RD Sharma Solutions Class 12 Mathematics Chapter 4 FBQ

# RD Sharma Solutions Class 12 Mathematics Chapter 4 FBQ

Edited By Kuldeep Maurya | Updated on Jan 20, 2022 01:30 PM IST

RD Sharma class 12th exercise FBQ can be your go-to NCERT solution when preparing for board exams. If you are in the 12th standard, you need to immediately start preparing for your exams so that you don't get overworked or stressed before you actually sit for your paper. Your paper might be quite challenging to solve but worry not as RD Sharma class 12 chapter 4 exercise FBQ will help make it easier for you.

Also Read - RD Sharma Solution for Class 9 to 12 Maths

## 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

(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 1st matrix is m X n and order of 2nd 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

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

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=AT\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

a = 3 and b = 4.
Hint:
Use basic concept of matrix multiplication
Given:
.$a
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

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

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

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

AB3×3 = 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

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

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

$\left[\begin{array}{ccc} 2 & 1 & -1 \\ -2 & -1 & -1 \\ 4 & 2 & -2 \end{array}\right]$
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

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

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

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

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

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

$A A^{T}=\left[\begin{array}{lll} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{array}\right]$
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

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

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

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

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
x = 5
Hint:
A = AT
Given:
$A=\left[\begin{array}{cc} 4 & x+2 \\ 2 x-3 & x+1 \end{array}\right]$
Solution:
A = AT
\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

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

3AT-2BT
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

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

Algebra of Matrices exercise Fill in the blank question 26

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 = ATB = BTAT
So,
AB = (AB)T = BTAT
From equation (1), (2)
AB = BA

Algebra of Matrices exercise Fill in the blank question 27

BT AB is symmetric Metrix.
Hint:
Use concept of symmetric Metrix
Given:
A is symmetric
Solution:
A = AT
Let’s take transpose of BT AB
BT AB = BABT
So, BT AB is symmetric Metrix.

Algebra of Matrices exercise Fill in the blank question 28

A2 is a symmetrical matrix.
Hint:
Use the basics of skew- symmetrical matrix
Given:
A is a skew symmetrical matrix.
Solution:
given
-A = AT
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

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

Algebra of Matrices exercise Fill in the blank question 30

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

(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

A-1 does 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-1 does not exist

Algebra of Matrices exercise Fill in the blank question 33

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

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

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

\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

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

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

-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

(A-1)T
Hint:
For this we can use basic concept of matrix singular
Given:
A is non singular matrix. Find (AT)-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

$A=\left[\begin{array}{ll} \frac{1}{3} & \frac{1}{3} \\ \\ \frac{2}{3} & \frac{1}{3} \end{array}\right]$
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}
\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

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}

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 1st has 3 columns and 2nd 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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