RD Sharma Solutions Class 12 Mathematics Chapter 4 FBQ

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RD Sharma Solutions Class 12 Mathematics Chapter 4 FBQ

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RD Sharma Class 12 Solutions Chapter 4 FBQ Algebra of Matrices- Other Exercise

Algebra of Matrices Excercise:FBQ

Algebra of Matrices exercise Fill in the blank question 1

Answer:
(a,b) = (2,4)
Hint:
We know that, here, we use the basic multiplication rules of two matrixes
Given:
A is order of a × 3 and B is order of 3 × b order of matrix.
Solution:
Here, we know that matrix multiplication is possible only if number of rows in second matrix are same number of columns in first matrix, i.e. order of 1^st matrix is m X n and order of 2^nd matrix is nXp.
And, the resultant matrix is order of m X p, I.e. number of rows=m and number of columns=p
So,
Here A is a × 3
B is 3 × b
And, AB is 2 × 4
So, A is 2 × 3
B is 3 × 4
So, (a,b) = (2,4)

Algebra of Matrices exercise Fill in the blank question 2

Answer:
PQ is 3p
Hint:
Use the basic concept of matrix multiplication.
Given:
P is 3 × n and
Q is n × p order of matrix.
Solution:
Here, number of P's columns and Q's rows are same
So, multiplication is possible.
∴PQ is order of 3p

Algebra of Matrices exercise Fill in the blank question 3

Answer:
x = 2 and y = 1 so, 2x + y = 2(2) + 1 = 5
Hint:
Use the basic concept of symmetric matrix.
Given:
$\text { Symmetric matrix } A=\left[\begin{array}{ccc} -1 & 2 & 3 x \\ 2 y & 4 & -1 \\ 6 & 5 & 0 \end{array}\right]$
Solution:
A=A^T $\begin{aligned} &\text { Here, } A=\left[\begin{array}{ccc} -1 & 2 & 3 x \\ 2 y & 4 & -1 \\ 6 & 5 & 0 \end{array}\right] \text { and } A^{T}=\left[\begin{array}{ccc} -1 & 2 y & 6 \\ 2 & 4 & 5 \\ 3 x & -1 & 0 \end{array}\right] \\ &{\left[\begin{array}{ccc} -1 & 2 & 3 x \\ 2 y & 4 & -1 \\ 6 & 5 & 0 \end{array}\right]=\left[\begin{array}{ccc} -1 & 2 y & 6 \\ 2 & 4 & 5 \\ 3 x & -1 & 0 \end{array}\right]} \end{aligned}$

By comparing respective elements,

$\begin{aligned} &\therefore 2 y=2 y=1 \\ &\text { And } 3 x=6 x=2 \\ &2 x+y=2(2)+1=5 \end{aligned}$

Algebra of Matrices exercise Fill in the blank question 4

Answer:
a = 3 and b = 4.
Hint:
Use basic concept of matrix multiplication
Given:
. $a<b \text { and }\left[\begin{array}{ll} a & b \end{array}\right]\left[\begin{array}{l} a \\ b \end{array}\right]=25$
Solution:
$\begin{aligned} &\left[\begin{array}{ll} a & b \end{array}\right]\left[\begin{array}{l} a \\ b \end{array}\right]=25 \\ &{[a \times a+b \times b]=25} \end{aligned}$
simplify the multiplication
$\begin{aligned} &{\left[a^{2}+b^{2}=25\right]} \\ &\text { Let's } a=3, b=4 \\ &\text { Then } L . H . S .=3^{2}+4^{2}=9+16=25=R . H . S . \\ &\text { So, } a=3, b=4 \end{aligned}$

Algebra of Matrices exercise Fill in the blank question 5

Answer:
x = 1
Hint:
Use the basic concept of identity matrix.
Given:
$A=\left[\begin{array}{cc} \frac{1}{3} & 2 \\ 0 & 2 x-3 \end{array}\right] \text { and } B=\left[\begin{array}{cc} 3 & 6 \\ 0 & -1 \end{array}\right],[A B]=[I]$
Find x
Solution:
$\begin{aligned} &\text { Here, }[A B]=[I] \\ &\text { So, }\left[\begin{array}{cc} \frac{1}{3} & 2 \\ 0 & 2 x-3 \end{array}\right]\left[\begin{array}{cc} 3 & 6 \\ 0 & -1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ &=\left[\begin{array}{ll} \frac{1}{3} \times 3+2 \times 0 & \frac{2}{3} \times 6+2 \times(-1) \\ 0 \times 3+0 \times 0 & 0+(2 x-3)(-1) \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \end{aligned}$
$=\left[\begin{array}{cc} 1 & 0 \\ 0 & 2 x+3 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$
So, here both sides have 2 x 2 matrices
Hence, by comparing respective elements,
$\begin{aligned} &-2 s+3=1 \\ &3-1=2 x \\ &2=2 x \\ &x=1 \end{aligned}$

Algebra of Matrices exercise Fill in the blank question 6

Answer:
x = ±1
Hint:
Use the basic concept of matrix multiplication
Given:
$A=\left[\begin{array}{cc} x & 1 \\ -1 & -x \end{array}\right] \text { and } A^{2}=0 \text { then } x=0$
Solution:
$\begin{aligned} &A \times A=\left[\begin{array}{cc} x & 1 \\ -1 & -x \end{array}\right] \times\left[\begin{array}{cc} x & 1 \\ -1 & -x \end{array}\right] \\ &=\left[\begin{array}{cc} x^{2}+(-1) & x+(-x) \\ -x+x & -1+x^{2} \end{array}\right] \\ &=\left[\begin{array}{cc} x^{2}-1 & 0 \\ 0 & x^{2}-1 \end{array}\right] \end{aligned}$
$\begin{aligned} &\text { Since, } A^{2}=0 \\ &{\left[\begin{array}{cc} x^{2}-1 & 0 \\ 0 & x^{2}-1 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]} \\ &x^{2}-1=0 \\ &x^{2}=1 \\ &x=\pm 1 \end{aligned}$

Algebra of Matrices exercise Fill in the blank question 7

Answer:
B is an n x m matrix.
Hint:
Use the basic concept of matrix order.
Given:
A is an m x n matrix and AB and BA are defined.
Solution:
Multiplication is possible, if first matrix has same number of columns as number of rows in second matrix.
AB is defined so number of rows in B is n
BA is defined so number of columns in B is m
Hence order of matrix B is
n x m

Algebra of Matrices exercise Fill in the blank question 8

Answer:
AB_3×3 = 4.
Hint:
Use the basic multiplication rules
Given:
$\left[\begin{array}{ccc} 0 & 2 & 0 \\ 0 & 0 & 3 \\ -2 & 2 & 0 \end{array}\right]\left[\begin{array}{ccc} 1 & 2 & 3 \\ 3 & 4 & 5 \\ 3 & -4 & 0 \end{array}\right]$
Solution:
$\begin{aligned} &A B=(-2 \times 3)+(2 \times 5)+(0 \times 0)=-6+10+0=10-6=4 \\ &A B_{3 \times 3}=4 \end{aligned}$

Algebra of Matrices exercise Fill in the blank question 9

Answer:
A⁴ = 81I
Hint:
Use the basic concept of matrix multiplication.
Given:
$A=\left[\begin{array}{lll} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{array}\right]$
Solution:
$\begin{aligned} &A=\left[\begin{array}{lll} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{array}\right]=3\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\\ &\text { Therefore, } A^{4}=(3 I)^{4}=81 I\\ &\text { Or }\\ &A^{4}=\left[\begin{array}{ccc} 81 & 0 & 0 \\ 0 & 81 & 0 \\ 0 & 0 & 81 \end{array}\right] \end{aligned}$

Algebra of Matrices exercise Fill in the blank question 10

Answer:
A²B = diag(-4,3,18)
Hint:
Use the concept of diagonal matrix.
Given:
A = diag(2,-1, 3) and B = diag(-1, 3, 2)
Solution:
$\begin{aligned} &A^{2}=\left[\begin{array}{ccc} 2 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 3 \end{array}\right] \times\left[\begin{array}{ccc} 2 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 3 \end{array}\right]=\left[\begin{array}{lll} 4 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 9 \end{array}\right] \\ &A^{2} \times B=\left[\begin{array}{lll} 4 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 9 \end{array}\right] \times\left[\begin{array}{ccc} -1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 2 \end{array}\right]=\left[\begin{array}{ccc} -4 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 16 \end{array}\right] \end{aligned}$

Algebra of Matrices exercise Fill in the blank question 11

Answer:
$\left[\begin{array}{ccc} 2 & 1 & -1 \\ -2 & -1 & -1 \\ 4 & 2 & -2 \end{array}\right]$
Hint:
Use the basic multiplication rules.
Given:
$\left[\begin{array}{c} 1 \\ -1 \\ 2 \end{array}\right]\left[\begin{array}{lll} 2 & 1 & -1 \end{array}\right]=?$
Solution:
$\begin{aligned} &{\left[\begin{array}{c} 1 \\ -1 \\ 2 \end{array}\right] \times\left[\begin{array}{ccc} 2 & 1 & -1 \end{array}\right]} \\ &=\left[\begin{array}{ccc} 1 \times 2 & 1 \times 1 & 1 \times-1 \\ -1 \times 2 & -1 \times 4 & -1 \times 1 \\ 2 \times 2 & 2 \times 1 & 2 \times-1 \end{array}\right]=\left[\begin{array}{ccc} 2 & 1 & -1 \\ -2 & -1 & -1 \\ 4 & 2 & -2 \end{array}\right] \end{aligned}$

Algebra of Matrices exercise Fill in the blank question 12

Answer:
x = 0
Hint:
Use the basics of identity matrix.
Given:
$A=\left[\begin{array}{ll} x & 1 \\ 1 & 0 \end{array}\right] \text { and } A^{2} \text { the identity matrix. }$
Solution:
$\begin{aligned} &{\left[\begin{array}{ll} x & 1 \\ 1 & 0 \end{array}\right] \times\left[\begin{array}{ll} x & 1 \\ 1 & 0 \end{array}\right]=I} \\ &{\left[\begin{array}{ll} x^{2}+1 & x+0 \\ x+0 & 1+0 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]} \\ &\text { So, } x+0=0 \\ &x=0 \end{aligned}$

Algebra of Matrices exercise Fill in the blank question 13

Answer:
x = (-1) and y = (-1)
Hint:
Use the basics of algebra.
Given:
$e\left[\begin{array}{ll} e^{x} & e^{y} \\ e^{y} & e^{x} \end{array}\right]=\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right]$
Solution:
$\begin{aligned} &{\left[\begin{array}{cc} e^{x} & e^{y} \\ e^{y} & e^{x} \end{array}\right]=\frac{1}{e}\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right]} \\ &{\left[\begin{array}{cc} e^{x} & e^{y} \\ e^{y} & e^{x} \end{array}\right]=\left[\begin{array}{cc} \frac{1}{e} & \frac{1}{e} \\ \frac{1}{e} & \frac{1}{e} \end{array}\right]} \end{aligned}$
So,
$e^{x}=\frac{1}{e} \therefore x=-1 e^{y}=1 \therefore y=-1$

Algebra of Matrices exercise Fill in the blank question 14

Answer:
k, a, b = (-6, -4, -9)
Hint:
Use the basics of scalar matrix.
Given:
$A=\left[\begin{array}{cc} 0 & 2 \\ 3 & -4 \end{array}\right], k A=\left[\begin{array}{cc} 0 & 39 \\ 2 b & 24 \end{array}\right]$
Solution:
Given
$\begin{aligned} &k A=k\left[\begin{array}{cc} 0 & 2 \\ 3 & -4 \end{array}\right]=\left[\begin{array}{cc} 0 & 2 k \\ 3 k & -4 k \end{array}\right] \\ &{\left[\begin{array}{cc} 0 & 39 \\ 2 b & 24 \end{array}\right]=\left[\begin{array}{cc} 0 & 2 k \\ 3 k & -4 k \end{array}\right]} \end{aligned}$
By the comparison,
$-4 k=24 k=-6$
And
$3 a=2 k a=2 \times \frac{(-6)}{3}=-4$
Also,
$2 b=3 k b=3 \times \frac{-6}{2}=-9$

Algebra of Matrices exercise Fill in the blank question 15

Answer:
$\begin{bmatrix} 1 &0 \\ 0 &1 \end{bmatrix}$
Hint:
Use the basic method of multiplication.
Given:
$A=\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$
Solution:
Calculate,
$A^{T}=\left[\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]$
So,
$\begin{aligned} &A \times A^{T}=\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right] \times\left[\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right] \\ &A \times A^{T}=\left[\begin{array}{cc} \cos ^{2} \theta+\sin ^{2} \theta & -\cos \theta \sin \theta+\cos \theta \sin \theta \\ -\sin \theta \cos \theta+\cos \theta \sin \theta & \sin ^{2} \theta+\cos ^{2} \theta \end{array}\right] \\ &A \times A^{T}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \end{aligned}$

Algebra of Matrices exercise Fill in the blank question 16

Answer:
Order 3 x 4
Hint:
Use the basic concept of order of matrix.
Given:
A is a 3 x 4 matrix and A^TB and B^TA are definied.
Solution:
Let’s assume order of B is m x n
A = 3 x 4
And
A^TB= (4 x 3) * (m x n)
So, m=3
Also B^TA=(m x n) * (4 x 3)
So,
n = 4
Hence,
Order B is 3 x 4

Algebra of Matrices exercise Fill in the blank question 17

Answer:
$A A^{T}=\left[\begin{array}{lll} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{array}\right]$
Hint:
Use the basics of transpose and multiplication of matrix
Given:
$A = \begin{bmatrix} 1\\ 2\\ 3 \end{bmatrix}$
Solution:
$A = \begin{bmatrix} 1\\ 2\\ 3 \end{bmatrix}, A^{T} = \begin{bmatrix} 1 &2 &3 \end{bmatrix}$
So,
$A A^{T}=\left[\begin{array}{lll} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{array}\right]$

Algebra of Matrices exercise Fill in the blank question 18

Answer:
z = x + y
Hint:
Use the basic of matrix.
Given:
F(x) F(y) = F(z)
Solution:
$\begin{aligned} &F(x) F(y)=\left[\begin{array}{cc} \cos x & \sin x \\ -\sin x & \cos x \end{array}\right]\left[\begin{array}{cc} \cos y & \sin y \\ -\sin y & \cos y \end{array}\right] \\ &=\left[\begin{array}{cc} \cos x \cos y+(-\sin x \sin y) & -\cos x \sin y-\sin x \cos y \\ \cos y \sin x+\cos x \sin y & -\sin x \sin y+\cos x \cos y \end{array}\right] \\ &=\left[\begin{array}{cc} \cos (x+y) & -\sin (x+y) \\ \sin (x+y) & \cos (x+y) \end{array}\right] \end{aligned}$
So,
F(x) F(y) = F(z) = F(x + y)
Hence,
z = x + y

Algebra of Matrices exercise Fill in the blank question 19

Answer:
m = q
Hint:
Use the basic concept of matrix order and multiplication
Given:
B_{[n x p]} , A_{[m x n]}C_{[p x q]}
Solution:
BC can be find because, B's columns and C's rows are same.
BC x A can be defined as,
BC_{[n x q]} x A_{[m x n]}
Hence, by the Rule of matrix multiplication,
m = q

Algebra of Matrices exercise Fill in the blank question 20

Answer:
l = 16
Hint:
Use the basic concept of identity matrix.
Given:
$A=\left[\begin{array}{lll} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array}\right]$
Solution:
A⁵= lA
Since,
$A=\left[\begin{array}{lll} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array}\right]=2\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=21\; \; \; \; \; \; \; \; \; \; ......(1)$
So,
A⁵= [2I]⁵
= 16 x 2I
=16 x A
Hence
I = 16

Algebra of Matrices exercise Fill in the blank question 21

Answer:
C = 0
Hint:
Use commutative property of matrix A X B = B X A
Given:
$A=\left[\begin{array}{ll} a & d \\ c & d \end{array}\right] \text { and } B=\left[\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right]$
Solution:
A X B = B X A
$\left[\begin{array}{ll} a & a+d \\ c & c+d \end{array}\right]=\left[\begin{array}{cc} a+c & b+d \\ c & d \end{array}\right]$
$\begin{aligned} &a=a+c \\ &a-a=c \\ &\text { So, } c=0 \end{aligned}$
C = 0

Algebra of Matrices exercise Fill in the blank question 22
Answer:

x = 5
Hint:
A = A^T
Given:
$A=\left[\begin{array}{cc} 4 & x+2 \\ 2 x-3 & x+1 \end{array}\right]$
Solution:
A = A^T
$\begin{aligned} &A=\left[\begin{array}{cc} 4 & x+2 \\ 2 x-3 & x+1 \end{array}\right] \\ &A^{T}=\left[\begin{array}{cc} 4 & 2 x-3 \\ x+2 & x+1 \end{array}\right] \\ &\text { So, } \\ &\qquad\left[\begin{array}{cc} 4 & x+2 \\ 2 x-3 & x+1 \end{array}\right]=\left[\begin{array}{cc} 4 & 2 x-3 \\ x+2 & x+1 \end{array}\right] \end{aligned}$
By comparing
x + 2 = 2x - 3
By solving, x = 5

Algebra of Matrices exercise Fill in the blank question 23

Answer:
BA = AB
Hint:
Use the basic concept of symmetric and skew symmetric matrix.
Given:
A and B are two skew symmetric matrixes.
Solution:
A and B are two skew symmetric matrixes, then
$\begin{aligned} &A=-A^{T} \quad \ldots(1) B=-B^{t} \quad \ldots \text { (2) }\\ &A B=(A B)^{T} A B=B^{T} A^{T}\\ &\text { By equations }(1),(2)\\ &A B=(-B)(-A)\\ &A B=B A \end{aligned}$

Algebra of Matrices exercise Fill in the blank question 24

Answer:
3A^T-2B^T
Hint:
Use concept of matrix order
Given:
A and B are same order of matrix.
Solution:
$\begin{aligned} &(3 A-2 B)^{T} \\ &=3 A^{T}-2 B^{T} \end{aligned}$

Algebra of Matrices exercise Fill in the blank question 25

Answer:
Same
Hint:
Addition of two matrices is defined if and only if order of the matrix is same
Given:
Addition of matrices is defined.
Solution:
Both has same order then addition is possible.

Algebra of Matrices exercise Fill in the blank question 26

Answer:
AB = BA
Hint:
Use the concept of symmetric matrix
Given:
A and B are symmetric metrices.
Solution:
Since A and B are symmetric so,
A = A^TB = B^TA^T
So,
AB = (AB)^T = B^TA^T
From equation (1), (2)
AB = BA

Algebra of Matrices exercise Fill in the blank question 27

Answer:
B^T AB is symmetric Metrix.
Hint:
Use concept of symmetric Metrix
Given:
A is symmetric
Solution:
A = A^T
Let’s take transpose of B^T AB
B^T AB = BAB^T
So, B^T AB is symmetric Metrix.

Algebra of Matrices exercise Fill in the blank question 28

Answer:
A² is a symmetrical matrix.
Hint:
Use the basics of skew- symmetrical matrix
Given:
A is a skew symmetrical matrix.
Solution:
given
-A = A^T
So,
$\begin{aligned} &A^{2}=\left(-A^{T}\right)^{2} \\ &A^{2}=\left(A^{T}\right)^{2} \\ &A^{2}=\left(A^{2}\right)^{T} \end{aligned}$
So, A² is a symmetrical matrix.

Algebra of Matrices exercise Fill in the blank question 29

Answer:
A³ is symmetric matrix.
Hint:
Use the basics of symmetric matrix.
Given:
A is symmetric matrix.
Solution:
Given
A is symmetric matrix.
A = A^T.....(1)
Take cube on both sides
A³ = (A^T)³ = (A³)^T
So, A³ is symmetric matrix.

Algebra of Matrices exercise Fill in the blank question 30

Answer:
kA is skew symmetric Metrix
Hint:
Use the concept of skew-symmetric matrix
Given:
A is skew symmetric Metrix
Solution:
Since A is skew symmetric Metrix
$\begin{aligned} &-A=A^{T} \ldots .(1) \\ &k A=-k\left(A^{T}\right) \\ &\text { Or, } \\ &-k A=k A^{T} \end{aligned}$
Hence, kA is skew symmetric matrix.

Algebra of Matrices exercise Fill in the blank question 31

Answer:
(AB - BA) is skew symmetric matrix
Hint:
Use the concept of symmetric and skew-symmetric Metrix.
Given:
A and B are symmetric.
Solution:
Since, A and B are symmetric.
$\begin{aligned} &(A B-B A)^{\prime}=(A B)^{\prime}-(B A)^{\prime}=B^{\prime} A^{\prime}-A^{\prime} B^{\prime} \\ &=B A-A B=-(A B-B A) \\ &(A B-B A)^{\prime}=-(A B-B A) \\ &A B-B A \end{aligned}$
So, (AB - BA) is skew symmetric matrix

Algebra of Matrices exercise Fill in the blank question 32

Answer:
A^-1does not exist
Hint:
We must be aware with elementary row operations
Given:
$A^{-1}=\frac{a d j A}{|A|},|A| \neq 0$
Solution:
A^-1 by elementary row operations we obtain all zero in one or more
So, if any one row or column has zero or all the element of a row or column is zero
So, |A| = 0
So, A^-1does not exist

Algebra of Matrices exercise Fill in the blank question 33

Answer:
zero
Hint:
We must be aware with scalar and null matrix
Given:
Product of matrix by scalar
Solution:
Null matrix is the matrix of which all element are zero
If we product of any matrix by the scalar zero then it is null

Algebra of Matrices exercise Fill in the blank question 34

Answer:
Rectangular
Hint:
We must know the types of matrix
Given:
A matrix is not square matrix
Solution:
If matrix is not square matrix, then it can be row matrix, column matrix or rectangular matrix
But row and column matrix is also part of rectangular matrix.

Algebra of Matrices exercise Fill in the blank question 35

Answer:
skew symmetric
Hint:
We must know the concept of skew symmetric matrix
Given:
The sum of two skew symmetric matrices is always
Solution:
let A and B are skew symmetric
$\begin{aligned} &A^{T}=-A \quad \text { and } \quad B^{T}=-B \\ &A^{T}+B^{T} \\ &-A-B=(A+B)^{T} \\ &-(A+B)=(A+B)^{T} \end{aligned}$
It is skew symmetric matrices

Algebra of Matrices exercise Fill in the blank question 36

Answer:
$\begin{aligned} &\text { i) } A B=B^{T} A^{T} \\ &\text { ii) } k A=k(A) \\ &\text { iii) } k(A-B)=k A-k B \end{aligned}$
Hint:
We must know the basic of square matrix
Given:
A and B are square matrix
Solution:
$\begin{aligned} &\text { i) } A B=B^{T} A^{T} \\ &\text { because }[A]_{m \times n}[B]_{m \times 9} \\ &\text { so } \\ &A B_{m \times 9} \end{aligned}$
$\begin{aligned} &\text { ii) } k A=k(A) \end{aligned}$
If k is scalar we can remove from A
$\begin{aligned} &\text { iii) } k(A-B)=k A-k B \end{aligned}$
If k is scalar we can multiply with A and B separately

Algebra of Matrices exercise Fill in the blank question 37

Answer:
Null matrix
Hint:
We should aware with symmetric and skew symmetric matrix
Given:
which matrix is symmetric and skew symmetric matrix
Solution:
Let A is symmetric and skew symmetric matrix
So,
$\begin{aligned} &A=A^{T} \; \; \; \; \; \; \; .......(i)\\ &-A=A^{T} \; \; \; \; \; \; \; .......(ii) \end{aligned}$
So,
$\begin{aligned} &A^{T}=A^{T} \\ &A=-A \\ &A+A=0 \\ &2 A=0 \\ &A=0 \end{aligned}$
A is Null matrix

Algebra of Matrices exercise Fill in the blank question 38

Answer:
Distributive
Hint:
To solve this, we should know the theory of matrix algebra
Given:
Matrix multiplication is ____ over matrix addition
Solution:
Matrix multiplication is distributive over matrix addition
Or
$\begin{aligned} &A(B+C)=A B+A C \\ &(A+B) C=A C+B C \end{aligned}$
A is Null matrix

Algebra of Matrices exercise Fill in the blank question 39

Answer:
-1
Hint:
We must be aware with basic algebra
Given:
How we can obtain negative matrix?
Solution:
we can obtain negative matrix by multiplying -1 with matrix

Algebra of Matrices exercise Fill in the blank question 40

Answer:
(A^-1)^T
Hint:
For this we can use basic concept of matrix singular
Given:
A is non singular matrix. Find (A^T)^-1
Solution:
For example,
$\begin{aligned} &\left(A^{T}\right)^{-1}=\left(A^{-1}\right)^{T} \\ &\text { So, we get }\left(A^{-1}\right)^{T} \end{aligned}$

Algebra of Matrices exercise Fill in the blank question 41

Answer:
$A=\left[\begin{array}{ll} \frac{1}{3} & \frac{1}{3} \\ \\ \frac{2}{3} & \frac{1}{3} \end{array}\right]$
Hint:
For that we must aware with basic operations of matrix
Given:
$\begin{aligned} &A+B=\left[\begin{array}{ll} 1 & 0 \\ 1 & 1 \end{array}\right] \\ &A-2 B=\left[\begin{array}{cc} -1 & 1 \\ 0 & -1 \end{array}\right] \end{aligned}$
Solution:
$\begin{aligned} &A+B=\left[\begin{array}{ll} 1 & 0 \\ 1 & 1 \end{array}\right]\; \; \; \; \; \; .....(i) \\ &A-2 B=\left[\begin{array}{cc} -1 & 1 \\ 0 & -1 \end{array}\right] \; \; \; \; \; \; .....(ii) \end{aligned}$
Lets add both matrices,
$\begin{aligned} &A+B+A-2 B=\left[\begin{array}{ll} 1 & 0 \\ 1 & 1 \end{array}\right]+\left[\begin{array}{cc} -1 & 1 \\ 0 & -1 \end{array}\right]\\ &2 A-B=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right] \; \; \; \; \; \; .....(ii) \end{aligned}$
Lets add both (iii) and (i) equation
$\begin{aligned} &3 A=\left[\begin{array}{ll} 1 & 1 \\ 2 & 1 \end{array}\right] \\ &A=\frac{1}{3}\left[\begin{array}{ll} 1 & 1 \\ 2 & 1 \end{array}\right] \\ &A=\left[\begin{array}{ll} \frac{1}{3} & \frac{1}{3} \\ \\ \frac{2}{3} & \frac{1}{3} \end{array}\right] \end{aligned}$

Algebra of Matrices exercise Fill in the blank question 42

Answer:
0
Hint:
For this we must aware with skew symmetric matrix
Given:
$A=\left[\begin{array}{ccc} 0 & a & 1 \\ -1 & b & 1 \\ -1 & c & 0 \end{array}\right] \text { is skew symmetric }$
Solution:
$\begin{aligned} &A^{T}=-A \\ &{\left[\begin{array}{ccc} 0 & -1 & -1 \\ a & b & c \\ 1 & 1 & 0 \end{array}\right]=-1\left[\begin{array}{ccc} 0 & a & 1 \\ -1 & b & 1 \\ -1 & c & 0 \end{array}\right]} \\ &{\left[\begin{array}{ccc} 0 & -1 & -1 \\ a & b & c \\ 1 & 1 & 0 \end{array}\right]=\left[\begin{array}{ccc} 0 & -a & -1 \\ 1 & -b & -1 \\ 1 & -c & 0 \end{array}\right]} \end{aligned}$
So both sides have 3 x 3 matrix
So,
$\begin{aligned} &a=1 \\ &b=-b \\ &b+b=0 \\ &2 b=0 \\ &c=-1 \\ &\text { So, }(a+b+c)^{2}=(-1+0+(-1))^{2}=(0)^{2}=0 \end{aligned}$

Algebra of Matrices exercise Fill in the blank question 43
Answer:

3
Hint:
For this we must aware with matrix multiplication concept
Given:
$\begin{aligned} &{\left[\begin{array}{lll} 3 & -2 & 0 \end{array}\right]\left[\begin{array}{c} 2 \\ k \\ -5 \end{array}\right]=0} \\ &\text { Find } k=\text { ? } \end{aligned}$
Solution:
$\begin{aligned} &{\left[\begin{array}{lll} 3 & -2 & 0 \end{array}\right]\left[\begin{array}{c} 2 \\ k \\ -5 \end{array}\right]=0} \\ \end{aligned}$
Here 1^st has 3 columns and 2^nd has 3 rows
So, matrix multiplication is possible
$\begin{aligned} &{\left[\begin{array}{lll} 3 & -2 & 0 \end{array}\right]\left[\begin{array}{c} 2 \\ k \\ -5 \end{array}\right]=0} \\ &{[6+(-2 k)+0]=0} \\ &6-2 k=0 \\ &k=3 \end{aligned}$

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