NCERT Solutions for Class 12 Maths Chapter 3 - Matrices

Question 1(i):In the matrix $A = \begin{bmatrix} 2& 5 &19 &-7 \\ 35 & -2 & \frac{5}{2} &12 \\ \sqrt3& 1 &-5 &17 \end{bmatrix}$, write:

The order of the matrix

Answer:

$A = \begin{bmatrix} 2& 5 &19 &-7 \\ 35 & -2 & \frac{5}{2} &12 \\ \sqrt3& 1 &-5 &17 \end{bmatrix}$

(i) The order of the matrix = number of row $\times$ number of columns $= 3\times 4$.

Question 1(ii): In the matrix $A = \begin{bmatrix}2&5&19&-7&\\ 35& -2&\frac{5}{2}&12\\\sqrt3&1&-5&17 \end{bmatrix}$, write:

The number of elements

Answer:

$A = \begin{bmatrix}2&5&19&-7&\\ 35& -2&\frac{5}{2}&12\\\sqrt3&1&-5&17 \end{bmatrix}$

(ii) The number of elements $3\times 4=12$.

Question 1(iii): In the matrix $A = \begin{bmatrix}2&5&19&-7&\\35&-2&\frac{5}{2}&12\\\sqrt3&1&-5&17 \end{bmatrix}$, write:

Write the elements a13, a21, a33, a24, a23

Answer:

$A = \begin{bmatrix}2&5&19&-7&\\35&-2&\frac{5}{2}&12\\\sqrt3&1&-5&17 \end{bmatrix}$

(iii) An element $a_{ij}$ implies the element in row number i and column number j.

$a_{13} = 19$, $a_{21} = 35$

$a_{33} = -5$, $a_{24} = 12$

$a_{23} = \frac{5}{2}$

Question 2: If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements?

Answer:

A matrix has 24 elements.

The possible orders are :

$1\times 24,24\times 1,2\times 12,12\times 2,3\times 8,8\times 3,4\times 6 \, \, and\, \, 6\times 4$.

If it has 13 elements, then possible orders are :

$1\times 13\, \, \, and \, \, \, \, 13\times 1$.

Question 3: If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?

Answer:

A matrix has 18 elements.

The possible orders are as follows

$1\times 18,18\times 1,2\times 9,9\times 2,3\times 6\, \, \, and\, \, \, \, 6\times 3$

If it has 5 elements, then possible orders are :

$1\times 5\, \, \, and \, \, \, \, 5\times 1$.

Question 4(i): Construct a 2 × 2 matrix, $A = [a_{ij} ]$ whose elements are given by:

$a_{ij} = \frac{(i + j)^2}{2}$

Answer:

$A = [a_{ij} ]$

(i) $a_{ij} = \frac{(i + j)^2}{2}$

Each element of this matrix is calculated as follows

$a_{11} = \frac{(1+1)^2}{2} = \frac{2^2}{2} = \frac{4}{2} = 2$, $a_{22} = \frac{(2+2)^2}{2} = \frac{4^2}{2} = \frac{16}{2} = 8$

$a_{12} = \frac{(1+2)^2}{2} = \frac{3^2}{2} = \frac{9}{2} = 4.5$, $a_{21} = \frac{(2+1)^2}{2} = \frac{3^2}{2} = \frac{9}{2} = 4.5$

Matrix A is given by

$A = \begin{bmatrix} 2&4.5 \\4.5 & 8 \end{bmatrix}$

Question 4(ii): Construct a 2 × 2 matrix, $A = [a_{ij} ]$, whose elements are given by:

$a_{ij} = \frac{i}{j}$

Answer:

A 2 × 2 matrix, $A = [a_{ij} ]$

(ii) $a_{ij} = \frac{i}{j}$

$a_{11} = \frac{1}{1} = 1$, $a_{22} = \frac{2}{2} = 1$

$a_{12} = \frac{1}{2}$, $a_{21} = \frac{2}{1} = 2$

Hence, the matrix is

$A = \begin{bmatrix} 1& \frac{1}{2} \\ 2 & 1 \end{bmatrix}$

Question 4(iii): Construct a 2 × 2 matrix, $A = [a_{ij} ]$, whose elements are given by:

$a_{ij} = \frac{(i+2j)^2}{2}$

Answer:

(iii)

$a_{ij} = \frac{(i + 2j)^2}{2}$

$a_{11} = \frac{(1 + (2 \times 1))^2}{2} = \frac{(1 + 2)^2}{2} = \frac{3^2}{2} = \frac{9}{2}$,

$a_{22} = \frac{(2 + (2 \times 2))^2}{2} = \frac{(2 + 4)^2}{2} = \frac{6^2}{2} = \frac{36}{2} = 18$,

$a_{21} = \frac{(2 + (2 \times 1))^2}{2} = \frac{(2 + 2)^2}{2} = \frac{4^2}{2} = \frac{16}{2} = 8$,

$a_{12} = \frac{(1 + (2 \times 2))^2}{2} = \frac{(1 + 4)^2}{2} = \frac{5^2}{2} = \frac{25}{2}$

Hence, the matrix is given by

$A = \begin{bmatrix} \frac{9}{2}& \frac{25}{2} \\ 8 & 18 \end{bmatrix}$

Question 5(i): Construct a 3 × 4 matrix whose elements are given by:

$a_{ij} = \frac{1}{2}|-3i + j|$

Answer:

(i)

$a_{ij} = \frac{1}{2} \left| -3i + j \right|$

$a_{11} = \frac{\left| -3 + 1 \right|}{2} = \frac{2}{2} = 1$,
$a_{12} = \frac{\left| (-3 \times 1) + 2 \right|}{2} = \frac{1}{2}$,
$a_{13} = \frac{\left| (-3 \times 1) + 3 \right|}{2} = 0$

$a_{21} = \frac{\left| (-3 \times 2) + 1 \right|}{2} = \frac{5}{2}$,
$a_{22} = \frac{\left| (-3 \times 2) + 2 \right|}{2} = \frac{4}{2} = 2$,
$a_{23} = \frac{\left| (-3 \times 2) + 3 \right|}{2} = \frac{\left| -6 + 3 \right|}{2} = \frac{\left| -3 \right|}{2} = \frac{3}{2}$

$a_{31} = \frac{\left| (-3 \times 3) + 1 \right|}{2} = \frac{8}{2} = 4$,
$a_{32} = \frac{\left| (-3 \times 3) + 2 \right|}{2} = \frac{7}{2}$,
$a_{33} = \frac{\left| (-3 \times 3) + 3 \right|}{2} = \frac{\left| -9 + 3 \right|}{2} = \frac{\left| -6 \right|}{2} = \frac{6}{2} = 3$

$a_{14} = \frac{\left| (-3 \times 1) + 4 \right|}{2} = \frac{\left| -3 + 4 \right|}{2} = \frac{\left| 1 \right|}{2} = \frac{1}{2}$,
$a_{24} = \frac{\left| (-3 \times 2) + 4 \right|}{2} = \frac{\left| -6 + 4 \right|}{2} = \frac{\left| -2 \right|}{2} = \frac{2}{2} = 1$,
$a_{34} = \frac{\left| (-3 \times 3) + 4 \right|}{2} = \frac{\left| -9 + 4 \right|}{2} = \frac{\left| -5 \right|}{2} = \frac{5}{2}$

Hence, the required matrix of the given order is

$A = \begin{bmatrix} 1& \frac{1}{2} & 0&\frac{1}{2} \\ \frac{5}{2} & 2&\frac{3}{2}&1 \\4&\frac{7}{2}&3&\frac{5}{2}\end{bmatrix}$

Question 5(ii): Construct a 3 × 4 matrix, whose elements are given by:

$a_{ij} = 2i - j$

Answer:

A 3 × 4 matrix,

(ii) $a_{ij} = 2i - j$

$a_{11} = 2 \times 1 - 1 = 2 - 1 = 1$, $a_{12} = 2 \times 1 - 2 = 2 - 2 = 0$, $a_{13} = 2 \times 1 - 3 = 2 - 3 = -1$

$a_{21} = 2 \times 2 - 1 = 4 - 1 = 3$, $a_{22} = 2 \times 2 - 2 = 4 - 2 = 2$, $a_{23} = 2 \times 2 - 3 = 4 - 3 = 1$

$a_{31} = 2 \times 3 - 1 = 6 - 1 = 5$, $a_{32} = 2 \times 3 - 2 = 6 - 2 = 4$, $a_{33} = 2 \times 3 - 3 = 6 - 3 = 3$

$a_{14} = 2 \times 1 - 4 = 2 - 4 = -2$, $a_{24} = 2 \times 2 - 4 = 4 - 4 = 0$, $a_{34} = 2 \times 3 - 4 = 6 - 4 = 2$

Hence, the matrix is

$A = \begin{bmatrix} 1 & 0& -1& -2 \\ \ 3 & 2&1& 0 \\5&4&3&2\end{bmatrix}$

Question 6(i): Find the values of x, y and z from the following equations:

$\begin{bmatrix}4&3\\x&5 \end{bmatrix} = \begin{bmatrix}y&z\\1&5 \end{bmatrix}$

Answer:

(i) $\begin{bmatrix}4&3\\x&5 \end{bmatrix} = \begin{bmatrix}y&z\\1&5 \end{bmatrix}$

If two matrices are equal, then their corresponding elements are also equal.

$\therefore$ $x=1\, \, \, ,\, \, \, y=4\, \, \, \, and\, \, \, \, z=3$

Question 6(ii): Find the values of x, y and z from the following equations:

$\begin{bmatrix} x +y & 2\\ 5 + z & xy \end{bmatrix} = \begin{bmatrix} 6 &2 \\ 5 & 8 \end{bmatrix}$

Answer:

(ii)

$\begin{bmatrix} x +y & 2\\ 5 + z & xy \end{bmatrix} = \begin{bmatrix} 6 &2 \\ 5 & 8 \end{bmatrix}$

If two matrices are equal, then their corresponding elements are also equal.

$\therefore$ $x+y=6$ $\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (i)$

$x=6-y$

$xy=8$ $\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (ii)$

Solving equations (i) and (ii),

$(6-y)y =8$

$6y-y^{2}=8$

$y^{2}-6y+8=0$

Solving this equation, we get,

$y=4 \, \, and\, \, y=2$

Putting the values of y, we get

$x=2 \, \, and\, \, x=4$

And also equating the first element of the second row

$5+z = 5$, $z=0$

Hence,

$x=2,y=4,z=0\, \, \, \, \, and\, \, \, \, \, \, x=4,y=2,z=0$

Question 6(iii): Find the values of x, y and z from the following equations

$\begin{bmatrix} x + y + z\\ x + z \\ y + z \end{bmatrix} = \begin{bmatrix} 9\\5 \\7 \end{bmatrix}$

Answer:

(iii)

$\begin{bmatrix} x + y + z\\ x + z \\ y + z \end{bmatrix} = \begin{bmatrix} 9\\5 \\7 \end{bmatrix}$

If two matrices are equal, then their corresponding elements are also equal

$x+y+z=9........(1)$

$x+z=5..............(2)$

$y+z=7..............(3)$

Subtracting (2) from (1), we will get y=4

Substituting the value of y in equation (3), we will get z=3

Now, substituting the value of z in equation (2), we will get x=2

therefore,

$x=2$, $y=4$ and $z=3$

Question 7: Find the value of a, b, c and d from the equation:

$\begin{bmatrix} a -b & 2a + c\\ 2a - b & 3c + d \end{bmatrix} = \begin{bmatrix} -1 & 5\\ 0 & 13 \end{bmatrix}$

Answer:

$\begin{bmatrix} a -b & 2a + c\\ 2a - b & 3c + d \end{bmatrix} = \begin{bmatrix} -1 & 5\\ 0 & 13 \end{bmatrix}$

If two matrices are equal, then their corresponding elements are also equal

$a-b=-1$ $.............................1$

$2a+c=5$ $.............................2$

$2a-b=0$ $.............................3$

$3c+d=13$ $.............................4$

Solving equations 1 and 3, we get

$a=1 \, \, \, \, and \, \, \, \, b=2$

Putting the value of a in equation 2, we get

$c=3$

Putting the value of c in equation 4, we get

$d=4$

Question 8: $A = [a_{ij}]_{m\times n}$ is a square matrix, if

(A) $m <n$

(B) $m >n$

(C) $m =n$

(D) None of these

Answer:

A square matrix has the number of rows and columns equal.

Thus, for $A = [a_{ij}]_{m\times n}$ to be a square matrix, m and n should be equal.

$\therefore m=n$

Option (c) is correct.

Question 9: Which of the given values of x and y make the following pair of matrices equal

$\begin{bmatrix} 3x + 7 &5 \\ y + 1 & 2 -3x \end{bmatrix}$, $\begin{bmatrix} 0 & y - 2 \\ 8 & 4 \end{bmatrix}$

(A) $x = \frac{-1}{3}, y = 7$

(B) Not possible to find

(C) $y =7, x = \frac{-2}{3}$

(D) $x = \frac{-1}{3}, y = \frac{-2}{3}$

Answer:

Given, $\begin{bmatrix} 3x + 7 &5 \\ y + 1 & 2 -3x \end{bmatrix}$ $=\begin{bmatrix} 0 & y - 2 \\ 8 & 4 \end{bmatrix}$

If two matrices are equal, then their corresponding elements are also equal

$3x+7=0\Rightarrow x=\frac{-7}{3}$

$y-2=5 \Rightarrow y=5+2=7$

$y+1=8\Rightarrow y=8-1=7$

$2-3x=4\Rightarrow 3x=2-4\Rightarrow 3x=-2\Rightarrow x=\frac{-2}{3}$

Here, the value of x is not unique, so option B is correct.

Question 10: The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is:

(A) 27
(B) 18
(C) 81
(D) 512

Answer:

Total number of elements in a 3 × 3 matrix

$=3\times 3=9$

If each entry is 0 or 1, then for every entry, there are 2 permutations.

The total permutations for 9 elements

$=2^{9}=512$

Thus, option (D) is correct.

Question 1(i): Let $A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$, $B = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$, $C = \begin{bmatrix} -2 &5 \\ 3 & 4 \end{bmatrix}$

Find each of the following:

A + B

Answer:

$A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ $B = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$

(i) A + B

The addition of a matrix can be done as follows

$A+B = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ $+ \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$

$A+B = \begin{bmatrix} 2+1 &4+3 \\ 3+(-2) & 2+5 \end{bmatrix}$

$A+B = \begin{bmatrix} 3 &7 \\ 1 & 7 \end{bmatrix}$

Question 1(ii): Let $A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$, $B = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$, $C = \begin{bmatrix} -2 &5 \\ 3 & 4 \end{bmatrix}$

Find each of the following:

A - B

Answer:

$A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ $B = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$

(ii) A - B

$A-B = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ $- \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$

$A-B = \begin{bmatrix} 2-1 &4-3 \\ 3-(-2) & 2-5 \end{bmatrix}$

$A-B = \begin{bmatrix} 1 &1 \\ 5 & -3 \end{bmatrix}$

Question 1(iii): Let $A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$, $B = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$, $C = \begin{bmatrix} -2 &5 \\ 3 & 4 \end{bmatrix}$

Find each of the following:

3A - C

Answer:

$A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ $C = \begin{bmatrix} -2 &5 \\ 3 & 4 \end{bmatrix}$

(iii) 3A - C

First multiply each element of A by 3 and then subtract C

$3A -C = 3\begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ $- \begin{bmatrix} -2 &5 \\ 3 & 4 \end{bmatrix}$

$3A -C = \begin{bmatrix} 6 &12 \\ 9 & 6 \end{bmatrix}$ $- \begin{bmatrix} -2 &5 \\ 3 & 4 \end{bmatrix}$

$3A -C = \begin{bmatrix} 6-(-2) &12-5 \\ 9-3 & 6-4 \end{bmatrix}$

$3A -C = \begin{bmatrix} 8 &7 \\ 6 & 2 \end{bmatrix}$

Question 1(iv): Let $A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$, $B = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$, $C = \begin{bmatrix} -2 &5 \\ 3 & 4 \end{bmatrix}$

Find each of the following:

AB

Answer:

$A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ $B = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$

(iv) AB

$AB = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ $\times \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$

$AB = \begin{bmatrix} 2\times 1+4\times -2 & \, \, \, 2\times 3+4\times 5 \\ 3\times 1+2\times -2 & \, \, \, 3\times 3+2 \times 5 \end{bmatrix}$

$AB = \begin{bmatrix} -6 &26 \\ -1 & 19 \end{bmatrix}$

Question 1(v): Let $A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$, $B = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$, $C = \begin{bmatrix} -2 &5 \\ 3 & 4 \end{bmatrix}$

Find each of the following:

BA

Answer:

The multiplication is performed as follows

$A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ ,$B = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$

$BA = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$ $\times \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$

$BA = \begin{bmatrix} 1\times 2+3\times 3 &1\times 4+3\times 2 \\ -2\times 2+5\times 3& -2\times 4+2\times 5 \end{bmatrix}$

$BA = \begin{bmatrix} 11 &10 \\ 11& 2 \end{bmatrix}$

Question 2(i): Compute the following:

$\begin{bmatrix} a &b \\ -b& a \end{bmatrix} + \begin{bmatrix} a &b \\ b& a \end{bmatrix}$

Answer:

(i) $\begin{bmatrix} a &b \\ -b& a \end{bmatrix} + \begin{bmatrix} a &b \\ b& a \end{bmatrix}$

$= \begin{bmatrix} a+a &b+b \\ -b+b & a+a \end{bmatrix}$

$= \begin{bmatrix} 2a &2b \\ 0 & 2a \end{bmatrix}$

Question 2(ii): Compute the following:

$\begin{bmatrix} a^2 + b^2& b^2+c^2\\ a^2 + c^2& a^2 + b^2 \end{bmatrix} + \begin{bmatrix} 2ab &2bc \\ -2ac & -2ab \end{bmatrix}$

Answer:

(ii) The addition operation can be performed as follows

$\begin{bmatrix} a^2 + b^2& b^2+c^2\\ a^2 + c^2& a^2 + b^2 \end{bmatrix} + \begin{bmatrix} 2ab &2bc \\ -2ac & -2ab \end{bmatrix}$

$=\begin{bmatrix} a^2 + b^2+2ab& b^2+c^2+2bc\\ a^2 + c^2-2ac& a^2 + b^2-2ab \end{bmatrix}$

$=\begin{bmatrix} (a+b)^2 & (b+c)^2\\ (a-c)^2 & (a-b)^2 \end{bmatrix}$

Question 2(iii): Compute the following:

$\begin{bmatrix} -1 & 4 & -6\\ 8 & 5 & 16\\ 2 & 8 & 5 \end{bmatrix} + \begin{bmatrix} 12 & 7 & 6\\ 8 & 0 &5 \\ 3 & 2 & 4 \end{bmatrix}$

Answer:

(iii) The addition of given three by three matrix is performed as follows

$\begin{bmatrix} -1 & 4 & -6\\ 8 & 5 & 16\\ 2 & 8 & 5 \end{bmatrix} + \begin{bmatrix} 12 & 7 & 6\\ 8 & 0 &5 \\ 3 & 2 & 4 \end{bmatrix}$

$=\begin{bmatrix} -1+12 & 4+7 & -6+6\\ 8+8 & 5+0 & 16+5\\ 2+3 & 8+2 & 5+4 \end{bmatrix}$

$=\begin{bmatrix} 11 & 11 & 0\\ 16 & 5 & 21\\ 5 & 10 & 9 \end{bmatrix}$

Question 2(iv): Compute the following:

$\begin{bmatrix} \cos^2 x &\sin^2 x\\ \sin^2 x & \cos^2x \end{bmatrix} + \begin{bmatrix} \sin^2 x &\cos^2 x\\ \cos^2 x & \sin^2x \end{bmatrix}$

Answer:

(iv) the addition is done as follows

$\begin{bmatrix} \cos^2 x &\sin^2 x\\ \sin^2 x & \cos^2x \end{bmatrix} + \begin{bmatrix} \sin^2 x &\cos^2 x\\ \cos^2 x & \sin^2x \end{bmatrix}$

$=\begin{bmatrix} \cos^2+ \sin^2 x &\sin^2 x+\cos^2 x\\ \sin^2 x+\cos^2 x & \cos^2x+ \sin^2 x \end{bmatrix}$ since $sin^2x+cos^2x=1$

$=\begin{bmatrix} 1 &1\\ 1 & 1 \end{bmatrix}$

Question 3(i): Compute the indicated products.

$\begin{bmatrix} a &b \\ -b &a \end{bmatrix} \begin{bmatrix} a & -b \\ b &a \end{bmatrix}$

Answer:

(i) The multiplication is performed as follows

$\begin{bmatrix} a &b \\ -b &a \end{bmatrix} \begin{bmatrix} a & -b \\ b &a \end{bmatrix}$

$=\begin{bmatrix} a &b \\ -b &a \end{bmatrix} \times \begin{bmatrix} a & -b \\ b &a \end{bmatrix}$

$=\begin{bmatrix} a\times a+b\times b &a\times -b+b\times a \\ -b\times a+a\times b &-b\times -b+a\times a \end{bmatrix}$

$=\begin{bmatrix} a^{2}+b^{2} & 0 \\ 0 & b^{2}+a^{2} \end{bmatrix}$

Question 3(ii): Compute the indicated products.

$\begin{bmatrix} 1\\ 2\\ 3 \end{bmatrix}\begin{bmatrix} 2 &3 & 4 \end{bmatrix}$

Answer:

(ii) the multiplication can be performed as follows

$\begin{bmatrix} 1\\ 2\\ 3 \end{bmatrix}\begin{bmatrix} 2 &3 & 4 \end{bmatrix}$

$=\begin{bmatrix} 1\times 2 &1\times 3&1\times 4\\ 2\times 2&2\times 3&2\times 4\\3\times 2&3\times 3&3\times 4 \end{bmatrix}$

$=\begin{bmatrix} 2 &3& 4\\ 4&6&8\\6&9&12 \end{bmatrix}$

Question 3(iii): Compute the indicated products.

$\begin{bmatrix} 1 & -2\\ 2 & 3 \end{bmatrix}\begin{bmatrix} 1 &2 &3\\ 2 & 3 & 1 \end{bmatrix}$

Answer:

(iii) The multiplication can be performed as follows

$\begin{bmatrix} 1 & -2\\ 2 & 3 \end{bmatrix}\begin{bmatrix} 1 &2 &3\\ 2 & 3 & 1 \end{bmatrix}$

$=\begin{bmatrix} 1\times 1+(-2)\times 2 & 1\times 2+(-2)\times 3&1\times 3+(-2)\times 1\\ 2\times 1+3\times 2 & 2\times 2+3\times 3&2\times 3+3\times 1 \end{bmatrix}$

Question 3(iv): Compute the indicated products.

$\begin{bmatrix} 2 & 3 & 4\\ 3 & 4 & 5\\ 4 & 5 & 6 \end{bmatrix} \begin{bmatrix} 1 & -3 & 5\\ 0& 2 & 4\\ 3 & 0 & 5 \end{bmatrix}$

Answer:

(iv) The multiplication is performed as follows

$\begin{bmatrix} 2 & 3 & 4\\ 3 & 4 & 5\\ 4 & 5 & 6 \end{bmatrix} \begin{bmatrix} 1 & -3 & 5\\ 0& 2 & 4\\ 3 & 0 & 5 \end{bmatrix}$

$=\begin{bmatrix} 2 & 3 & 4\\ 3 & 4 & 5\\ 4 & 5 & 6 \end{bmatrix}\times \begin{bmatrix} 1 & -3 & 5\\ 0& 2 & 4\\ 3 & 0 & 5 \end{bmatrix}$

$=\begin{bmatrix} 2\times 1+3\times 0+4\times 3 \, \, & 2\times (-3)+3\times 2+4\times 0 \, \, & 2\times 5+3\times 4+4\times 5 \\ 3\times 1+4\times 0+5\times 3 \, \, & 3\times (-3)+4\times 2+5\times 0 & 3\times 5+4\times 4+5\times 5 \\ 4\times 1+5\times 0+6\times 3 \, \, & 4\times (-3)+5\times 2+6\times 0\, \, & 4\times 5+5\times 4+6\times 5 \end{bmatrix}$

$= \begin{bmatrix} 14 & 0 & 42\\ 18 & -1 & 56\\ 22 & -2 & 70 \end{bmatrix}$

Question 3(v): Compute the indicated products.

$\begin{bmatrix} 2 &1 \\ 3 & 2\\ -1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 1\\ -1 &2 & 1 \end{bmatrix}$

Answer:

(v) The product can be computed as follows

$\begin{bmatrix} 2 &1 \\ 3 & 2\\ -1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 1\\ -1 &2 & 1 \end{bmatrix}$

$=\begin{bmatrix} 2 &1 \\ 3 & 2\\ -1 & 1 \end{bmatrix}\times \begin{bmatrix} 1 & 0 & 1\\ -1 &2 & 1 \end{bmatrix}$

$=\begin{bmatrix} 2\times 1+1\times (-1) &2\times 0+1\times (2) & 2\times 1+1\times (1) \\ 3\times 1+2\times (-1) & 3\times 0+2\times (2) &3\times 1+2\times (1) \\ (-1)\times 1+1\times (-1) & (-1)\times 0+1\times (2) & (-1)\times 1+1\times (1) \end{bmatrix}$

$=\begin{bmatrix} 1 &2&3 \\ 1 & 4&5\\ -2 & 2&0 \end{bmatrix}$

Question 3(vi): Compute the indicated products.

$\begin{bmatrix} 3 & -1 & 3\\ -1 & 0 & 2 \end{bmatrix}\begin{bmatrix} 2 & -3\\ 1 & 0\\ 3 & 1 \end{bmatrix}$

Answer:

(vi) The given product can be computed as follows

$\begin{bmatrix} 3 & -1 & 3\\ -1 & 0 & 2 \end{bmatrix}\begin{bmatrix} 2 & -3\\ 1 & 0\\ 3 & 1 \end{bmatrix}$

$=\begin{bmatrix} 3 & -1 & 3\\ -1 & 0 & 2 \end{bmatrix}\times \begin{bmatrix} 2 & -3\\ 1 & 0\\ 3 & 1 \end{bmatrix}$

$=\begin{bmatrix} 3 \times 2+(-1)\times 1+3\times 3\, \, \, & 3 \times (-3)+(-1)\times 0+3\times 1 \\ (-1) \times 2+ 0 \times 1+2\times 3 \, \, \, & (-1) \times -3+0\times 0+2\times 1 \end{bmatrix}$

$=\begin{bmatrix} 14 & -6 \\ 4 & 5 \end{bmatrix}$

Question 4: If $A = \begin{bmatrix} 1 &2 &-3 \\ 5 &0 &2 \\ 1 & -1 &1 \end{bmatrix}$, $B = \begin{bmatrix} 3 &-1 &2 \\ 4 &2 &5 \\ 2 & 0 &3 \end{bmatrix}$ and $C = \begin{bmatrix} 4 &1 &2 \\ 0 &3 &2 \\ 1 & -2 &3 \end{bmatrix}$, then compute (A+B) and (B-C). Also verify that A + (B - C) = (A + B) - C

Answer:

$A = \begin{bmatrix} 1 &2 &-3 \\ 5 &0 &2 \\ 1 & -1 &1 \end{bmatrix}$, $B = \begin{bmatrix} 3 &-1 &2 \\ 4 &2 &5 \\ 2 & 0 &3 \end{bmatrix}$ and $C = \begin{bmatrix} 4 &1 &2 \\ 0 &3 &2 \\ 1 & -2 &3 \end{bmatrix}$

$A+B = \begin{bmatrix} 1 &2 &-3 \\ 5 &0 &2 \\ 1 & -1 &1 \end{bmatrix}$ $+ \begin{bmatrix} 3 &-1 &2 \\ 4 &2 &5 \\ 2 & 0 &3 \end{bmatrix}$

$A+B = \begin{bmatrix} 1+3 &2+(-1) &-3+2 \\ 5+4 &0+2 &2+5 \\ 1+2 & -1+0 &1+3 \end{bmatrix}$

$A+B = \begin{bmatrix} 4 &1 &-1 \\ 9 &2 &7 \\ 3 & -1 &4 \end{bmatrix}$

$B-C = \begin{bmatrix} 3 &-1 &2 \\ 4 &2 &5 \\ 2 & 0 &3 \end{bmatrix}$ $-\begin{bmatrix} 4 &1 &2 \\ 0 &3 &2 \\ 1 & -2 &3 \end{bmatrix}$

$B-C = \begin{bmatrix} 3-4 &-1-1 &2-2 \\ 4-0 &2-3 &5-2 \\ 2-1 & 0-(-2) &3-3 \end{bmatrix}$

$B-C = \begin{bmatrix} -1 &-2 &0 \\ 4 &-1 &3 \\ 1 & 2 &0 \end{bmatrix}$

Now, to prove A + (B - C) = (A + B) - C

$L.H.S\, \, :\, A+(B-C)$

$A+(B-C)=\begin{bmatrix} 1 &2 &-3 \\ 5 &0 &2 \\ 1 & -1 &1 \end{bmatrix}$ $+ \begin{bmatrix} -1 &-2 &0 \\ 4 &-1 &3 \\ 1 & 2 &0 \end{bmatrix}$ (Puting value of $B-C$ from above)

$A+(B-C)=\begin{bmatrix} 1-1 &2-2 &-3+0 \\ 5+4 &0+(-1) &2+3 \\ 1+1 & -1+2 &1+0 \end{bmatrix}$

$A+(B-C)=\begin{bmatrix} 0 &0 &-3 \\ 9 &-1 &5 \\ 2 & 1 &1 \end{bmatrix}$

$R.H.S\, \, :\, (A+B)-C$

$(A+B)-C = \begin{bmatrix} 4 &1 &-1 \\ 9 &2 &7 \\ 3 & -1 &4 \end{bmatrix}$ $- \begin{bmatrix} 4 &1 &2 \\ 0 &3 &2 \\ 1 & -2 &3 \end{bmatrix}$

$(A+B)-C = \begin{bmatrix} 4-4 &1-1 &-1-2 \\ 9-0 &2-3 &7-2 \\ 3-1 & -1-(-2) &4-3 \end{bmatrix}$

$(A+B)-C = \begin{bmatrix} 0 &0 &-3 \\ 9 &-1 &5 \\ 2 & 1 &1 \end{bmatrix}$

Hence, we can see L.H.S = R.H.S = $\begin{bmatrix} 0 &0 &-3 \\ 9 &-1 &5 \\ 2 & 1 &1 \end{bmatrix}$

Question 5: If $A = \begin{bmatrix} \frac{2}{3} & 1 & \frac{5}{3}\\ \frac{1}{3} & \frac{2}{3} &\frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3} \end{bmatrix}$ and $B = \begin{bmatrix} \frac{2}{5} & \frac{3}{5}&1\\ \frac{1}{5} & \frac{2}{5} &\frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5} \end{bmatrix}$, then compute 3A - 5B

Answer:

$A = \begin{bmatrix} \frac{2}{3} & 1 & \frac{5}{3}\\ \frac{1}{3} & \frac{2}{3} &\frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3} \end{bmatrix}$ and $B = \begin{bmatrix} \frac{2}{5} & \frac{3}{5}&1\\ \frac{1}{5} & \frac{2}{5} &\frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5} \end{bmatrix}$

$3A-5B = 3\times \begin{bmatrix} \frac{2}{3} & 1 & \frac{5}{3}\\ \frac{1}{3} & \frac{2}{3} &\frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3} \end{bmatrix}$ $-5\times \begin{bmatrix} \frac{2}{5} & \frac{3}{5}&1\\ \frac{1}{5} & \frac{2}{5} &\frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5} \end{bmatrix}$

$3A-5B = \begin{bmatrix} 2 & 3 & 5\\ 1 & 2 &4 \\ 7 & 6 & 2 \end{bmatrix}$ $- \begin{bmatrix} 2 & 3 & 5\\ 1 & 2 &4 \\ 7 & 6 & 2 \end{bmatrix}$

$3A-5B = \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}$

$3A-5B = 0$

Question 6: Simplify $\cos\theta\begin{bmatrix} \cos\theta & \sin\theta\\ -\sin\theta & \cos\theta \end{bmatrix} + \sin\theta\begin{bmatrix} \sin\theta & -\cos\theta\\ \cos\theta & \sin\theta \end{bmatrix}$.

Answer:

The simplification is explained in the following step

$\cos\theta\begin{bmatrix} \cos\theta & \sin\theta\\ -\sin\theta & \cos\theta \end{bmatrix} + \sin\theta\begin{bmatrix} \sin\theta & -\cos\theta\\ \cos\theta & \sin\theta \end{bmatrix}$

$= \begin{bmatrix} \cos^{2}\theta & \sin\theta \cos\theta \\ -\sin\theta \cos\theta & \cos^{2}\theta \end{bmatrix} +\begin{bmatrix} \sin^{2}\theta & - \sin\theta \cos\theta\\ \sin\theta\cos\theta & \sin^{2}\theta \end{bmatrix}$

$= \begin{bmatrix} \cos^{2}\theta+\sin^{2}\theta & \sin\theta \cos\theta - \sin\theta \cos\theta \\ -\sin\theta \cos\theta + \sin\theta \cos\theta & \cos^{2}\theta + \sin^{2}\theta\end{bmatrix}$

$= \begin{bmatrix} 1&0 \\ 0 & 1\end{bmatrix} =I$

the final answer is an identity matrix of order 2

Question 7(i): Find X and Y, if

$X + Y = \begin{bmatrix} 7 &0 \\ 2 &5 \end{bmatrix}$ and $X - Y = \begin{bmatrix} 3 &0 \\ 0 &3 \end{bmatrix}$

Answer:

(i) The given matrices are

$X + Y = \begin{bmatrix} 7 &0 \\ 2 &5 \end{bmatrix}$ and $X - Y = \begin{bmatrix} 3 &0 \\ 0 &3 \end{bmatrix}$

$X + Y = \begin{bmatrix} 7 &0 \\ 2 &5 \end{bmatrix}.............................1$

$X - Y = \begin{bmatrix} 3 &0 \\ 0 &3 \end{bmatrix}.............................2$

Adding equation 1 and 2, we get

$2 X = \begin{bmatrix} 7 &0 \\ 2 &5 \end{bmatrix}$ $+ \begin{bmatrix} 3 &0 \\ 0 &3 \end{bmatrix}$

$2 X = \begin{bmatrix} 7+3 &0+0 \\ 2+0 &5+3 \end{bmatrix}$

$2 X = \begin{bmatrix} 10 &0 \\ 2 &8 \end{bmatrix}$

$X = \begin{bmatrix} 5 &0 \\ 1 &4 \end{bmatrix}$

Putting the value of X in equation 1, we get

$\begin{bmatrix} 5 &0 \\ 1 &4 \end{bmatrix}$ $+Y = \begin{bmatrix} 7 &0 \\ 2 &5 \end{bmatrix}$

$Y = \begin{bmatrix} 7 &0 \\ 2 &5 \end{bmatrix} -$ $\begin{bmatrix} 5 &0 \\ 1 &4 \end{bmatrix}$

$Y = \begin{bmatrix} 7-5 &0-0 \\ 2-1 &5-4 \end{bmatrix}$

$Y = \begin{bmatrix} 2 &0 \\ 1 &1 \end{bmatrix}$

Question 7(ii): Find X and Y, if

$2X + 3Y = \begin{bmatrix} 2 &3 \\ 4 & 0 \end{bmatrix}$ and $3X + 2Y = \begin{bmatrix} 2 &-2 \\ -1 & 5 \end{bmatrix}$

Answer:

(ii) $2X + 3Y = \begin{bmatrix} 2 &3 \\ 4 & 0 \end{bmatrix}$ and $3X + 2Y = \begin{bmatrix} 2 &-2 \\ -1 & 5 \end{bmatrix}$

$2X + 3Y = \begin{bmatrix} 2 &3 \\ 4 & 0 \end{bmatrix}..........................1$

$3X + 2Y = \begin{bmatrix} 2 &-2 \\ -1 & 5 \end{bmatrix}......................2$

Multiply equation 1 by 3 and equation 2 by 2 and subtract them,

$3(2X + 3Y)-2(3X+2Y) = 3 \times \begin{bmatrix} 2 &3 \\ 4 & 0 \end{bmatrix}$ $- \, \, \, 2\times \begin{bmatrix} 2 &-2 \\ -1 & 5 \end{bmatrix}$

$6X + 9Y-6X-4Y= \begin{bmatrix} 6 &9 \\ 12 & 0 \end{bmatrix}$ $- \begin{bmatrix} 4 &-4 \\ -2 & 10 \end{bmatrix}$

$9Y-4Y= \begin{bmatrix} 6-4 &9-(-4) \\ 12-(-2) & 0-10 \end{bmatrix}$

$5Y= \begin{bmatrix} 2 &13 \\ 14 & -10 \end{bmatrix}$

$Y= \begin{bmatrix} \frac{2}{5} &\frac{13}{5} \\ \frac{14}{5} & -2 \end{bmatrix}$

Putting value of Y in equation 1 , we get

$2X + 3Y = \begin{bmatrix} 2 &3 \\ 4 & 0 \end{bmatrix}$

$2X + 3 \begin{bmatrix} \frac{2}{5} &\frac{13}{5} \\ \frac{14}{5} & -2 \end{bmatrix} = \begin{bmatrix} 2 &3 \\ 4 & 0 \end{bmatrix}$

$2X + \begin{bmatrix} \frac{6}{5} &\frac{39}{5} \\ \frac{42}{5} & -6 \end{bmatrix} = \begin{bmatrix} 2 &3 \\ 4 & 0 \end{bmatrix}$

$2X = \begin{bmatrix} 2 &3 \\ 4 & 0 \end{bmatrix} - \begin{bmatrix} \frac{6}{5} &\frac{39}{5} \\ \frac{42}{5} & -6 \end{bmatrix}$

$2X = \begin{bmatrix} 2-\frac{6}{5} &3-\frac{39}{5} \\ 4-\frac{42}{5} & 0 -(-6)\end{bmatrix}$

$2X = \begin{bmatrix} \frac{4}{5} &-\frac{24}{5} \\ -\frac{22}{5} & 6\end{bmatrix}$

$X = \begin{bmatrix} \frac{2}{5} &-\frac{12}{5} \\ -\frac{11}{5} & 3\end{bmatrix}$

Question 8: Find X, if $Y = \begin{bmatrix} 3 &2 \\ 1 & 4 \end{bmatrix}$ and $2X+ Y = \begin{bmatrix} 1 &0 \\ -3 & 2 \end{bmatrix}$

Answer:

$Y = \begin{bmatrix} 3 &2 \\ 1 & 4 \end{bmatrix}$

$2X+ Y = \begin{bmatrix} 1 &0 \\ -3 & 2 \end{bmatrix}$

Substituting the value of Y in the above equation

$2X+ \begin{bmatrix} 3 &2 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} 1 &0 \\ -3 & 2 \end{bmatrix}$

$2X = \begin{bmatrix} 1 &0 \\ -3 & 2 \end{bmatrix}- \begin{bmatrix} 3 &2 \\ 1 & 4 \end{bmatrix}$

$2X = \begin{bmatrix} 1-3 &0-2 \\ -3-1 & 2-4 \end{bmatrix}$

$2X = \begin{bmatrix} -2 &-2 \\ -4 & -2 \end{bmatrix}$

$X = \begin{bmatrix} -1 &-1 \\ -2 & -1 \end{bmatrix}$

Question 9: Find x and y, if $2\begin{bmatrix} 1 & 3\\ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0\\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6\\ 1 & 8\end{bmatrix}$

Answer:

$2\begin{bmatrix} 1 & 3\\ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0\\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6\\ 1 & 8\end{bmatrix}$

$\begin{bmatrix} 2 & 6\\ 0 & 2x \end{bmatrix} + \begin{bmatrix} y & 0\\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6\\ 1 & 8\end{bmatrix}$

$\begin{bmatrix} 2+y & 6+0\\ 0+1 & 2x+2 \end{bmatrix} = \begin{bmatrix} 5 & 6\\ 1 & 8\end{bmatrix}$

$\begin{bmatrix} 2+y & 6\\ 1 & 2x+2 \end{bmatrix} = \begin{bmatrix} 5 & 6\\ 1 & 8\end{bmatrix}$

Now equating LHS and RHS we can write the following equations

$2+y=5$ $2x+2=8$

$y=5-2$ $2x=8-2$

$y=3$ $2x=6$

$x=3$

Question 10: Solve the equation for x, y, z and t, if $2\begin{bmatrix}x & z \\ y &t \end{bmatrix} + 3\begin{bmatrix} 1 & -1\\ 0 & 2 \end{bmatrix} = 3\begin{bmatrix} 3 & 5\\ 4 & 6 \end{bmatrix}$

Answer:

$2\begin{bmatrix}x & z \\ y &t \end{bmatrix} + 3\begin{bmatrix} 1 & -1\\ 0 & 2 \end{bmatrix} = 3\begin{bmatrix} 3 & 5\\ 4 & 6 \end{bmatrix}$

Multiplying with constant terms and rearranging we can rewrite the matrix as

$\begin{bmatrix}2x &2 z \\ 2y &2t \end{bmatrix} = \begin{bmatrix} 9 &15\\ 12 & 18 \end{bmatrix} - 3\begin{bmatrix} 1& -1\\ 0 & 2 \end{bmatrix}$

$\begin{bmatrix}2x &2 z \\ 2y &2t \end{bmatrix} = \begin{bmatrix} 9 &15\\ 12 & 18 \end{bmatrix} - \begin{bmatrix} 3& -3\\ 0 & 6 \end{bmatrix}$

$\begin{bmatrix}2x &2 z \\ 2y &2t \end{bmatrix} = \begin{bmatrix} 9-3 &15-(-3)\\ 12-0 & 18-6 \end{bmatrix}$

$\begin{bmatrix}2x &2 z \\ 2y &2t \end{bmatrix} = \begin{bmatrix} 6 &18\\ 12 & 12 \end{bmatrix}$

Dividing by 2 on both sides

$\begin{bmatrix}x & z \\ y &t \end{bmatrix} = \begin{bmatrix} 3 &9\\ 6 & 6 \end{bmatrix}$

$x=3,y=6,z=9\, \, and\, \, t=6$

Question 11: If $x\begin{bmatrix}2\\3 \end{bmatrix} + y\begin{bmatrix} -1\\1 \end{bmatrix} = \begin{bmatrix} 10\\5 \end{bmatrix}$, find the values of x and y.

Answer:

$x\begin{bmatrix}2\\3 \end{bmatrix} + y\begin{bmatrix} -1\\1 \end{bmatrix} = \begin{bmatrix} 10\\5 \end{bmatrix}$

$\begin{bmatrix}2x\\3x \end{bmatrix} + \begin{bmatrix} -y\\y \end{bmatrix} = \begin{bmatrix} 10\\5 \end{bmatrix}$

Adding both the matrices in the LHS and rewriting

$\begin{bmatrix}2x-y\\3x+y \end{bmatrix} = \begin{bmatrix} 10\\5 \end{bmatrix}$

$2x-y=10........................1$

$3x+y=5........................2$

Adding equations 1 and 2, we get

$5x=15$

$x=3$

Put the value of x in equation 2, we have

$3x+y=5$

$3\times 3+y=5$

$9+y=5$

$y=5-9$

$y=-4$

Question 12: Given $3\begin{bmatrix}x & y \\ z & w \end{bmatrix} = \begin{bmatrix} x & 6 \\ -1 & 2w \end{bmatrix} + \begin{bmatrix} 4 &x + y \\ z + w & 3 \end{bmatrix}$, find the values of x, y, z and w.

Answer:

$3\begin{bmatrix}x & y \\ z & w \end{bmatrix} = \begin{bmatrix} x & 6 \\ -1 & 2w \end{bmatrix} + \begin{bmatrix} 4 &x + y \\ z + w & 3 \end{bmatrix}$

$\begin{bmatrix}3x &3 y \\3 z & 3w \end{bmatrix} = \begin{bmatrix} x+4 & 6+x+y \\ -1+z+w & 2w+3 \end{bmatrix}$

If two matrices are equal than corresponding elements are also equal.

Thus, we have

$3x=x+4$

$3x-x=4$

$2x=4$

$x=2$

$3y=6+x+y$

Put the value of x

$3y-y=6+2$

$2y=8$

$y=4$

$3w=2w+3$

$3w-2w=3$

$w=3$

$3z=-1+z+w$

$3z-z=-1+3$

$2z=2$

$z=1$

Hence, we have $x=2,y=4,z=1\, \, and\, \, w=3.$

Question 13: If $F(x) = \begin{bmatrix} \cos x & -\sin x& 0\\\sin x &\cos x & 0 \\ 0 &0&1\end{bmatrix}$, show that $F(x) F(y) = F(x + y)$.

Answer:

$F(x) = \begin{bmatrix} \cos x & -\sin x& 0\\\sin x &\cos x & 0 \\ 0 &0&1\end{bmatrix}$

To prove : $F(x) F(y) = F(x + y)$

$R.H.S : F(x + y)$

$F(x+y) = \begin{bmatrix} \cos (x+y) & -\sin (x+y)& 0\\\sin (x+y) &\cos (x+y) & 0 \\ 0 &0&1\end{bmatrix}$

$L.H.S : F(x) F(y)$

$F(x)F(y) = \begin{bmatrix} \cos x & -\sin x& 0\\\sin x &\cos x & 0 \\ 0 &0&1\end{bmatrix}\times \begin{bmatrix} \cos y & -\sin y& 0\\\sin y &\cos y & 0 \\ 0 &0&1\end{bmatrix}$

$F(x)F(y) = \begin{bmatrix} \cos x \cos y- \sin x\sin y+0 & -\cos x \sin y-\sin x\cos y+0& 0+0+0\\\ sin x\cos y+\cos x \sin y+0 & - \sin x\sin y+\cos x \cos y+0 &0+0+0 \\ 0+0+0 &0+0+0&0+0+1\end{bmatrix}$

$F(x) F(y)= \begin{bmatrix} \cos (x+y) & -\sin (x+y)& 0\\\sin (x+y) &\cos (x+y) & 0 \\ 0 &0&1\end{bmatrix}$

Hence, we have L.H.S. = R.H.S i.e. $F(x) F(y) = F(x + y)$.

Question 14(i): Show that

$\begin{bmatrix}5&-1\\6&7 \end{bmatrix}\begin{bmatrix} 2 & 1\\ 3 & 4 \end{bmatrix}\neq \begin{bmatrix} 2 & 1\\ 3 & 4 \end{bmatrix}\begin{bmatrix}5&-1\\6&7 \end{bmatrix}$

Answer:

To prove:

$\begin{bmatrix}5&-1\\6&7 \end{bmatrix}\begin{bmatrix} 2 & 1\\ 3 & 4 \end{bmatrix}\neq \begin{bmatrix} 2 & 1\\ 3 & 4 \end{bmatrix}\begin{bmatrix}5&-1\\6&7 \end{bmatrix}$

$L.H.S : \begin{bmatrix}5&-1\\6&7 \end{bmatrix}\begin{bmatrix} 2 & 1\\ 3 & 4 \end{bmatrix}$

$= \begin{bmatrix}5\times 2+(-1)\times 3 &5\times 1+(-1)\times 4\\6\times 2+7\times 3&6\times 1+7\times 4 \end{bmatrix}$

$= \begin{bmatrix}7 &1\\33&34 \end{bmatrix}$

$R.H.S : \begin{bmatrix} 2 & 1\\ 3 & 4 \end{bmatrix}\begin{bmatrix}5&-1\\6&7 \end{bmatrix}$

$= \begin{bmatrix} 2\times 5+1\times 6 & 2\times (-1)+1\times 7\\ 3\times 5+4\times 6 & 3\times (-1)+4\times 7 \end{bmatrix}$

$= \begin{bmatrix} 16 & 5\\ 39 & 25 \end{bmatrix}$

Hence, the right-hand side is not equal to the left-hand side, that is

Question 14(ii): Show that

$\begin{bmatrix}1& 2&3\\0&1&0\\1&1&0 \end{bmatrix} \begin{bmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4 \end{bmatrix} \neq \begin{bmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix}1& 2&3\\0&1&0\\1&1&0 \end{bmatrix}$

Answer:

To prove the following multiplication of three by three matrices are not equal

$\begin{bmatrix}1& 2&3\\0&1&0\\1&1&0 \end{bmatrix} \begin{bmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4 \end{bmatrix} \neq \begin{bmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix}1& 2&3\\0&1&0\\1&1&0 \end{bmatrix}$

$L.H.S: \begin{bmatrix}1& 2&3\\0&1&0\\1&1&0 \end{bmatrix} \begin{bmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4 \end{bmatrix}$

$= \begin{bmatrix}1\times(-1)+2\times 0+3\times 2 \, \, \, & 1\times(1)+2\times (-1)+3\times 3\, \, \, &1\times(0)+2\times 1+3\times 4\\0\times(-1)+1\times 0+0\times 2\, \, \, &0\times(1)+1\times (-1)+0\times 3\, \, \, &0\times(0)+1\times 1+0\times 4\\1\times(-1)+1\times 0+0\times 2\, \, \, &1\times(1)+1\times (-1)+0\times 3\, \, \, &1\times(0)+1\times 1+0\times 4 \end{bmatrix}$

$= \begin{bmatrix}5& 8&14\\0&-1&1\\-1&0&1\end{bmatrix}$

$R.H.S : \begin{bmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix}1& 2&3\\0&1&0\\1&1&0 \end{bmatrix}$

$= \begin{bmatrix}-1\times(1)+1\times 0+0\times 1 \, \, \, & -1\times(2)+1\times (1)+0\times 1\, \, \, &-1\times(3)+1\times 0+0\times 0\\0\times(1)+-(1)\times 0+1\times 1\, \, \, &0\times(2)+(-1)\times (1)+1\times 1\, \, \, &0\times(3)+(-1)\times 0+1\times 0\\2\times(1)+3\times 0+4\times 1\, \, \, &2\times(2)+3\times (1)+4\times 1\, \, \, &2\times(3)+3\times 0+4\times 0 \end{bmatrix}$

$= \begin{bmatrix}-1& -1&-3\\1&0&0\\6&11&6\end{bmatrix}$

Hence, $L.H.S \neq R.H.S$ i.e. $\begin{bmatrix}1& 2&3\\0&1&0\\1&1&0 \end{bmatrix} \begin{bmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4 \end{bmatrix} \neq \begin{bmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix}1& 2&3\\0&1&0\\1&1&0 \end{bmatrix}$.

Question 15: Find$A^2 -5A + 6I$, if

$A = \begin{bmatrix} 2 & 0 & 1\\ 2 & 1 &3 \\ 1 & -1 & 0 \end{bmatrix}$

Answer:

$A = \begin{bmatrix} 2 & 0 & 1\\ 2 & 1 &3 \\ 1 & -1 & 0 \end{bmatrix}$

First, we will find out the value of the square of matrix A

$A\times A = \begin{bmatrix} 2 & 0 & 1\\ 2 & 1 &3 \\ 1 & -1 & 0 \end{bmatrix}\times \begin{bmatrix} 2 & 0 & 1\\ 2 & 1 &3 \\ 1 & -1 & 0 \end{bmatrix}$

$A^{2} = \begin{bmatrix} 2\times 2+0\times 2+1\times 1 & 2\times 0+0\times 1+1\times -1 & 2\times 1+0\times 3+1\times 0\\ 2\times 2+1\times 2+3\times 1& 2\times 0+1\times 1+3\times -1 &2\times 1+1\times 3+3\times 0 \\ 1\times 2+(-1)\times 2+0\times 1 & 1\times 0+(-1)\times 1+0\times -1 & 1\times 1+(-1)\times 3+0\times 0 \end{bmatrix}$

$A^{2} = \begin{bmatrix} 5 & -1 & 2\\ 9 & -2 &5 \\ 0 & -1 & -2 \end{bmatrix}$

$I= \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 &0 \\ 0 & 0 & 1 \end{bmatrix}$

$\therefore$ $A^2 -5A + 6I$

$= \begin{bmatrix} 5 & -1 & 2\\ 9 & -2 &5 \\ 0 & -1 & -2 \end{bmatrix}$ $-5 \begin{bmatrix} 2 & 0 & 1\\ 2 & 1 &3 \\ 1 & -1 & 0 \end{bmatrix}$$+6 \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 &0 \\ 0 & 0 & 1 \end{bmatrix}$

$= \begin{bmatrix} 5 & -1 & 2\\ 9 & -2 &5 \\ 0 & -1 & -2 \end{bmatrix}$$- \begin{bmatrix} 10 & 0 & 5\\ 10 & 5 &15 \\ 5 & -5 & 0 \end{bmatrix}$$+\begin{bmatrix} 6 & 0 & 0\\ 0 & 6 &0 \\ 0 & 0 & 6 \end{bmatrix}$

$= \begin{bmatrix} 5-10+6 & -1-0+0 & 2-5+0\\ 9-10+0 & -2-5+6 &5-15+0 \\ 0-5+0 & -1-(-5)+0 & -2-0+6 \end{bmatrix}$

$= \begin{bmatrix} 1 & -1 & -3\\ -1 & -1 &-10 \\ -5 & 4 & 4 \end{bmatrix}$

Question 16: If $A = \begin{bmatrix}1&0&2\\0&2&1\\2&0&3 \end{bmatrix}$ prove that $A^3 - 6A^2 + 7A + 2I = 0$.

Answer:

$A = \begin{bmatrix}1&0&2\\0&2&1\\2&0&3 \end{bmatrix}$

First, find the square of matrix A and then multiply it with A to get the cube of matrix A

$A\times A = \begin{bmatrix}1&0&2\\0&2&1\\2&0&3 \end{bmatrix}$$\times \begin{bmatrix}1&0&2\\0&2&1\\2&0&3 \end{bmatrix}$

$A^{2} = \begin{bmatrix}1+0+4&0+0+0&2+0+6\\0+0+2&0+4+0&0+2+3\\2+0+6&0+0+0&4+0+9 \end{bmatrix}$

$A^{2} = \begin{bmatrix}5&0&8\\2&4&5\\8&0&13 \end{bmatrix}$

$A^{3}=A^{2}\times A$

$A^{2}\times A = \begin{bmatrix}5&0&8\\2&4&5\\8&0&13 \end{bmatrix}$ $\times \begin{bmatrix}1&0&2\\0&2&1\\2&0&3 \end{bmatrix}$

$A^{3} = \begin{bmatrix}5+0+16&0+0+0&10+0+24\\2+0+10&0+8+0&4+4+15\\8+0+26&0+0+0&16+0+39 \end{bmatrix}$

$A^{3} = \begin{bmatrix}21&0&34\\12&8&23\\34&0&55 \end{bmatrix}$

$I= \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 &0 \\ 0 & 0 & 1 \end{bmatrix}$

$\therefore$ $A^3 - 6A^2 + 7A + 2I = 0$

L.H.S :

$\begin{bmatrix}21&0&34\\12&8&23\\34&0&55 \end{bmatrix}$$- 6\begin{bmatrix}5&0&8\\2&4&5\\8&0&13 \end{bmatrix}$$+7 \begin{bmatrix}1&0&2\\0&2&1\\2&0&3 \end{bmatrix}$$+2 \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 &0 \\ 0 & 0 & 1 \end{bmatrix}$

$=\begin{bmatrix}21&0&34\\12&8&23\\34&0&55 \end{bmatrix}$ $- \begin{bmatrix}30&0&48\\12&24&30\\48&0&78 \end{bmatrix}$ $+ \begin{bmatrix}7&0&14\\0&14&7\\14&0&21 \end{bmatrix}$ $+ \begin{bmatrix} 2 & 0 & 0\\ 0 & 2 &0 \\ 0 & 0 & 2 \end{bmatrix}$

$=\begin{bmatrix}21-30+7+2&0-0+0+0&34-48+14+0\\12-12+0+0&8-24+14+2&23-30+7+0\\34-48+14+0&0-0+0+0&55-78+21+2 \end{bmatrix}$

$=\begin{bmatrix}30-30&0&48-48\\12-12&24-24&30-30\\48-48&0&78-78 \end{bmatrix}$

$= \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 &0 \\ 0 & 0 & 0 \end{bmatrix}=0$

Hence, L.H.S = R.H.S

i.e.$A^3 - 6A^2 + 7A + 2I = 0$.

Question 17: If $A = \begin{bmatrix}3 &-2\\4&-2 \end{bmatrix}$ and $I = \begin{bmatrix}1 &0\\0&1 \end{bmatrix}$, find k so that $A^{2} = kA - 2I$.

Answer:

$A = \begin{bmatrix}3 &-2\\4&-2 \end{bmatrix}$

$I = \begin{bmatrix}1 &0\\0&1 \end{bmatrix}$

$A \times A= \begin{bmatrix}3 &-2\\4&-2 \end{bmatrix}$$\times \begin{bmatrix}3 &-2\\4&-2 \end{bmatrix}$

$A^{2} = \begin{bmatrix}9-8 &-6+4\\12-8&-8+4 \end{bmatrix}$

$A^{2} = \begin{bmatrix}1 &-2\\4&-4 \end{bmatrix}$

$A^{2} = kA - 2I$

$\begin{bmatrix}1 &-2\\4&-4 \end{bmatrix}=$$k\begin{bmatrix}3 &-2\\4&-2 \end{bmatrix} -$$2 \begin{bmatrix}1 &0\\0&1 \end{bmatrix}$

$\begin{bmatrix}1 &-2\\4&-4 \end{bmatrix}=$$k\begin{bmatrix}3 &-2\\4&-2 \end{bmatrix} -$$\begin{bmatrix}2 &0\\0&2 \end{bmatrix}$

$\begin{bmatrix}1 &-2\\4&-4 \end{bmatrix}+$ $\begin{bmatrix}2 &0\\0&2 \end{bmatrix}$ $=k\begin{bmatrix}3 &-2\\4&-2 \end{bmatrix}$

$\begin{bmatrix}1+2 &-2+0\\4+0&-4+2 \end{bmatrix}$$=\begin{bmatrix}3k&-2k\\4k&-2k \end{bmatrix}$

$\begin{bmatrix}3 &-2\\4&-2 \end{bmatrix}$ $=\begin{bmatrix}3k&-2k\\4k&-2k \end{bmatrix}$

We have,$3=3k$

$k=\frac{3}{3}=1$

Hence, the value of k is 1.

Question 18: If $A = \begin{bmatrix} 0&-\tan\frac{\alpha}{2}\\\tan\frac{\alpha}{2}& 0\end{bmatrix}$ and I is the identity matrix of order 2, show that$I + A = (I- A)\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$

Answer:

$A = \begin{bmatrix} 0&-\tan\frac{\alpha}{2}\\\tan\frac{\alpha}{2}& 0\end{bmatrix}$

$I = \begin{bmatrix}1 &0\\0&1 \end{bmatrix}$

To prove : $I + A = (I- A)\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$

L.H.S : $I+A$

$I+A = \begin{bmatrix}1 &0\\0&1 \end{bmatrix}$$+ \begin{bmatrix} 0&-\tan\frac{\alpha}{2}\\\tan\frac{\alpha}{2}& 0\end{bmatrix}$

$I+A = \begin{bmatrix} 1+0&0-\tan\frac{\alpha}{2}\\0+\tan\frac{\alpha}{2}&1+ 0\end{bmatrix}$

$I+A = \begin{bmatrix} 1&-\tan\frac{\alpha}{2}\\\tan\frac{\alpha}{2}&1\end{bmatrix}$

R.H.S : $(I- A)\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$

$(I- A)\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$$= (\begin{bmatrix}1 &0\\0&1 \end{bmatrix}-$ $\begin{bmatrix} 0&-\tan\frac{\alpha}{2}\\\tan\frac{\alpha}{2}& 0\end{bmatrix})$$\times \begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$

$(I- A)\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$ $=\begin{bmatrix} 1-0&0-(-\tan\frac{\alpha}{2})\\0-\tan\frac{\alpha}{2}&1- 0\end{bmatrix}$ $\times \begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$

$(I- A)\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$$=\begin{bmatrix} 1&\tan\frac{\alpha}{2}\\-\tan\frac{\alpha}{2}&1\end{bmatrix}$ $\times \begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$

$=\begin{bmatrix} \cos\alpha + \sin \alpha\tan\frac{\alpha}{2} &- \sin \alpha+ \cos \alpha \tan\frac{\alpha}{2}\\-\tan\frac{\alpha}{2} \cos\alpha + \sin \alpha &\tan\frac{\alpha}{2} \sin\alpha + \cos \alpha \end{bmatrix}$

$=\begin{bmatrix} 1-2\sin^{2}\frac{\alpha }{2} + 2\sin\frac{\alpha }{2} \ cos \frac{\alpha }{2}\tan\frac{\alpha}{2} &- 2\sin\frac{\alpha }{2} \ cos \frac{\alpha }{2}+ (2\cos^{2} \frac{\alpha }{2} -1)\tan\frac{\alpha}{2}\\-\tan\frac{\alpha}{2} (2\cos^{2} \frac{\alpha }{2} -1) + 2\sin\frac{\alpha }{2} \ cos \frac{\alpha }{2} &\tan\frac{\alpha}{2} 2\sin\frac{\alpha } {2} \ cos \frac{\alpha }{2} + 1-2\sin^{2}\frac{\alpha }{2} \end{bmatrix}$

$=\begin{bmatrix} 1-2\sin^{2}\frac{\alpha }{2} + 2\sin^{2}\frac{\alpha }{2} &- 2\sin\frac{\alpha }{2} \ cos \frac{\alpha }{2}+ 2\sin\frac{\alpha }{2} \ cos \frac{\alpha }{2} -\tan\frac{\alpha}{2}\\-2\sin\frac{\alpha }{2} \ cos \frac{\alpha }{2}+\tan\frac{\alpha}{2} + 2\sin\frac{\alpha }{2} \ cos \frac{\alpha }{2} & 2\sin^{2}\frac{\alpha } {2} + 1-2\sin^{2}\frac{\alpha }{2} \end{bmatrix}$

$= \begin{bmatrix} 1&-\tan\frac{\alpha}{2}\\\tan\frac{\alpha}{2}&1\end{bmatrix}$

Hence, we can see L.H.S = R.H.S

i.e. $I + A = (I- A)\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$.

Question 19(i): A trust fund has Rs. 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide Rs 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of:

Rs. 1800

Answer:

Let Rs. x be invested in the first bond.

Money invested in second bond = Rs (3000-x)

The first bond pays 5% interest per year and the second bond pays 7% interest per year.

To obtain an annual total interest of Rs. 1800, we have

$\begin{bmatrix}x &(30000-x) \end{bmatrix}$ $\begin{bmatrix} \frac{5}{100} \\ \frac{7}{100} \end{bmatrix}$ $=1800$ (simple interest for 1 year $=\frac{pricipal\times rate}{100}$ )

$\frac{5}{100}x+\frac{7}{100}(30000-x) = 1800$

$5x+210000-7x=180000$

$210000-180000=7x-5x$

$30000=2x$

$x=15000$

Thus, to obtain an annual total interest of Rs. 1800, the trust fund should invest Rs 15000 in the first bond and Rs 15000 in the second bond.

Question 19(ii): A trust fund has Rs. 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide Rs 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of: Rs. 2000

Answer:

Let Rs. x be invested in the first bond.

Money invested in second bond = Rs (3000-x)

The first bond pays 5% interest per year, and the second bond pays 7% interest per year.

To obtain an annual total interest of Rs. 1800, we have

$\begin{bmatrix} x & (30000 - x) \end{bmatrix} \begin{bmatrix} \frac{5}{100} \\ \frac{7}{100} \end{bmatrix} = 2000$

(Simple interest for 1 year = $\frac{\text{Principal} \times \text{Rate}}{100}$)

$\frac{5}{100}x + \frac{7}{100}(30000 - x) = 2000$

$\frac{5x + 210000 - 7x}{100} = 2000$

$\frac{210000 - 2x}{100} = 2000$

$210000 - 2x = 200000$

$210000 - 200000 = 2x$

$10000 = 2x$

$x = 5000$

Thus, to obtain an annual total interest of Rs. 2000, the trust fund should invest Rs 5000 in the first bond and Rs 25000 in the second bond.

Question 20: The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are Rs 80, Rs 60 and Rs 40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.

Answer:

The bookshop has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books.

Their selling prices are Rs 80, Rs 60 and Rs 40 each, respectively.

The total amount the bookshop will receive from selling all the books:

$12$$\begin{bmatrix}10 &8&10 \end{bmatrix}$ $\begin{bmatrix}80\\60\\40 \end{bmatrix}$

$=12(10\times 80+8\times 60+10\times 40)$

$= 12(800+480+ 400)$

$= 12(1680)$

$=20160$

The total amount the bookshop will receive from selling all the books is 20160.

Question 21: Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k, respectively. Choose the correct answer in Exercises 21 and 22.

Q21. The restrictions on $n, k$ and $p$ so that $P Y+W Y$ will be defined are:
(A) $k=3, p=n$
(B) k is arbitrary, $p=2$
(C) p is arbitrary, $k=3$
(D) $k=2, p=3$

Answer:
P and Y are of order $p \times k$ and $\mathbf{3} \times k$ respectively.
Therefore, $P Y$ will be defined only if $k=3$, i.e., the order of $P Y$ is $p \times k$.
W and Y are of order $n \times 3$ and $3 \times k$ respectively.
Therefore, $W Y$ is defined because the number of columns of $W$ is equal to the number of rows of $Y$, which is 3, i.e., the order of $W Y$ is $n \times k$.

Matrices $P Y$ and $W Y$ can only be added if they both have the same order, i.e., $p \times k=n \times k \Rightarrow p=n$.

Therefore, $k=3, p=n$ are restrictions on $n, k$, and $p$ so that $P Y+W Y$ will be defined.
Option (A) is correct.

Question 22: Assume X, Y, Z, W, and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k,
respectively. Choose the correct answer in Exercises 21 and 22. If n = p, then the order of the matrix $7X - 5Z$ is:
(A) p × 2
(B) 2 × n
(C) n × 3
(D) p × n

Answer:

$X$ has order $2 \times n$.

Therefore, $7X$ also has order $2 \times n$.

$Z$ has order $2 \times p$.

Therefore, $5Z$ also has order $2 \times p$.

Matrices $7X$ and $5Z$ can only be subtracted if they both have the same order, i.e., $2 \times n = 2 \times p$, and it is given that $p = n$.

We can say that both matrices have order $2 \times n$.

Therefore, the order of $7X - 5Z$ is $2 \times n$.

Option (B) is correct.

Question 1(i). Find the transpose of each of the following matrices:

$\begin{bmatrix} 5\\ \frac{1}{2} \\-1 \end{bmatrix}$

Answer:

$A=\begin{bmatrix} 5\\ \frac{1}{2} \\-1 \end{bmatrix}$

The transpose of the given matrix is

$A^{T}=\begin{bmatrix} 5& \frac{1}{2} &-1 \end{bmatrix}$

Question 1(ii). Find the transpose of each of the following matrices:

$\begin{bmatrix} 1 & -1\\ 2 & 3 \end{bmatrix}$

Answer:

$A=\begin{bmatrix} 1 & -1\\ 2 & 3 \end{bmatrix}$

Interchanging the rows and columns of the matrix A, we get

$A^{T}=\begin{bmatrix} 1 & 2\\ -1 & 3 \end{bmatrix}$

Question 1(iii) Find the transpose of each of the following matrices:

$\begin{bmatrix} -1 & 5 & 6\\ \sqrt3& 5 &6 \\ 2 &3 &-1 \end{bmatrix}$

Answer:

$A = \begin{bmatrix} -1 & 5 & 6\\ \sqrt3& 5 &6 \\ 2 &3 &-1 \end{bmatrix}$

Transpose is obtained by interchanging the rows and columns of a matrix

$A^{T} = \begin{bmatrix} -1 & \sqrt3 & 2\\ 5& 5 &3 \\ 6 &6 &-1 \end{bmatrix}$

Question 2(i). If $A = \begin{bmatrix} -1 & 2 & 3\\ 5 &7 &9 \\ -2 & 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} -4 & 1 & -5\\ 1 &2 &0 \\ 1 & 3 & 1 \end{bmatrix}$, then verify

$(A + B)' = A' + B'$

Answer:

$A = \begin{bmatrix} -1 & 2 & 3\\ 5 &7 &9 \\ -2 & 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} -4 & 1 & -5\\ 1 &2 &0 \\ 1 & 3 & 1 \end{bmatrix}$

$(A + B)' = A' + B'$

L.H.S : $(A + B)'$

$A+B = \begin{bmatrix} -1 & 2 & 3\\ 5 &7 &9 \\ -2 & 1 & 1 \end{bmatrix}$ $+ \begin{bmatrix} -4 & 1 & -5\\ 1 &2 &0 \\ 1 & 3 & 1 \end{bmatrix}$

$A+B = \begin{bmatrix} -1+(-4) & 2+1 & 3+(-5)\\ 5+1 &7+2 &9+0 \\ -2+1 & 1+3 & 1+1 \end{bmatrix}$

$A+B = \begin{bmatrix} -5 & 3 & -2\\ 6 &9 &9 \\ -1 & 4 & 2 \end{bmatrix}$

$(A+B)' = \begin{bmatrix} -5 & 6 & -1\\ 3 &9 &4 \\ -2 & 9 & 2 \end{bmatrix}$

R.H.S : $A' + B'$

$A'+B' = \begin{bmatrix} -1 & 5 & -2\\ 2 &7 &1 \\ 3 & 9 & 1 \end{bmatrix}$ $+ \begin{bmatrix} -4 & 1 & 1\\ 1 &2 &3\\ -5 & 0 & 1 \end{bmatrix}$

$A'+B' = \begin{bmatrix} -1+(-4) & 5+1 & -2+1\\ 2+1 &7+2 &1+3 \\ 3+(-5) & 9+0 & 1+1 \end{bmatrix}$

$A'+B' = \begin{bmatrix} -5 & 6 & -1\\ 3 &9 &4 \\ -2 & 9 & 2 \end{bmatrix}$

Thus, we find that the LHS is equal to the RHS and hence verified.

Question 2(ii). If $A = \begin{bmatrix} -1 & 2 & 3\\ 5 &7 &9 \\ -2 & 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} -4 & 1 & -5\\ 1 &2 &0 \\ 1 & 3 & 1 \end{bmatrix}$, then verify

$(A - B)' = A' - B'$

Answer:

$A = \begin{bmatrix} -1 & 2 & 3\\ 5 &7 &9 \\ -2 & 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} -4 & 1 & -5\\ 1 &2 &0 \\ 1 & 3 & 1 \end{bmatrix}$

$(A - B)' = A' - B'$

L.H.S : $(A - B)'$

$A-B = \begin{bmatrix} -1 & 2 & 3\\ 5 &7 &9 \\ -2 & 1 & 1 \end{bmatrix}$ $- \begin{bmatrix} -4 & 1 & -5\\ 1 &2 &0 \\ 1 & 3 & 1 \end{bmatrix}$

$A-B = \begin{bmatrix} -1-(-4) & 2-1 & 3-(-5)\\ 5-1 &7-2 &9-0 \\ -2-1 & 1-3 & 1-1 \end{bmatrix}$

$A-B = \begin{bmatrix} 3 & 1 & 8\\ 4 &5 &9 \\ -3 & -2& 0 \end{bmatrix}$

$(A-B)' = \begin{bmatrix} 3 & 4 & -3\\ 1 &5 &-2 \\ 8 & 9& 0 \end{bmatrix}$

R.H.S : $A' - B'$

$A'-B' = \begin{bmatrix} -1 & 5 & -2\\ 2 &7 &1 \\ 3 & 9 & 1 \end{bmatrix}$ $- \begin{bmatrix} -4 & 1 & 1\\ 1 &2 &3\\ -5 & 0 & 1 \end{bmatrix}$

$A'-B' = \begin{bmatrix} -1-(-4) & 5-1 & -2-1\\ 2-1 &7-2 &1-3 \\ 3-(-5) & 9-0 & 1-1 \end{bmatrix}$

$A'-B' = \begin{bmatrix} 3 & 4 & -3\\ 1 &5 &-2 \\ 8 & 9& 0 \end{bmatrix}$

Hence, L.H.S = R.H.S. so verified that

$(A - B)' = A' - B'$.

Question 3(i). If $A' = \begin{bmatrix} 3 & 4\\ -1 &2 \\ 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} -1 & 2 & 1\\ 1 &2 &3 \end{bmatrix}$, then verify

$(A + B)' = A' + B'$

Answer:

$A' = \begin{bmatrix} 3 & 4\\ -1 &2 \\ 0 & 1 \end{bmatrix}$ $B = \begin{bmatrix} -1 & 2 & 1\\ 1 &2 &3 \end{bmatrix}$

$A=(A')' = \begin{bmatrix} 3 & -1&0\\ 4 &2 & 1 \end{bmatrix}$

To prove: $(A + B)' = A' + B'$

$L.H.S : (A + B)' =$

$A+B = \begin{bmatrix} 3 & -1&0\\ 4 &2 & 1 \end{bmatrix}$ $+ \begin{bmatrix} -1 & 2 & 1\\ 1 &2 &3 \end{bmatrix}$

$A+B = \begin{bmatrix} 3+(-1) & -1+(-1)&0+1\\ 4+1 &2+2 & 1+3 \end{bmatrix}$

$A+B = \begin{bmatrix} 2 & -2&1\\ 5 &4 & 4 \end{bmatrix}$

$\therefore \, \, \, (A+B)' = \begin{bmatrix} 2 & 5\\ 1 &4\\1 & 4 \end{bmatrix}$

R.H.S: $A' + B'$

$A'+B' = \begin{bmatrix} 3 & 4\\ -1 &2 \\ 0 & 1 \end{bmatrix}$ $+ \begin{bmatrix} -1 & 1\\ 2 &2 \\ 1 & 3 \end{bmatrix}$

$A'+B' = \begin{bmatrix} 2 & 5\\ 1 &4 \\ 1 & 4 \end{bmatrix}$

Hence, L.H.S = R.H.S i.e. $(A + B)' = A' + B'$.

Question 3(ii). If $A = \begin{bmatrix} 3 & 4\\ -1 &2 \\ 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} -1 & 2 & 1\\ 1 &2 &3 \end{bmatrix}$, then verify

$(A - B)' = A' - B'$

Answer:

$A' = \begin{bmatrix} 3 & 4\\ -1 &2 \\ 0 & 1 \end{bmatrix}$ $B = \begin{bmatrix} -1 & 2 & 1\\ 1 &2 &3 \end{bmatrix}$

$A=(A')' = \begin{bmatrix} 3 & -1&0\\ 4 &2 & 1 \end{bmatrix}$

To prove: $(A - B)' = A' - B'$

$L.H.S : (A - B)' =$

$A-B = \begin{bmatrix} 3 & -1&0\\ 4 &2 & 1 \end{bmatrix}$ $- \begin{bmatrix} -1 & 2 & 1\\ 1 &2 &3 \end{bmatrix}$

$A-B = \begin{bmatrix} 3-(-1) & -1-(2)&0-1\\ 4-1 &2-2 & 1-3 \end{bmatrix}$

$A-B = \begin{bmatrix} 4 & -3&-1\\ 3 &0 & -2 \end{bmatrix}$

$\therefore \, \, \, (A-B)' = \begin{bmatrix} 4 & 3\\ -3 &0\\-1 & -2 \end{bmatrix}$

R.H.S: $A' - B'$

$A'-B' = \begin{bmatrix} 3 & 4\\ -1 &2 \\ 0 & 1 \end{bmatrix}$ $- \begin{bmatrix} -1 & 1\\ 2 &2 \\ 1 & 3 \end{bmatrix}$

$A'-B' = \begin{bmatrix} 4 & 3\\ -3 &0 \\ -1 & -2 \end{bmatrix}$

Hence, L.H.S = R.H.S i.e. $(A - B)' = A' - B'$.

Question 4. If $A' = \begin{bmatrix} -2 & 3\\ 1 & 2 \end{bmatrix}$ and $B= \begin{bmatrix} -1 & 0\\ 1 & 2 \end{bmatrix}$, then find $(A + 2B)'$

Answer:

$B= \begin{bmatrix} -1 & 0\\ 1 & 2 \end{bmatrix}$

$A' = \begin{bmatrix} -2 & 3\\ 1 & 2 \end{bmatrix}$

$A=(A')' = \begin{bmatrix} -2 & 1\\ 3 & 2 \end{bmatrix}$

$(A + 2B)'$ :

$A+2B = \begin{bmatrix} -2 & 1\\ 3 & 2 \end{bmatrix}$$+2 \begin{bmatrix} -1 & 0\\ 1 & 2 \end{bmatrix}$

$A+2B = \begin{bmatrix} -2 & 1\\ 3 & 2 \end{bmatrix}$$+ \begin{bmatrix} -2 & 0\\ 2 & 4 \end{bmatrix}$

$A+2B = \begin{bmatrix} -2+(-2) & 1+0\\ 3+2 & 2+4 \end{bmatrix}$

$A+2B = \begin{bmatrix} -4 & 1\\ 5 & 6 \end{bmatrix}$

Transpose is obtained by interchanging rows and columns and the transpose of A+2B is

$(A+2B)' = \begin{bmatrix} -4 & 5\\ 1 & 6 \end{bmatrix}$

Question 5(i) For the matrices A and B, verify that $(AB)' = B'A'$, where

$A = \begin{bmatrix} 1\\-4 \\3 \end{bmatrix}$, $B = \begin{bmatrix} -1& 2 &1 \end{bmatrix}$

Answer:

$A = \begin{bmatrix} 1\\-4 \\3 \end{bmatrix}$, $B = \begin{bmatrix} -1& 2 &1 \end{bmatrix}$

To prove : $(AB)' = B'A'$

$L.H.S : (AB)'$

$AB = \begin{bmatrix} 1\\-4 \\3 \end{bmatrix}$$\begin{bmatrix} -1& 2 &1 \end{bmatrix}$

$AB = \begin{bmatrix} -1&2&1\\4&-8&-4 \\-3 &6&3\end{bmatrix}$

$(AB)' = \begin{bmatrix} -1&4&-3\\2&-8&6 \\1 &-4&3\end{bmatrix}$

$R.H.S : B'A'$

$B' = \begin{bmatrix} -1\\2 \\1 \end{bmatrix}$

$A' = \begin{bmatrix} 1& -4 &3 \end{bmatrix}$

$B'A' = \begin{bmatrix} -1\\2 \\1 \end{bmatrix}$$\begin{bmatrix} 1& -4 &3 \end{bmatrix}$

$B'A' = \begin{bmatrix} -1&4&-3\\2&-8&6 \\1&-4&3 \end{bmatrix}$

Hence, L.H.S =R.H.S

so it is verified that $(AB)' = B'A'$.

Question 5(ii) For the matrices A and B, verify that $(AB)' = B'A'$, where

$A = \begin{bmatrix} 0\\ 1\\ 2 \end{bmatrix}$, $B = \begin{bmatrix} 1 & 5&7 \end{bmatrix}$

Answer:

$A = \begin{bmatrix} 0\\ 1\\ 2 \end{bmatrix}$, $B = \begin{bmatrix} 1 & 5&7 \end{bmatrix}$

To prove : $(AB)' = B'A'$

$L.H.S : (AB)'$

$AB = \begin{bmatrix} 0\\1 \\2 \end{bmatrix}$$\begin{bmatrix} 1& 5 &7 \end{bmatrix}$

$AB = \begin{bmatrix} 0&0&0\\1&5&7 \\2 &10&14\end{bmatrix}$

$(AB)' = \begin{bmatrix} 0&1&2\\0&5&10 \\0 &7&14\end{bmatrix}$

$R.H.S : B'A'$

$B' = \begin{bmatrix} 1\\5 \\7 \end{bmatrix}$

$A' = \begin{bmatrix} 0& 1 &2 \end{bmatrix}$

$B'A' = \begin{bmatrix} 1\\5 \\7 \end{bmatrix}$$\begin{bmatrix} 0& 1 &2 \end{bmatrix}$

$B'A' = \begin{bmatrix} 0&1&2\\0&5&10 \\0&7&14 \end{bmatrix}$

Hence, L.H.S =R.H.S i.e.$(AB)' = B'A'$.

Question 6(i). If $A = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}$, then verify that $A'A =I$

Answer:

$A = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}$

By interchanging rows and columns, we get the transpose of A

$A' = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}$

To prove: $A'A =I$

L.H.S :$A'A$

$A'A = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}$ $\begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}$

$A'A = \begin{bmatrix} \cos^{2}\alpha + \sin ^{2}\alpha & \sin\alpha \cos \alpha - \sin \alpha \ cos \alpha \\ \sin\alpha \cos \alpha - \sin \alpha \cos \alpha & \ sin^{2}\alpha +\cos^{2}\alpha \end{bmatrix}$

$A'A = \begin{bmatrix} 1& 0 \\ 0& 1 \end{bmatrix}=I=R.H.S$

Question 6(ii). If $A = \begin{bmatrix} \sin\alpha & \cos\alpha \\ -\cos\alpha & \sin\alpha \end{bmatrix}$, then verify that $A'A = I$

Answer:

$A = \begin{bmatrix} \sin\alpha & \cos\alpha \\ -\cos\alpha & \sin\alpha \end{bmatrix}$

By interchanging columns and rows of the matrix A we get the transpose of A

$A' = \begin{bmatrix} \sin\alpha & -\cos\alpha \\ \cos\alpha & \sin\alpha \end{bmatrix}$

To prove: $A'A =I$

L.H.S :$A'A$

$A'A = \begin{bmatrix} \sin\alpha & -\cos\alpha \\ \cos\alpha & \sin\alpha \end{bmatrix}$ $\begin{bmatrix} \sin\alpha & \cos\alpha \\ -\cos\alpha & \sin\alpha \end{bmatrix}$

$A'A = \begin{bmatrix} \cos^{2}\alpha + \sin ^{2}\alpha & \sin\alpha \cos \alpha - \sin \alpha \ cos \alpha \\ \sin\alpha \cos \alpha - \sin \alpha \cos \alpha & \ sin^{2}\alpha +\cos^{2}\alpha \end{bmatrix}$

$A'A = \begin{bmatrix} 1& 0 \\ 0& 1 \end{bmatrix}=I=R.H.S$

Question 7(i). Show that the matrix $A = \begin{bmatrix} 1 &- 1& 5\\ -1 & 2 & 1\\ 5 & 1 & 3 \end{bmatrix}$ is a symmetric matrix.

Answer:

$A = \begin{bmatrix} 1 &- 1& 5\\ -1 & 2 & 1\\ 5 & 1 & 3 \end{bmatrix}$

The transpose of A is

$A' = \begin{bmatrix} 1 &- 1& 5\\ -1 & 2 & 1\\ 5 & 1 & 3 \end{bmatrix}$

Since $ A'' = A$, so given matrix is a symmetric matrix.

Question 7(ii) Show that the matrix $A = \begin{bmatrix} 0 & 1 & -1\\ -1 & 0 &1 \\ 1 & -1 &0 \end{bmatrix}$ is a skew-symmetric matrix.

Answer:

$A = \begin{bmatrix} 0 & 1 & -1\\ -1 & 0 &1 \\ 1 & -1 &0 \end{bmatrix}$

The transpose of A is

$A' = \begin{bmatrix} 0 & -1 & 1\\ 1 & 0 &-1 \\- 1 & 1 &0 \end{bmatrix}$

$A' =- \begin{bmatrix} 0 & 1 & -1\\ -1 & 0 &1 \\ 1 & -1 &0 \end{bmatrix}$

$A' =- A$

Since $ A'=-A$ so given matrix is a skew-symmetric matrix.

Question 8(i). For the matrix $A = \begin{bmatrix} 1 & 5\\ 6 & 7 \end{bmatrix}$, verify that

$(A + A')$ is a symmetric matrix.

Answer:

$A = \begin{bmatrix} 1 & 5\\ 6 & 7 \end{bmatrix}$

$A' = \begin{bmatrix} 1 & 6\\ 5 & 7 \end{bmatrix}$

$A + A'= \begin{bmatrix} 1 & 5\\ 6 & 7 \end{bmatrix}$ $+ \begin{bmatrix} 1 & 6\\ 5 & 7 \end{bmatrix}$

$A + A'= \begin{bmatrix} 1+1 & 5+6\\ 6+5 & 7+7 \end{bmatrix}$

$A + A'= \begin{bmatrix}2 & 11\\ 11& 14 \end{bmatrix}$

$(A + A')'= \begin{bmatrix}2 & 11\\ 11& 14 \end{bmatrix}$

We have $A+A'=(A + A')'$

Hence, $(A + A')$ is a symmetric matrix.

Question 8(ii) For the matrix $A = \begin{bmatrix} 1 & 5\\ 6 & 7 \end{bmatrix}$, verify that

$(A - A')$ is a skew symmetric matrix.

Answer:

$A = \begin{bmatrix} 1 & 5\\ 6 & 7 \end{bmatrix}$

$A' = \begin{bmatrix} 1 & 6\\ 5 & 7 \end{bmatrix}$

$A - A'= \begin{bmatrix} 1 & 5\\ 6 & 7 \end{bmatrix}$ $- \begin{bmatrix} 1 & 6\\ 5 & 7 \end{bmatrix}$

$A - A'= \begin{bmatrix} 1-1 & 5-6\\ 6-5 & 7-7 \end{bmatrix}$

$A - A'= \begin{bmatrix}0 & -1\\ 1& 0 \end{bmatrix}$

$(A - A')'= \begin{bmatrix}0 & 1\\ -1& 0 \end{bmatrix}=-(A-A')$

We have $A-A'=-(A - A')'$

Hence, $(A-A')$ is a skew-symmetric matrix.

Question 9. Find $\frac{1}{2}(A+A')$ and $\frac{1}{2}(A-A')$, when $A = \begin{bmatrix} 0 & a & b\\ -a & 0 & c\\ -b & -c & 0 \end{bmatrix}$

Answer:

$A = \begin{bmatrix} 0 & a & b\\ -a & 0 & c\\ -b & -c & 0 \end{bmatrix}$

The transpose of the matrix is obtained by interchanging rows and columns

$A' = \begin{bmatrix} 0 & -a & -b\\ a & 0 & -c\\ b & c & 0 \end{bmatrix}$

$\frac{1}{2}(A+A') = \frac{1}{2}(\begin{bmatrix} 0 & a & b\\ -a & 0 & c\\ -b & -c & 0 \end{bmatrix}$ $+\begin{bmatrix} 0 & -a & -b\\ a & 0 & -c\\ b & c & 0 \end{bmatrix})$

$\frac{1}{2}(A+A') = \frac{1}{2}(\begin{bmatrix} 0+0 & a+(-a) & b+(-b)\\ -a+a & 0+0 & c+(-c)\\ -b+b & -c+c & 0+0 \end{bmatrix})$

$\frac{1}{2}(A+A') = \frac{1}{2}\begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}$

$\frac{1}{2}(A+A') = \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}$

$\frac{1}{2}(A+A') = 0$

$\frac{1}{2}(A-A') = \frac{1}{2}(\begin{bmatrix} 0 & a & b\\ -a & 0 & c\\ -b & -c & 0 \end{bmatrix}$$- \begin{bmatrix} 0 & -a & -b\\ a & 0 & -c\\ b & c & 0 \end{bmatrix})$

$\frac{1}{2}(A-A') = \frac{1}{2}(\begin{bmatrix} 0-0 & a-(-a) & b-(-b)\\ -a-a & 0-0 & c-(-c)\\ -b-b & -c-c & 0-0 \end{bmatrix})$

$\frac{1}{2}(A-A') = \frac{1}{2}\begin{bmatrix} 0 & 2a &2 b\\ -2a & 0 & 2c\\ -2b & -2c & 0 \end{bmatrix}$

$\frac{1}{2}(A-A') = \begin{bmatrix} 0 & a & b\\ -a & 0 & c\\ -b & -c & 0 \end{bmatrix}$

Question 10(i). Express the following matrices as the sum of a symmetric and a skew-symmetric matrix:

$\begin{bmatrix} 3 & 5\\ 1 & -1 \end{bmatrix}$

Answer:

$A=\begin{bmatrix} 3 & 5\\ 1 & -1 \end{bmatrix}$

$A'=\begin{bmatrix} 3 & 1\\ 5 & -1 \end{bmatrix}$

$A+A'=\begin{bmatrix} 3 & 5\\ 1 & -1 \end{bmatrix}$$+\begin{bmatrix} 3 & 1\\ 5 & -1 \end{bmatrix}$

$A+A'=\begin{bmatrix} 6 & 6\\ 6 & -2 \end{bmatrix}$

Let

$B=\frac{1}{2}(A+A')=\frac{1}{2}\begin{bmatrix} 6 & 6\\ 6 & -2 \end{bmatrix}$$=\begin{bmatrix} 3 & 3\\ 3 & -1 \end{bmatrix}$

$B'=\begin{bmatrix} 3 & 3\\ 3 & -1 \end{bmatrix}=B$

Thus, $\frac{1}{2}(A+A')$ is a symmetric matrix.

$A-A'=\begin{bmatrix} 3 & 5\\ 1 & -1 \end{bmatrix}$$-\begin{bmatrix} 3 & 1\\ 5 & -1 \end{bmatrix}$

$A-A'=\begin{bmatrix} 0 & 4\\ -4 & 0 \end{bmatrix}$

Let

$C= \frac{1}{2}(A-A')=\frac{1}{2}\begin{bmatrix} 0 & 4\\ -4 & 0 \end{bmatrix}$$= \begin{bmatrix} 0 & 2\\ -2 & 0 \end{bmatrix}$

$C'= \begin{bmatrix} 0 & -2\\ 2 & 0 \end{bmatrix}$

$C=-C'$

Thus, $\frac{1}{2}(A-A')$ is a skew symmetric matrix.

Represent A as the sum of B and C.

$B+C = \begin{bmatrix} 3 & 3\\ 3 & -1 \end{bmatrix}$ $+ \begin{bmatrix} 0 & 2\\ -2 & 0 \end{bmatrix}$ $= \begin{bmatrix} 3 & 5\\ 1 & -1\end{bmatrix}=A$

Question:10(ii). Express the following matrices as the sum of a symmetric and a skew-symmetric matrix:

$\begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}$

Answer:

$A=\begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}$

$A'=\begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}$

$A+A'=\begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}$$+ \begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}$

$A+A'=\begin{bmatrix} 12 & -4 & 4\\ -4 & 6 & -2\\ 4 & -2 & 6 \end{bmatrix}$

Let

$B= \frac{1}{2}(A+A')=\frac{1}{2}\begin{bmatrix} 12 & -4 & 4\\ -4 & 6 & -2\\ 4 & -2 & 6 \end{bmatrix}$$= \begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}$

$B'= \begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}=B$

Thus, $\frac{1}{2}(A+A')$ is a symmetric matrix.

$A-A'=\begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}$$- \begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}$

$A-A'=\begin{bmatrix} 0 & 0&0\\ 0 & 0&0 \\0&0&0\end{bmatrix}$

Let

$C= \frac{1}{2}(A-A')=\frac{1}{2}\begin{bmatrix} 0 & 0&0\\ 0&0 & 0\\0&0&0 \end{bmatrix}$$=\begin{bmatrix} 0 & 0&0\\ 0&0 & 0\\0&0&0 \end{bmatrix}$

$C'=\begin{bmatrix} 0 & 0&0\\ 0&0 & 0\\0&0&0 \end{bmatrix}$

$C=-C'$

Thus, $\frac{1}{2}(A-A')$ is a skew-symmetric matrix.

Represent A as the sum of B and C.

$B+C= \begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}$ $+\begin{bmatrix} 0 & 0&0\\ 0&0 & 0\\0&0&0 \end{bmatrix}$ $= \begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}=A$

Question 10(iii). Express the following matrices as the sum of a symmetric and a skew-symmetric matrix:

$\begin{bmatrix} 3 & 3 & -1\\ -2 & -2 & 1\\ -4 & -5 & 2 \end{bmatrix}$

Answer:

$A=\begin{bmatrix} 3 & 3 & -1\\ -2 & -2 & 1\\ -4 & -5 & 2 \end{bmatrix}$

$A'=\begin{bmatrix} 3 & -2 & -4\\ 3 & -2 & -5\\ -1 & 1 & 2 \end{bmatrix}$

$A+A'=\begin{bmatrix} 3 & 3 & -1\\ -2 & -2 & 1\\ -4 & -5 & 2 \end{bmatrix}$$+\begin{bmatrix} 3 & -2 & -4\\ 3 & -2 & -5\\ -1 & 1 & 2 \end{bmatrix}$

$A+A'=\begin{bmatrix} 6 & 1 & -5\\ 1& -4 & -4\\ -5 & -4 & 4 \end{bmatrix}$

Let

$B= \frac{1}{2}(A+A')=\frac{1}{2}\begin{bmatrix} 6 & 1 & -5\\ 1 & -4 & -4\\ -5 & -4 & 4 \end{bmatrix}$$= \begin{bmatrix} 3 & \frac{1}{2} & -\frac{5}{2}\\ \frac{1}{2} & -2 & -2\\ \frac{-5}{2} & -2 & 2 \end{bmatrix}$

$B'= \begin{bmatrix} 3 & \frac{1}{2} & -\frac{5}{2}\\ \frac{1}{2} & -2 & -2\\ \frac{-5}{2} & -2 & 2 \end{bmatrix}=B$

Thus, $\frac{1}{2}(A+A')$ is a symmetric matrix.

$A-A'=\begin{bmatrix} 3 & 3 & -1\\ -2 & -2 & 1\\ -4 & -5 & 2 \end{bmatrix}$$-\begin{bmatrix} 3 & -2 & -4\\ 3 & -2 & -5\\ -1 & 1 & 2 \end{bmatrix}$

$A-A'=\begin{bmatrix} 0 & 5&3\\ -5 & 0&6 \\-3&-6&0\end{bmatrix}$

Let

$C= \frac{1}{2}(A-A')=\frac{1}{2}\begin{bmatrix} 0 & 5&3\\ -5&0 & 6\\-3&-6&0 \end{bmatrix}$$=\begin{bmatrix} 0 & \frac{5}{2}&\frac{3}{2}\\ -\frac{5}{2}&0 & 3\\\frac{-3}{2}&-3&0 \end{bmatrix}$

$C'=\begin{bmatrix} 0 &- \frac{5}{2}&-\frac{3}{2}\\ \frac{5}{2}&0 &- 3\\\frac{3}{2}&3&0 \end{bmatrix}$

$C=-C'$

Thus, $\frac{1}{2}(A-A')$ is a skew-symmetric matrix.

Represent A as the sum of B and C.

$B+C= \begin{bmatrix} 3 & \frac{1}{2} & -\frac{5}{2}\\ \frac{1}{2} & -2 & -2\\ \frac{-5}{2} & -2 & 2 \end{bmatrix}$ $+\begin{bmatrix} 0 & \frac{5}{2}&\frac{3}{2}\\ -\frac{5}{2}&0 & 3\\\frac{-3}{2}&-3&0 \end{bmatrix}$ $=\begin{bmatrix} 3 & 3 & -1\\ -2 & -2 & 1\\ -4 & -5 & 2 \end{bmatrix}=A$

Question 10(iv). Express the following matrices as the sum of a symmetric and a skew-symmetric matrix:

$\begin{bmatrix} 1 & 5\\ -1 & 2 \end{bmatrix}$

Answer:

$A =\begin{bmatrix} 1 & 5\\ -1 & 2 \end{bmatrix}$

$A'=\begin{bmatrix} 1 & -1\\ 5 & 2 \end{bmatrix}$

$A+A'=\begin{bmatrix} 1 & 5\\ -1 & 2 \end{bmatrix}$$+\begin{bmatrix} 1 & -1\\ 5 & 2 \end{bmatrix}$

$A+A'=\begin{bmatrix} 2 & 4\\ 4 & 4 \end{bmatrix}$

Let

$B=\frac{1}{2}(A+A')=\frac{1}{2}\begin{bmatrix} 2 & 4\\ 4 & 4 \end{bmatrix}$$=\begin{bmatrix} 1 & 2\\ 2 & 2 \end{bmatrix}$

$B'=\begin{bmatrix} 1 & 2\\ 2 & 2 \end{bmatrix}=B$

Thus, $\frac{1}{2}(A+A')$ is a symmetric matrix.

$A-A'=\begin{bmatrix} 1 & 5\\ -1 & 2 \end{bmatrix}$$-\begin{bmatrix} 1 & -1\\ 5 & 2 \end{bmatrix}$

$A-A'=\begin{bmatrix} 0 & 6\\ -6 & 0 \end{bmatrix}$

Let

$C= \frac{1}{2}(A-A')=\frac{1}{2}\begin{bmatrix} 0 & 6\\ -6 & 0 \end{bmatrix}$$= \begin{bmatrix} 0 & 3\\ -3 & 0 \end{bmatrix}$

$C'= \begin{bmatrix} 0 & -3\\ 3 & 0 \end{bmatrix}$

$C=-C'$

Thus, $\frac{1}{2}(A-A')$ is a skew-symmetric matrix.

Represent A as the sum of B and C.

$B+C=\begin{bmatrix} 1 & 2\\ 2 & 2 \end{bmatrix}$ $- \begin{bmatrix} 0 & -3\\ 3 & 0 \end{bmatrix}$ $= \begin{bmatrix} 1 & 5\\ -1 & 2\end{bmatrix}=A$

Question 11: Choose the correct answer in Exercises 11 and 12.

If A, B are symmetric matrices of the same order, then AB – BA is a

(A) Skew-symmetric matrix
(B) Symmetric matrix
(C) Zero matrix
(D) Identity matrix

Answer:

If A and B are symmetric matrices, then

$A'=A$ and $B' = B$

we have, $\left ( AB-BA \right )'=\left ( AB \right )'-\left ( BA \right )'=B'A'-A'B'$

$=BA-AB$

$= -(AB-BA)$

Hence, we have $(AB-BA) = -(AB-BA)'$

Thus,( AB-BA)' is skew-symmetric.

Option A is correct.

Question 12: Choose the correct answer in Exercises 11 and 12.

If $A = \begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha& \cos\alpha \end{bmatrix}$ and $A+A' =I$, then the value of $\alpha$ is

(A) $\frac{\pi}{6}$

(B) $\frac{\pi}{3}$

(C) $\pi$

(D) $\frac{3\pi}{2}$

Answer:

$A = \begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha& \cos\alpha \end{bmatrix}$

$A' = \begin{bmatrix} \cos\alpha & \sin\alpha\\ -\sin\alpha& \cos\alpha \end{bmatrix}$

$A+A' = \begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha& \cos\alpha \end{bmatrix}$$+ \begin{bmatrix} \cos\alpha & \sin\alpha\\ -\sin\alpha& \cos\alpha \end{bmatrix}$$= \begin{bmatrix} 1& 0\\ 0 & 1 \end{bmatrix}$

$A+A' = \begin{bmatrix} 2\cos\alpha & 0\\ 0 & 2\cos\alpha \end{bmatrix}$$= \begin{bmatrix} 1& 0\\ 0 & 1 \end{bmatrix}$

$2 cos \alpha=1$

$cos \alpha=\frac{1}{2}$

$\alpha=\frac{\pi}{3}$

Option B is correct.

Question 1: Matrices A and B will be inverse of each other only if

(A) AB=BA

(B) AB=BA=0

(C) AB=0,BA=I

(D) AB=BA=I

Answer:

We know that if A is a square matrix of order n and there is another matrix B of the same order n, such that $AB=BA=I$, then B is the inverse of matrix A.

In this case, it is clear that A is the inverse of B.

Hence, matrices A and B will be inverse of each other only if $AB=BA=I$.

Option D is correct.

Question:1 Let $A = \begin{bmatrix} 0 &1 \\ 0 & 0 \end{bmatrix}$, show that $(aI + bA)^n = a^n I + na^{n-1} bA$, where I is the identity matrix of order 2 and $n \in N$.

Answer:

Given :

$A = \begin{bmatrix} 0 &1 \\ 0 & 0 \end{bmatrix}$

To prove : $(aI + bA)^n = a^n I + na^{n-1} bA$

For n=1, $aI + bA = a I + a^{0} bA =a I + bA$

The result is true for n=1.

Let the result be true for n=k,

$(aI + bA)^k = a^k I + ka^{k-1} bA$

Now, we prove that the result is true for n=k+1,

$(aI + bA)^{k+1} = (aI + bA)^k (aI + bA)$

$= (a^k I + ka^{k-1} bA)$$(aI + bA)$

$=a^{k+1}I+Ka^{k}bAI+a^{k}bAI+ka^{k-1}b^{2}A^{2}$

$=a^{k+1}I+(k+1)a^{k}bAI+ka^{k-1}b^{2}A^{2}$

$A^{2} = \begin{bmatrix} 0 &1 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} 0 &1 \\ 0 & 0 \end{bmatrix}$

$A^{2} = \begin{bmatrix} 0 &0 \\ 0 & 0 \end{bmatrix}=0$

Put the value of $A^{2}$ in above equation,

$(aI + bA)^{k+1}$$=a^{k+1}I+(k+1)a^{k}bAI+ka^{k-1}b^{2}A^{2}$

$(aI + bA)^{k+1}$$=a^{k+1}I+(k+1)a^{k}bAI+0$

$=a^{k+1}I+(k+1)a^{k}bAI$

Hence, the result is true for n=k+1.

Thus, we have $(aI + bA)^n = a^n I + na^{n-1} bA$ where $A = \begin{bmatrix} 0 &1 \\ 0 & 0 \end{bmatrix}$,$n \in N$.

Question 2. If $A = \begin{bmatrix} 1 & 1 & 1\\ 1& 1& 1\\ 1& 1& 1 \end{bmatrix}$ then show that $A^n =\begin{bmatrix} 3^{n-1} & 3^{n-1} &3^{n-1} \\ 3^{n-1}& 3^{n-1} & 3^{n-1}\\ 3^{n-1} & 3^{n-1}& 3^{n-1} \end{bmatrix}$, $n\in N$.

Answer:

Given :

$A = \begin{bmatrix} 1 & 1 & 1\\ 1& 1& 1\\ 1& 1& 1 \end{bmatrix}$

To prove:

$A^n =\begin{bmatrix} 3^{n-1} & 3^{n-1} &3^{n-1} \\ 3^{n-1}& 3^{n-1} & 3^{n-1}\\ 3^{n-1} & 3^{n-1}& 3^{n-1} \end{bmatrix}$

For n=1, we have

$A^1 =\begin{bmatrix} 3^{1-1} & 3^{1-1} &3^{1-1} \\ 3^{1-1}& 3^{1-1} & 3^{1-1}\\ 3^{1-1} & 3^{1-1}& 3^{1-1} \end{bmatrix}$$=\begin{bmatrix} 3^{0} & 3^{0} &3^{0} \\ 3^{0}& 3^{0} & 3^{0}\\ 3^{0} & 3^{0}& 3^{0} \end{bmatrix}$$= \begin{bmatrix} 1 & 1 & 1\\ 1& 1& 1\\ 1& 1& 1 \end{bmatrix}=A$

Thus, the result is true for n=1.

Now, take n=k,

$A^k =\begin{bmatrix} 3^{k-1} & 3^{k-1} &3^{k-1} \\ 3^{k-1}& 3^{k-1} & 3^{k-1}\\ 3^{k-1} & 3^{k-1}& 3^{k-1} \end{bmatrix}$

For n=k+1,

$A^{K+1}=A.A^K$

$= \begin{bmatrix} 1 & 1 & 1\\ 1& 1& 1\\ 1& 1& 1 \end{bmatrix}$$\begin{bmatrix} 3^{k-1} & 3^{k-1} &3^{k-1} \\ 3^{k-1}& 3^{k-1} & 3^{k-1}\\ 3^{k-1} & 3^{k-1}& 3^{k-1} \end{bmatrix}$

$=\begin{bmatrix}3. 3^{k-1} & 3.3^{k-1} &3.3^{k-1} \\3. 3^{k-1}& 3.3^{k-1} & 3.3^{k-1}\\3. 3^{k-1} & 3.3^{k-1}&3. 3^{k-1} \end{bmatrix}$

$=\begin{bmatrix} 3^{(K+1)-1} &3^{(K+1)-1} &3^{(K+1)-1}\\ 3^{(K+1)-1}&3^{(K+1)-1} &3^{(K+1)-1}\\ 3^{(K+1)-1} & 3^{(K+1)-1}& 3^{(K+1)-1}\end{bmatrix}$

Thus, the result is true for n=k+1.

Hence, we have $A^n =\begin{bmatrix} 3^{n-1} & 3^{n-1} &3^{n-1} \\ 3^{n-1}& 3^{n-1} & 3^{n-1}\\ 3^{n-1} & 3^{n-1}& 3^{n-1} \end{bmatrix}$, $n\in N$ where $A = \begin{bmatrix} 1 & 1 & 1\\ 1& 1& 1\\ 1& 1& 1 \end{bmatrix}$.

Question 3. If $A = \begin{bmatrix} 3 & -4\\ 1& -1 \end{bmatrix}$, then prove that $A^n = \begin{bmatrix} 1+2n & -4n \\ n & 1-2n \end{bmatrix}$, where n is any positive integer.

Answer:

Given :

$A = \begin{bmatrix} 3 & -4\\ 1& -1 \end{bmatrix}$

To prove:

$A^n = \begin{bmatrix} 1+2n & -4n \\ n & 1-2n \end{bmatrix}$

For n=1, we have

$A^1 = \begin{bmatrix} 1+2\times 1 & -4\times 1\\ 1 & 1-2\times 1 \end{bmatrix}$$= \begin{bmatrix} 3 & -4\\ 1 & -1 \end{bmatrix}=A$

Thus, the result is true for n=1.

Now, the result is true for n=k,

$A^k = \begin{bmatrix} 1+2k & -4k\\ k & 1-2k \end{bmatrix}$

For n=k+1,

$A^{K+1}=A.A^K$

$= \begin{bmatrix} 3 & -4\\ 1& -1 \end{bmatrix}$$\begin{bmatrix} 1+2k & -4k\\ k & 1-2k \end{bmatrix}$

$=\begin{bmatrix} 3(1+2k)-4k & -12k-4(1-2k)\\ (1+2k)-k &-4k-(1-2k) \end{bmatrix}$

$=\begin{bmatrix} 3+6k-4k & -12k-4k+8k\\ 1+k &-4k-1+2k \end{bmatrix}$

$=\begin{bmatrix} 3+2k & -4k-4k\\ 1+k &-2k-1 \end{bmatrix}$

$=\begin{bmatrix} 1+2(k+1)& -4(k+1)\\ 1+k &1-2(k+1) \end{bmatrix}$

Thus, the result is true for n=k+1.

Hence, we have $A^n = \begin{bmatrix} 1+2n & -4n \\ n & 1-2n \end{bmatrix}$, where $A = \begin{bmatrix} 3 & -4\\ 1& -1 \end{bmatrix}$.

Question 4. If A and B are symmetric matrices, prove that $AB - BA$ is a skew-symmetric matrix.

Answer:

If A and B are symmetric matrices, then

$A'=A$ and $B' = B$

we have, $\left ( AB-BA \right )'=\left ( AB \right )'-\left ( BA \right )'=B'A'-A'B'$

$=BA-AB$

$= -(AB-BA)$

Hence, we have $(AB-BA) = -(AB-BA)'$

Thus,( AB-BA)' is skew-symmetric.

Question 5. Show that the matrix B′AB is symmetric or skew-symmetric according as A is symmetric or skew-symmetric.

Answer:

Let be a symmetric matrix, then $A'=A$

Consider, $(B'AB)' ={B'(AB)}'$

$={(AB)}'(B')'$

$= B'A'(B)$

$= B'(A'B)$

Replace $A'$ by $A$

$=B'(AB)$

i.e. $(B'AB)'$ $=B'(AB)$

Thus, if A is a symmetric matrix than $B'(AB)$ is a symmetric matrix.

Now, let A be a skew-symmetric matrix, then $A'=-A$.

$(B'AB)' ={B'(AB)}'$

$={(AB)}'(B')'$

$= B'A'(B)$

$= B'(A'B)$

Replace $A'$ by -$A$,

$=B'(-AB)$

$= - B'AB$

i.e. $(B'AB)'$ $= - B'AB$.

Thus, if A is a skew-symmetric matrix, then $- B'AB$ is a skew-symmetric matrix.

Hence, the matrix B′AB is symmetric or skew-symmetric according to whether A is symmetric or skew-symmetric.

Question 6. Find the values of x, y, z if the matrix $A = \begin{bmatrix} 0 & 2y & z\\ x & y & -z\\ x & -y &z \end{bmatrix}$ satisfy the equation $A'A = I$

Answer:

$A = \begin{bmatrix} 0 & 2y & z\\ x & y & -z\\ x & -y &z \end{bmatrix}$

$A' = \begin{bmatrix} 0 & x & x\\ 2y & y & -y\\ z & -z &z \end{bmatrix}$

$A'A = I$

$\begin{bmatrix} 0 & x & x\\ 2y & y & -y\\ z & -z &z \end{bmatrix}$$\begin{bmatrix} 0 & 2y & z\\ x & y & -z\\ x & -y &z \end{bmatrix}$$= \begin{bmatrix} 1 & 0& 0\\ 0 & 1 & 0\\ 0 & 0 &1\end{bmatrix}$

$\begin{bmatrix} x^{2}+x^{2} & xy-xy& -xz+xz\\ xy-xy& 4y^{2}+y^{2}+y^{2} & 2yz-yz-yz\\ -zx+zx & 2yz-yz-yz &z^{2}+z^{2}+z^{2}\end{bmatrix}$$= \begin{bmatrix} 1 & 0& 0\\ 0 & 1 & 0\\ 0 & 0 &1\end{bmatrix}$

$\begin{bmatrix} 2x^{2} & 0& 0\\ 0& 6y^{2} & 0\\ 0 & 0 &3z^{2}\end{bmatrix}$$= \begin{bmatrix} 1 & 0& 0\\ 0 & 1 & 0\\ 0 & 0 &1\end{bmatrix}$

Thus equating the terms element-wise

$2x^{2} = 1$ $6y^{2} = 1$ $3z^{2} = 1$

$x^{2} = \frac{1}{2}$ $y^{2} = \frac{1}{6}$ $z^{2}=\frac{1}{3}$

$x = \pm \frac{1}{\sqrt{2}}$ $y= \pm \frac{1}{\sqrt{6}}$ $z=\pm \frac{1}{\sqrt{3}}$

Question 7. For what values of x: $\begin{bmatrix} 1 & 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 0\\ 2 & 0 &1 \\ 1& 0& 2\end{bmatrix}\begin{bmatrix} 0\\ 2 \\ x \end{bmatrix} = O$?

Answer:

$\begin{bmatrix} 1 & 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 0\\ 2 & 0 &1 \\ 1& 0& 2\end{bmatrix}\begin{bmatrix} 0\\ 2 \\ x \end{bmatrix} = O$

$\begin{bmatrix} 1+4+1 & 2+0+0 & 0+2+2 \end{bmatrix} \begin{bmatrix} 0\\ 2 \\ x \end{bmatrix} = O$

$\begin{bmatrix} 6& 2& 4 \end{bmatrix} \begin{bmatrix} 0\\ 2 \\ x \end{bmatrix} = O$

$\begin{bmatrix} 0+4+4x \end{bmatrix} = O$

$4+4x=0$

$4x=-4$

$x=-1$

Thus, the value of x is -1.

Question 8. If $A = \begin{bmatrix} 3 &1 \\ -1 & 2 \end{bmatrix}$, show that $A^2 -5A + 7I= 0$.

Answer:

$A = \begin{bmatrix} 3 &1 \\ -1 & 2 \end{bmatrix}$

$A^{2} = \begin{bmatrix} 3 &1 \\ -1 & 2 \end{bmatrix}$$\begin{bmatrix} 3 &1 \\ -1 & 2 \end{bmatrix}$

$A^{2} = \begin{bmatrix} 9-1 &3+2 \\ -3-2 & -1+4 \end{bmatrix}$

$A^{2} = \begin{bmatrix} 8 &5 \\ -5 & 3 \end{bmatrix}$

$I= \begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}$

To prove: $A^2 -5A + 7I= 0$

L.H.S : $A^2 -5A + 7I$

$= \begin{bmatrix} 8 &5 \\ -5 & 3 \end{bmatrix}$$-5 \begin{bmatrix} 3 &1 \\ -1 & 2 \end{bmatrix}$$+ 7 \begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}$

$=\begin{bmatrix} 8-15+7 &5-5+0 \\ -5+5+0& 3-10+7 \end{bmatrix}$

$=\begin{bmatrix} 0 &0 \\ 0& 0 \end{bmatrix} =0=R.H.S$

Hence, we proved that

$A^2 -5A + 7I= 0$.

Question 9. Find x, if $\begin{bmatrix} x & -5 & -1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 2\\ 0 & 2 & 1\\ 2 & 0 & 3 \end{bmatrix} \begin{bmatrix} x\\ 4\\ 1 \end{bmatrix} = 0$.

Answer:

$\begin{bmatrix} x & -5 & -1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 2\\ 0 & 2 & 1\\ 2 & 0 & 3 \end{bmatrix} \begin{bmatrix} x\\ 4\\ 1 \end{bmatrix} = 0$

$\begin{bmatrix} x +0-2& 0-10+0 & 2x-5-3 \end{bmatrix} \begin{bmatrix} x\\ 4\\ 1 \end{bmatrix} = 0$

$\begin{bmatrix} x -2& -10 & 2x-8 \end{bmatrix} \begin{bmatrix} x\\ 4\\ 1 \end{bmatrix} = 0$

$\begin{bmatrix}x (x -2)-40+(2x-8) \end{bmatrix} = 0$

$\begin{bmatrix}x ^{2}-2x-40+2x-8\end{bmatrix} = 0$

$\therefore \, \, x ^{2}-48= 0$

$x ^{2}=48$

thus the value of x is

$x =\pm 4\sqrt{3}$

Question 10(a) A manufacturer produces three products, x, y, z, which he sells in two markets.
Annual sales are indicated below:

Market Products
I 10,000 2,000 18,000
II 6,000 20,000 8,000

If unit sale prices of x, y and z are ` 2.50, ` 1.50 and ` 1.00, respectively, find the total revenue in each market with the help of matrix algebra.

Answer:

The unit sale prices of x, y and z are ` 2.50, ` 1.50 and ` 1.00, respectively.

The total revenue in the market I, with the help of matrix algebra, can be represented as :

$\begin{bmatrix} 10000& 2000 & 18000 \end{bmatrix} \begin{bmatrix} 2.50\\ 1.50\\ 1.00 \end{bmatrix}$

$= 10000\times 2.50+2000\times 1.50+18000\times 1.00$

$= 25000+3000+18000$

$= 46000$

The total revenue in market II, with the help of matrix algebra, can be represented as :

$\begin{bmatrix} 6000& 20000 & 8000 \end{bmatrix} \begin{bmatrix} 2.50\\ 1.50\\ 1.00 \end{bmatrix}$

$= 6000\times 2.50+20000\times 1.50+8000\times 1.00$

$= 15000+30000+8000$

$= 53000$

Hence, total revenue in the market I is 46000, and total revenue in market II is 53000.

Question 10(b). A manufacturer produces three products x, y, z, which he sells in two markets.
Annual sales are indicated below:

Market Products
I 10,000 2,000 18,000
II 6,000 20,000 8,000

If the unit costs of the above three commodities are ` 2.00, ` 1.00 and 50 paise respectively. Find the gross profit.

Answer:

The unit costs of the above three commodities are ` 2.00, ` 1.00 and 50 paise respectively.

The total cost price in market I, with the help of matrix algebra, can be represented as :

$\begin{bmatrix} 10000& 2000 & 18000 \end{bmatrix} \begin{bmatrix} 2.00\\ 1.00\\ 0.50 \end{bmatrix}$

$= 10000\times 2.00+2000\times 1.00+18000\times 0.50$

$= 20000+2000+9000$

$= 31000$

Total revenue in the market I is 46000, gross profit in the market is $= 46000-31000$$=Rs. 15000$

The total cost price in market II, with the help of matrix algebra, can be represented as :

$\begin{bmatrix} 6000& 20000 & 8000 \end{bmatrix} \begin{bmatrix} 2.00\\ 1.00\\ 0.50 \end{bmatrix}$

$= 6000\times 2.0+20000\times 1.0+8000\times 0.50$

$= 12000+20000+4000$

$= 36000$

Total revenue in market II is 53000, gross profit in the market is$= 53000-36000= Rs. 17000$

Question 11. Find the matrix X so that $X\begin{bmatrix} 1 & 2 &3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9\\ 2 &4 & 6 \end{bmatrix}$

Answer:

$X\begin{bmatrix} 1 & 2 &3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9\\ 2 &4 & 6 \end{bmatrix}$

The matrix given on R.H.S is $2\times 3$ matrix and on LH.S is $2\times 3$ matrix.Therefore, X has to be $2\times 2$ matrix.

Let X be $\begin{bmatrix} a & c\\ b & d \end{bmatrix}$

$\begin{bmatrix} a & c\\ b & d \end{bmatrix}$$\begin{bmatrix} 1 & 2 &3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9\\ 2 &4 & 6 \end{bmatrix}$

$\begin{bmatrix} a+4c & 2a+5c &3a+6c \\ b+4d & 2b+5d & 3b+6d \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9\\ 2 &4 & 6 \end{bmatrix}$

$a+4c=-7$ $2a+5c=-8$ $3a+6c=-9$

$b+4d=2$ $2b+5d=4$ $3b+6d=6$

Taking, $a+4c=-7$

$a=-4c-7$

$2a+5c=-8$

$-8c-14+5c=-8$

$-3c=6$

$c=-2$

$a=-4\times -2-7$

$a=8-7=1$

$b+4d=2$

$b=-4d+2$

$2b+5d=4$

$\Rightarrow$ $-8d+4+5d=4$

$\Rightarrow -3d=0$

$\Rightarrow d=0$

$b=-4d+2$

$\Rightarrow b=-4\times 0+2=2$

Hence, we have $a=1, b=2,c=-2,d=0$

Matrix X is $\begin{bmatrix} 1 & -2\\ 2 & 0 \end{bmatrix}$.

Question 12. If A and B are square matrices of the same order such that $AB = BA$, then prove by induction that $AB^n = B^n A$. Further, prove that $(AB)^n = A^n B^n$for all $n \in N$.

Answer:

A and B are square matrices of the same order such that $AB = BA$,

To prove : $AB^n = B^n A$, $n \in N$

For n=1, we have $AB^1 = B^1 A$

Thus, the result is true for n=1.

Let the result be true for n=k,then we have $AB^k = B^k A$

Now, taking n=k+1 , we have $AB^{k+1} = AB^k .B$

$AB^{k+1} = (B^kA) .B$

$AB^{k+1} = (B^k) .AB$

$AB^{k+1} = (B^k) .BA$

$AB^{k+1} = (B^k.B) .A$

$AB^{k+1} = (B^{k+1}) .A$

Thus, the result is true for n=k+1.

Hence, we have $AB^n = B^n A$, $n \in N$.

To prove: $(AB)^n = A^n B^n$

For n=1, we have $(AB)^1 = A^1 B^1$

Thus, the result is true for n=1.

Let the result be true for n=k,then we have $(AB)^k = A^k B^k$

Now, taking n=k+1 , we have $(AB)^{k+1} = (A B)^k.(AB)$

$(AB)^{k+1} = A^k B^k.(AB)$

$(AB)^{k+1} = A^{K}( B^kA)B$

$(AB)^{k+1} = A^{K}( AB^k)B$

$(AB)^{k+1} = (A^{K}A)(B^kB)$

$(AB)^{k+1} = (A^{k+1})(B^{k+1})$

Thus, the result is true for n=k+1.

Hence, we have $AB^n = B^n A$ and $(AB)^n = A^n B^n$for all $n \in N$.

Question 13 Choose the correct answer in the following questions:

If $A = \begin{bmatrix} \alpha &\beta\\ \gamma &-\alpha \end{bmatrix}$ is such that $A^2 = I$

(A) $1 + \alpha^2 + \beta \gamma = 0$

(B) $1 - \alpha^2 + \beta \gamma = 0$

(C) $1 - \alpha^2 - \beta \gamma = 0$

(D) $1 + \alpha^2 - \beta \gamma = 0$

Answer:

$A = \begin{bmatrix} \alpha &\beta\\ \gamma &-\alpha \end{bmatrix}$

$A^2 = I$

$\begin{bmatrix} \alpha &\beta\\ \gamma &-\alpha \end{bmatrix}$$\begin{bmatrix} \alpha &\beta\\ \gamma &-\alpha \end{bmatrix}$$= \begin{bmatrix} 1 &0\\0&1 \end{bmatrix}$

$\begin{bmatrix} \alpha^{2} +\beta \gamma&\alpha \beta-\alpha \beta\\\alpha \gamma-\alpha \gamma&\beta \gamma+\alpha^{2} \end{bmatrix}$$= \begin{bmatrix} 1 &0\\0&1 \end{bmatrix}$

$\begin{bmatrix} \alpha^{2} +\beta \gamma&0\\0&\beta \gamma+\alpha^{2} \end{bmatrix}$$= \begin{bmatrix} 1 &0\\0&1 \end{bmatrix}$

Thus, we obtained that

$\alpha^{2} +\beta \gamma=1$

$\Rightarrow 1-\alpha^{2} -\beta \gamma=0$

Option C is correct.

Question 14. If the matrix A is both symmetric and skew-symmetric, then

(A) A is a diagonal matrix
(B) A is a zero matrix
(C) A is a square matrix
(D) None of these

Answer:

If the matrix A is both symmetric and skew-symmetric, then

$A'=A$ and $A'=-A$

$A'=A'$

$\Rightarrow \, \, \, \, \, \, \, A=-A$

$\Rightarrow \, \, \, \, \, \, \, A+A=0$

$\Rightarrow \, \, \, \, \, \, \, 2A=0$

$\Rightarrow \, \, \, \, \, \, \, A=0$

Hence, A is a zero matrix.

Option B is correct.

Question 15. If A is square matrix such that $A^{2}=A$, then $(I + A)^3 - 7 A$ is equal to

(A) A
(B) I – A
(C) I
(D) 3A

Answer:

A is a square matrix such that $A^{2}=A$

$(I + A)^3 - 7 A$

$=I^{3}+A^{3}+3I^{2}A+3IA^{2}-7A$

$=I+A^{2}.A+3A+3A^{2}-7A$

$=I+A.A+3A+3A-7A$ (Replace $A^{2}$ by $A$)

$=I+A^{2}+6A-7A$

$=I+A-A$

$=I$

Hence, we have $(I + A)^3 - 7 A=I$

Option C is correct.