p× m.
We can see (AB)’ and B’ A’ are of the same order p× m.
Hence proved, (AB)’ = B’ A’
(ii) (kA)’ = ________. (k is any scalar)
If a scalar “k” is multiplied by any matrix the new matrix becomes
K times of the old matrix.
$\begin{array}{l} \text { Eg: } A=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right] \\ 2 A=2\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]=\left[\begin{array}{ll} 2 & 4 \\ 6 & 8 \end{array}\right] \\ (2 A)=\left[\begin{array}{ll} 2 & 6 \\ 4 & 8 \end{array}\right] \\ A^{\prime}=\left[\begin{array}{ll} 1 & 3 \\ 2 & 4 \end{array}\right] \end{array}$
Now 2A’ = $2\left[\begin{array}{ll} 1 & 3 \\ 2 & 4 \end{array}\right]=\left[\begin{array}{ll} 2 & 6 \\ 4 & 8 \end{array}\right]$
Hence (2A)’ =2A’
Hence (kA)’ = k(A)’
(iii) [k (A - B)]’ = ________.
$\begin{aligned} &A=\left[\begin{array}{ll} 5 & 7 \\ 7 & 6 \end{array}\right]\\ &A'=\left[\begin{array}{ll} 5 & 7 \\ 7 & 6 \end{array}\right]\\ &2A'=\left[\begin{array}{ll} 10 & 14 \\ 14 & 12 \end{array}\right]\\ &B=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]\\ &B^{\prime}=\left[\begin{array}{ll} 1 & 3 \\ 2 & 4 \end{array}\right]\\ &2 B^{\prime}={2}\left[\begin{array}{ll} 1 & 3 \\ 2 & 4 \end{array}\right]\\ &=\left[\begin{array}{ll} 2 & 6 \\ 4 & 8 \end{array}\right]\\ &A-B=\left[\begin{array}{ll} 4 & 5 \\ 4 & 2 \end{array}\right] \end{aligned}$
$\begin{array}{l} \text { Now Let } k=2 \\ 2(A-B)=2\left[\begin{array}{ll} 4 & 5 \\ 4 & 2 \end{array}\right]=\left[\begin{array}{ll} 8 & 10 \\ 8 & 4 \end{array}\right] \\ {[2(A-B)]=\left[\begin{array}{cc} 8 & 8 \\ 10 & 4 \end{array}\right]} \\ 2 A^{\prime}-2 B'=\left[\begin{array}{ll} 10 & 14 \\ 14 & 12 \end{array}\right]-\left[\begin{array}{ll} 2 & 6 \\ 4 & 8 \end{array}\right]=\left[\begin{array}{cc} 8 & 8 \\ 10 & 4 \end{array}\right] \\ A^{\prime}-B'=\left[\begin{array}{ll} 5 & 7 \\ 7 & 6 \end{array}\right]-\left[\begin{array}{ll} 1 & 3 \\ 2 & 4 \end{array}\right]=\left[\begin{array}{ll} 4 & 4 \\ 5 & 3 \end{array}\right] \\ 2\left(A^{\prime}-B^{\prime}\right)=2\left[\begin{array}{ll} 4 & 4 \\ 5 & 3 \end{array}\right]=\left[\begin{array}{cc} 8 & 8 \\ 10 & 6 \end{array}\right] \\ \text { Hence we can see }[k(A-B)]^{\prime}=k(A)^{\prime}-k(B)^{\prime}=k\left(A^{\prime}-B^{\prime}\right) \end{array}$
Question:77
Fill in the blanks in each of the following:
If A is skew-symmetric, then kA is a ______. (k is any scalar)
Answer:
A skew-symmetric matrix.
We are given that, A’=-A
⇒ (kA)’=k(A)’=k(-A)
⇒ (kA)’=-(kA)
Question:78
Fill in the blanks in each of the following:
If A and B are symmetric matrices, then
(i) AB - BA is a _________.
(ii) BA - 2AB is a _________.
Answer:
(i) AB - BA is a Skew Symmetric matrix
We are given that A’=A and B’=B
⇒ (AB-BA)’=(AB)’-(BA)’
⇒ (AB)’-(BA)’=B’A’-A’B’
⇒ B’A’-A’B’=BA-AB=-(AB-BA)
⇒ (AB-BA)’=-(AB-BA) (skew symmetric matrix)
$\begin{aligned} &\text { For example, Let }\\ &A=\left[\begin{array}{ll} 1 & 3 \\ 3 & 2 \end{array}\right]\\ &B=\left[\begin{array}{ll} 1 & 2 \\ 2 & 1 \end{array}\right]\\ &\Rightarrow \mathrm{AB}=\left[\begin{array}{ll} 7 & 5 \\ 7 & 8 \end{array}\right] \text { and } \mathrm{BA}=\left[\begin{array}{ll} 7 & 7 \\ 5 & 8 \end{array}\right]\\ &\Rightarrow A B-B A=\left[\begin{array}{cc} 0 & -2 \\ 2 & 0 \end{array}\right]\\ &\Rightarrow(A B-B A)^{\prime}=\left[\begin{array}{cc} 0 & 2 \\ -2 & 0 \end{array}\right]\\ &\Rightarrow=(A B-B A)=\left[\begin{array}{cc} 0 & 2 \\ -2 & 0 \end{array}\right] \end{aligned}$
(ii) BA - 2AB is neither a Symmetric nor Skew Symmetric matrix
Given A’=A and B’=B
⇒ (BA-2AB)’=(BA)’-(2AB)’
⇒ (BA)’-(2AB)’=A’B’-2B’A’
⇒ A’B’-2B’A’=AB-2BA=-(2BA-AB)
⇒ (BA-2AB)’=-(2BA-AB)
$\begin{aligned} &\text { For example Let }\\ &A=\left[\begin{array}{ll} 1 & 3 \\ 3 & 2 \end{array}\right]\\ &B=\left[\begin{array}{ll} 1 & 2 \\ 2 & 1 \end{array}\right]\\ &\Rightarrow \mathrm{AB}=\left[\begin{array}{ll} 7 & 5 \\ 7 & 8 \end{array}\right] \text { and } B A=\left[\begin{array}{ll} 7 & 7 \\ 5 & 8 \end{array}\right]\\ &\Rightarrow B A-2 A B=\left[\begin{array}{cc} 7 & -3 \\ -9 & 8 \end{array}\right] \end{aligned}$
Question:79
Fill in the blanks in each of the following:
If A is a symmetric matrix, then B’AB is _______.
Answer:
B’AB is a symmetric matrix.
Solution:
Given A is a symmetric matrix.
⇒ A’=A ..(1)
Now in B’AB,
Let AB=C ..(2)
⇒ B’AB=B’C
Now Using Property (AB)’=B’A’
⇒ (B’C)’=C’(B’)’ (As (B’)’=B)
⇒ C’(B’)’=C’B
⇒ C’B=(AB)’B (Using Property (AB)’=B’A’)
⇒ (AB)’ B=B’A’B (Using (1))
⇒ B’A’B= B’AB
⇒ Hence (B’AB)’= B’AB
Question:80
Answer:
Given A and B are symmetric matrices,
⇒ A’=A ..(1)
⇒ B’=B ..(2)
Let AB be a Symmetric matrix:-
⇒ (AB)’=AB
Using Property (AB)’=B’A’
⇒ B’A’=AB
⇒ Now using (1) and (2)
⇒ BA=AB
Hence, A and B matrices commute.
Question:81
Fill in the blanks in each of the following:
In applying one or more now operations while finding $A^{-1}$ by elementary row operations, we obtain all zeros in one or more, then $A^{-1}$ ______.
Answer:
$A^{-1}$ Does not exist,
$A^{-1}=\frac{1}{|A|} a d j(A)$
And |A|=0 if there is one or more rows or columns with all zero elements.
Question:82
Which of the following statements are True or False
A matrix denotes a number.
Answer:
False
A matrix is an ordered rectangular array of numbers of functions.
Only a matrix of order (1×1) denotes a number.
For example, $[8]_{1\times 1}=8$
Question:83
Which of the following statements are True or False
Matrices of any order can be added.
Answer:
False
Matrices having the same order can be added.
For example
$\begin{array}{l} \text { Let } A=\left[\begin{array}{ll} 1 & 3 \\ 3 & 2 \end{array}\right] \\ B=\left[\begin{array}{ll} 1 & 2 \\ 2 & 1 \end{array}\right] \\ \Rightarrow A+B=\left[\begin{array}{ll} 2 & 5 \\ 5 & 3 \end{array}\right] \end{array}$
Question:84
Answer:
False
Two matrices are equal if they have the same number of rows and the same number of columns and corresponding elements within each matrix are equal or identical.
For example:
$\Rightarrow A=\left[\begin{array}{ll} 2 & 5 \\ 5 & 3 \end{array}\right], B=\left[\begin{array}{ll} 2 & 5 \\ 5 & 3 \end{array}\right]$
Here both matrices have two rows and two columns.
Also, they both have the same elements.
Question:85
Which of the following statements are True or False
Matrices of different order cannot be subtracted.
Answer:
True
Matrices of only the same order can be added or subtracted.
Let A = $\begin{bmatrix} 1 &3 \\3 &2 \end{bmatrix}$
B= $\begin{bmatrix} 1 & 0 \end{bmatrix}$
⇒ A-B= Not possible
Question:86
Answer:
True
1. A+B=B+A (commutative)
2. (A+B)+C= A+(B+C) (associative)
Question:87
Which of the following statements are True or False
Matrix multiplication is commutative.
Answer:
False
In general matrix multiplication is not commutative.
But it’s associative.
⇒ (AB)C=A(BC)
Question:88
Answer:
False
A square matrix where every element of the leading diagonal is unity and the rest elements are zero is called an identity matrix.
i.e $I=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$
Question:89
Answer:
True
If A and B are two square matrices of the same order, then A + B = B + A ( Property of square matrix)
For example,
$\begin{array}{l} \text {Let } A=\left[\begin{array}{lll} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 3 & 4 & 1 \end{array}\right] \text { and } B=\left[\begin{array}{lll} 4 & 5 & 7 \\ 5 & 5 & 6 \\ 7 & 6 & 9 \end{array}\right] \\ A+B=\left[\begin{array}{lll} 5 & 7 & 10 \\ 7 & 6 & 10 \\ 10 & 10 & 10 \end{array}\right] \\ \Rightarrow \quad B+A=\left[\begin{array}{lll} 5 & 7 & 10 \\ 7 & 6 & 10 \\ 10 & 10 & 10 \end{array}\right] \\ \end{array}$
Question:90
Answer:
False
If A and B are two matrices of the same order,
then A - B = -(B - A)
For example,
$\begin{aligned} &\text { Let } A=\left[\begin{array}{lll} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 3 & 4 & 1 \end{array}\right] \text { and } B=\left[\begin{array}{lll} 4 & 5 & 7 \\ 5 & 5 & 6 \\ 7 & 6 & 9 \end{array}\right]\\ &A-B=\left[\begin{array}{lll} -3 & -3 & -4 \\ -3 & -4 & -2 \\ -4 & -2 & -8 \end{array}\right]\\ &B-A=\left[\begin{array}{lll} 3 & 3 & 4 \\ 3 & 4 & 2 \\ 4 & 2 & 8 \end{array}\right]\\ &\Rightarrow-(B-A)=\left[\begin{array}{ccc} -3 & -3 & -4 \\ -3 & -4 & -2 \\ -4 & -2 & -8 \end{array}\right] \end{aligned}$
Question:91
Answer:
False
It's not necessary that for the multiplication of matrices A and B to be 0 one of them has to be a null matrix.
For example,
$\begin{array}{c} \text {Let } A=\left[\begin{array}{ccc} 1 & 1 & -1 \\ 1 & 1 & -1 \\ 1 & 1 & -1 \end{array}\right] \text { and } B=\left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 2 & 2 & 2 \end{array}\right] \\ A \times B=\left[\begin{array}{ccc} 1 & 1 & -1 \\ 1 & 1 & -1 \\ 1 & 1 & -1 \end{array}\right]\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 2 & 2 & 2 \end{array}\right]=\left[\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \end{array}$
Question:92
Which of the following statements are True or False
Transpose of a column matrix is a column matrix.
Answer:
False
Transpose of a column matrix is a Row matrix and vice-versa.
$\begin{array}{l} \text { Let } A=\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right] \text { (Column Matrix) } \\ \Rightarrow A^{\prime}=\left[\begin{array}{lll} 1 & 2 & 3 \end{array}\right] \text { (Row Matrix) } \end{array}$
Question:93
Answer:
False
Matrix multiplication is not commutative.
For example,
$\begin{array}{l} \text { Let } A=\left[\begin{array}{ll} 1 & 3 \\ 3 & 2 \end{array}\right] \\ B=\left[\begin{array}{ll} 1 & 2 \\ 2 & 1 \end{array}\right] \\ \Rightarrow A B=\left[\begin{array}{ll} 7 & 5 \\ 7 & 8 \end{array}\right] \text { and } B A=\left[\begin{array}{ll} 7 & 7 \\ 5 & 8 \end{array}\right] \\ \Rightarrow A B \neq B A \end{array}$
Question:94
Answer:
True
For example,
$\begin{array}{l} \text { Let } A=\left[\begin{array}{lll} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 3 & 4 & 1 \end{array}\right], B=\left[\begin{array}{lll} 4 & 5 & 7 \\ 5 & 5 & 6 \\ 7 & 6 & 9 \end{array}\right] \end{array}$ and $C = \begin{bmatrix} 3 &9 &1 \\ 9 &2 &8 \\1 &8 &5 \end{bmatrix}$
$\Rightarrow A+B+C=\left[\begin{array}{lll} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 3 & 4 & 1 \end{array}\right]+\left[\begin{array}{lll} 4 & 5 & 7 \\ 5 & 5 & 6 \\ 7 & 6 & 9 \end{array}\right]+\left[\begin{array}{lll} 3 & 9 & 1 \\ 9 & 2 & 8 \\ 1 & 8 & 5 \end{array}\right]$
$A+B+C=\left[\begin{array}{lll} (1+4+3) & (2+5+9) & (3+7+1) \\ (2+5+9) & (1+5+2) & (4+6+8) \\ (3+7+1) & (4+6+8) & (1+9+5) \end{array}\right]$
$A+B+C=\left[\begin{array}{ccc} 8 & 16 & 11 \\ 16 & 8 & 18 \\ 11 & 18 & 15 \end{array}\right]$
Question:95
Answer:
False
If A and B are any two matrices for which AB is defined, then
(AB)’=B’A’.
Question:96
Answer:
Let $\mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right]_{\mathrm{m} \times \mathrm{n}}$ and $\mathrm{B}=\left[\mathrm{b}_{\mathrm{ij}}\right]_{\mathrm{p} \times \mathrm{q}}$
$A B$ is defined when $n=P$
$\therefore$ Order of $\mathrm{AB}=\mathrm{m} \times \mathrm{q}$
$\Rightarrow \operatorname{Order}$ of $(\mathrm{AB})^{\prime}=\mathrm{q} \times \mathrm{m}$
Order of $\mathrm{B}^{\prime}$ is $\mathrm{q} \times \mathrm{p}$ and order of $\mathrm{A}^{\prime}$ is $\mathrm{n} \times \mathrm{m}$
$\therefore \mathrm{B}^{\prime} \mathrm{A}^{\prime}$ is defined when $\mathrm{P}=\mathrm{n}$
And the order of $\mathrm{B}^{\prime} \mathrm{A}^{\prime}$ is $\mathrm{q} \times \mathrm{m}$
Hence, order of $(A B)^{\prime}=$ Order of $B^{\prime} A^{\prime}$ i.e., $q \times m$
Hence, the given statement is true.
Question:97
Answer:
False
Let $A=\left[\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right]$
$B=\left[\begin{array}{ll}0 & 0 \\ 2 & 0\end{array}\right]$
And $C=\left[\begin{array}{ll}0 & 0 \\ 3 & 4\end{array}\right]$
$\therefore \mathrm{AB}=\left[\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right]\left[\begin{array}{ll}0 & 0 \\ 2 & 0\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
$\mathrm{AC}=\left[\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right]\left[\begin{array}{ll}0 & 0 \\ 3 & 4\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
Here $\mathrm{AB}=\mathrm{AC}=0$ but $\mathrm{B} \neq \mathrm{C}$.
Question:98
Answer:
True
(AA’)’=(A’)’A’
As we know (A’)’ = A
(AA’)’=AA’ (Condition of the symmetric matrix)
Question:99
Answer:
False
Here A has an order (2×3) and B has an order (3×2),
Hence AB is defined and will give an output matrix of order (2×2)
And BA is also defined but will give an output matrix of order (3×3).
⇒ AB ≠ BA
Question:100
Answer:
True
For skew-symmetric matrix A’=-A
$\begin{gathered}\Rightarrow\left(A^2\right)^{\prime}=(A A)^{\prime}=A^{\prime} A^{\prime} \\ \Rightarrow A^{\prime} A^{\prime}=(-A)(-A)=A^2 \\ \left.\Rightarrow\left(A^2\right)^{\prime}=A^2 \quad \text { (since } A \text { is symmetric }\right)\end{gathered}$
This equation shows that the transpose of $A^2$ is equal to $A^2$ itself when $A$ is a symmetric matrix (i.e., $\left.A^{\prime}=A\right)$.
Question 101:
Answer:
True
Given:
$\begin{aligned} & A B=B A \\ & \begin{aligned}(A B)\left(A^{-1} B^{-1}\right) & =(B A)\left(A^{-1} B^{-1}\right) \\ & =B\left(A A^{-1}\right) B^{-1} \\ & =B \cdot I \cdot B^{-1}=B \cdot B^{-1}=I \\ (A B)^{-1} & =A^{-1} B^{-1}\end{aligned}\end{aligned}$
Students can make use of the NCERT Exemplar Class 12 Maths solutions chapter 3 pdf download to access it offline. We will help the students understand the matrices and their functions and operations by solving the questions given in the NCERT.
NCERT exemplar solutions for Class 12 Maths Chapter 3 Matrices Sub-topics covered
The Sub-Topics That Are Covered Under The Class 12 Maths NCERT Exemplar Solutions Chapter 3 Matrices Are:
- Introduction
- Matrix
- Order of matrix
- Types of matrices
- Operations on matrices
- Addition of matrices
- Multiplication of a matrix by a scalar
- Properties of matrix addition
- Properties of scalar multiplication of a matrix
- Multiplication of matrices
- Properties of multiplication of matrices
- Transpose of matrix
- Properties of the transpose of matrices
- Symmetric and skew-symmetric matrices
- Elementary operation of matrix
- Invertible matrices
- The inverse of matrices by elementary operations
NCERT Exemplar Class 12 Maths Solutions
Chapter 3 Matrices |
Importance of NCERT Exemplar Class 12 Maths Solutions Chapter 3
- NCERT exemplar solutions for Class 12 Math Chapter 3 will help the students understand the basics. It will help in developing the concepts.
- From this article, a student will be able to solve the problems based on applications of matrices.
- NCERT Exemplar Class 12 Maths Solutions Chapter 3 covers all types of problems, which enable a student to solve higher-order problems.
- NCERT Exemplar Class 12 Maths chapter 3 solutions are prepared in a simple language only so that it is easier for the student to understand the solution and hence develop the right approach for a particular question.
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- NCERT Solution for Class 12 Physics
- NCERT Solution for Class 12 Chemistry
- NCERT Solution for Class 12 Maths
- NCERT Solution for Class 12 Biology
Read more NCERT Notes subject-wise
Given below are the subject-wise NCERT Notes of class 12 :
- NCERT Notes for Class 12 Physics
- NCERT Notes for Class 12 Chemistry
- NCERT Notes for Class 12 Maths
- NCERT Notes for Class 12 Biology
Also, Check NCERT Books and NCERT Syllabus here
Here are some useful links for NCERT books and the NCERT syllabus for class 12:
Frequently Asked Questions (FAQs)
Q: What are the properties of matrix multiplication?
A:
Properties of Matrix Multiplication
Matrix multiplication has unique properties different from regular multiplication:
1. Associative Property: (A B) C=A(B C), when the order is compatible.
2. Distributive Property: A(B+C)=A B+A C and (A+B) C=A C+B C.
3. Non-Commutative: In general, AB is not equal to BA, even if both products are defined.
4. Multiplicative Identity: AI=IA=A, where I is the identity matrix of suitable order.
5. Zero Product Property Doesn't Hold: AB=0 does not imply A=0 or B=0.
6. Compatibility: Matrix multiplication is only defined when the number of columns in the first matrix equals the number of rows in the second.
Q: What are the properties of matrix addition?
A:
Properties of Matrix Addition
Matrix addition has several important properties, similar to regular number addition:
1. Commutative Property: A+B=B+A, if A and B are of the same order.
2. Associative Property: (A+B)+C=A+(B+C), for matrices of the same order.
3. Additive Identity: There exists a zero matrix O such that A+O=A.
4. Additive Inverse: For every matrix A, there exists a matrix -A such that A+(-A)=O.
5. Closure Property: The sum of two matrices of the same order is also a matrix of the same order.
Q: What are the different types of matrices?
A:
Matrices come in various types based on their elements and structure. A row matrix has only one row, while a column matrix has only one column. A square matrix has the same number of rows and columns. A diagonal matrix has non-zero elements only on its main diagonal, and if all diagonal elements are 1, it's called an identity matrix. A zero matrix has all elements as zero. A symmetric matrix is equal to its transpose, while a skew-symmetric matrix has its transpose equal to its negative. Matrices are also classified as upper or lower triangular based on zero entries.
Q: What is the significance of determinants in matrices?
A:
Determinants play a crucial role in matrix algebra and have wide applications in solving mathematical and real-world problems. The determinant of a square matrix is a scalar value that helps determine whether a matrix is invertible; if the determinant is zero, the matrix is singular and non-invertible. Determinants are essential in solving systems of linear equations using Cramer’s Rule, finding the area or volume in geometry, and analyzing linear transformations. In physics and engineering, they help study stability, force systems, and transformations. Thus, determinants are key tools in understanding and applying matrix concepts effectively.
Q: What is the condition for a matrix to be invertible?
A:
Condition for a Matrix to Be Invertible
A square matrix is invertible (also called non-singular) if there exists another matrix, called its inverse, such that A-1 A=A A-1=I, where I is the identity matrix of the same order.
The key condition for a matrix to be invertible is that its determinant must not be zero, i.e., {det}(A) not equal to 0.
If {det}(A)=0, the matrix is called singular and does not have an inverse.
Invertible matrices are essential in solving systems of linear equations, finding matrix equations, and various applications in linear algebra.
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Yes, if you’re not satisfied with your marks even after the improvement exam, many education boards allow you to reappear as a private candidate next year to improve your scores. This means you can register independently, study at your own pace, and take the exams without attending regular classes. It’s a good option to improve your results and open up more opportunities for higher studies or careers. Just make sure to check the specific rules and deadlines of your education board so you don’t miss the registration window. Keep your focus, and you will do better next time.
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Yes, in the case that you appeared for the 2025 improvement exam and your roll number is different from what was on the previous year’s marksheet, the board will usually release a new migration certificate. This is because the migration certificate will reflect the most recent exam details, roll number and passing year. You can apply to get it from your board using the process prescribed by them either online or through your school/college.
Yes, if you miss the 1st CBSE exam due to valid reasons, then you can appear for the 2nd CBSE compartment exam.
From the academic year 2026, the board will conduct the CBSE 10th exam twice a year, while the CBSE 12th exam will be held once, as per usual. For class 10th, the second phase exam will act as the supplementary exam. Check out information on w hen the CBSE first exam 2026 will be conducted and changes in 2026 CBSE Board exam by clicking on the link .
If you want to change your stream to humanities after getting a compartment in one subject in the CBSE 12th Board Exam , you actually have limited options to qualify for your board exams. You can prepare effectively and appear in the compartment examination for mathematics again. If you do not wish to continue with the current stream, you can take readmission in the Humanities stream and start from Class 11th again, and continue studying for two more years to qualify for the 12th examination.
The GUJCET Merit List is prepared based on the Class 12th marks and GUJCET marks received by the students. CBSE students who are not from the Gujarat board can definitely compete with GSEB students, as their eligibility is decided based on the combined marks scored by them in GUJCET and the 12th board. The weightage of the GUJCET score is 40% and the weightage of the class 12 scores is 60%.
Popular CBSE Class 12th Questions
A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is
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An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range
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A particle is projected at 600 to the horizontal with a kinetic energy . The kinetic energy at the highest point
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In the reaction,
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How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?
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0.02 |
Option 2)
3.125 × 10-2 |
| Option 3)
1.25 × 10-2 |
Option 4)
2.5 × 10-2 |
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