1192 + Interview Questions in MATRICES IN GENERAL AWARENESS Page 16 InterviewSolution

751.	If `A=[{:( 0,-x),(x,0):}].B=[{:(0,1),(1,0):}]` and `x^(2)=-1` , then show that `(A+B)^(2)=A^(2)+B^(2)`.
Answer» We have `A=[{:(0,-x),(x,0):}] B=[{:(0,1),(1,0):}]` and `x^(2)=-1` `therefore (A+B)=[{:(0,-2+1),( x+1,0):}]` and `(A+B)^(2)[{:(0,-x+1),(x+1,0):}][{:(0,-x+1),(x+ 1,0):}]` `=[{:(1-x^(2),0),( 0,1-x^(2)):}] ` Also, `A^(2)=A.A=[ {:(0,-x),(x,0):}][{:(0,-x),(x,0):}]=[{:(-x^(2),0),(0,-x^(2)):}]` and `B^(2)=B.B=[{:( 0,1) ,(1,0):}][{:(0,1),(1,0):}]=[{:(1,0),(0,1):}]` Now, `A^(2)+B^(2)=[{:(-x^(2)+1,0),(0,-x^(2)+1):}]=[{:(1-x^(2),0),(0,1-x^(2)):}]` `=(A+B)^(2)`

Discussion

752.	If\(\begin{vmatrix}x&3x-y\\zx+z&3y-w\end{vmatrix} = \begin{vmatrix}3&2\\4&7\end{vmatrix}\), then ___________[(x, 3x-y) (zx+z, 3y-w)] = [(3, 2) (4, 7)](a) x = 3, y = 7, z = 1, w = 14 (b) x = 3, y = -5, z = -1, w = -4 (c) x = 3, y = 6, z = 2, w = 7 (d) x = -3, y = -7, z = -1, w = -14
Answer» Correct option is : (a) x = 3, y = 7, z = 1, w = 14

Discussion

753.	Let A and B are two matrices such that `AB = BA,` then for every `n in N`A. `A^(n) B = BA^(n)`B. `(AB)^(n) = A^(n)B^(n)`C. `(A+B)^(n) = ""^(n)C_(0) A^(n) + ""^(n)C_(1) A^(n-1) B+...+ ""^(n) C_(n) B^(n) `D. `A^(2n) - B ^(2n) = (A^(n)-B^(n) ) (A^(n)+B^(n))`
Answer» Correct Answer - A::C::D `because A^(2) B = A (AB) = A(BA) = (AB)A = (BA)A= BA^(2)` Similarly, `A^(3) B = BA^(3) ` In general, `A^(n) B = BA^(n) , AA n ge 1 ` and `(A +B) ^(n) = ""^(n) C_(0)A^(n) + ""^(n)C_(1)A^(n-1) B +""^(n)C_(2) A^(n-2) B^(2) +...+""^(n) C_(n) B^(n)` Also, `(A^(n) - B^(n)) (A^(n) + B^(n) ) = A^(n) A^(n) + A^(n) B^(n) - B^(n) A^(n) - B^(n) B^(n) ` `= A^(2n)-B^(2n) [ because AB = BA]`

Discussion

754.	`A=[[2,1],[4,1]]; B=[[3,4],[2,3]]` & `c=[[3,-4],[-2,3]]`, `tr(A)+tr[(ABC)/2]+tr[(A(BC)^2)/4]+tr[(A(BC)^2)/8]+.....oo` is:A. 6B. 9C. 12D. none of these
Answer» Correct Answer - A `BC=[(3,4),(2,3)][(3,-4),(-2,3)]impliesB=[(1,0),(0,1)]` `implies tr(A)+tr((ABC)/2)+tr((A(BC)^(2))/4)+tr ((A(BC)^(3))/8)+...+ oo` `=tr (A)+tr (A/2)+tr (A/2^(2))+...` `=tr (A)+1/2 tr (A)+1/2^(2) tr (A)...` `=(tr (A))/(1-(1//2))` `=2tr (A)=2(2+1)=6`

Discussion

755.	If M is the matrix `[(1,-3),(-1,1)]` then find matrix `sum_(r=0)^(oo) ((-1)/3)^(r) M^(r+1)`
Answer» Let `N=M-M^(2)/3+M^(3)/9-M^(4)/27 +...` `:. -(MN)/(3) =- M^(2)/3+M^(3)/9-M^(4)/27 ... oo` Subtracting, we get `N+(MN)/3=M` `implies (I+M/3) N=M` `implies (3I+M)N=3M` `implies N=[(3,-9),(-3,3)][(4,-3),(-1,4)]^(-1)` `=1/13 [(3,-9),(-3,3)][(4,-3),(-1,4)]^(-1)` `=1/13 [(3,-27),(-9,3)]` `=3/13 [(1,-9),(-3,1)]`

Discussion

756.	If A and B are `3xx3` matrices and `abs(A)ne 0,` which of the following are true?A. `abs(AB)=0rArr abs(B) = 0`B. `abs(AB)=0rArr B = 0`C. `abs(A^(-1) ) = abs(A)^-1`D. `abs(A+A) = 2 abs(A)`
Answer» Correct Answer - A::C `abs(AB) = 0 rArr abs(A) abs(B) = 0` `therefore abs(B) = 0 " as "abs(A) ne 0` Also, `abs(A^(-1)) = abs(A)^(-1)`

Discussion

757.	For suitable matrices A, B, the false statement is ___________(a) (AB)T = AT BT (b) (AT)T = A (c) (A – B)T = AT – BT (d) (A + B)T = AT + BT
Answer» Correct option is : (a) (AB)^T = A^T B^T (AB)^T = B^T A^T

757.

For suitable matrices A, B, the false statement is ___________(a) (AB)T = AT BT (b) (AT)T = A (c) (A – B)T = AT – BT (d) (A + B)T = AT + BT

Answer»

Correct option is : (a) (AB)^T = A^T B^T

(AB)^T = B^T A^T

Discussion

758.	If `A=[{:(cos q,sin q),(-sin q, cos q):}]` , then variefy that `A^(2)=[{:( cos 2q, sin 2q),(-sin 2q,cos 2 q):}]`
Answer» We have, `A=[{:(cos q,sin q),(-sin q,cos q):}]` `therefore A^(2)=A.A=[{:(cos q,sin q),(-sin q,cos q):}].[{:(cos q,sin q),(-sin q,cos q):}]` `[{:(cos^(2)q-sin^(2)q,cos q sin q+sin q cos q),(-sin q cos q-cos q sin q,-sin^(2)q+cos^(2)q):}]` `=[{:(cos2q,2sinq cosq),(-2sinq, cos 2q):}] [because cos^(2) theta-sin^(2) theta=cos2 theta]` `=[{:(cos2q,sin 2q),(-sin2q,cos2q):}][because sin2 theta=2 sin theta-cos theta]`

Discussion

759.	If `A=[{:(,1,a),(,0,1):}]` then find `underset(n-oo)(lim)(1)/(n)A^(n)`A. `{:[(0,a),(0,0)]:}`B. `{:[(0,0),(0,0)]:}`C. `{:[(0,1),(0,0)]:}`D. none of these
Answer» Correct Answer - A

Discussion

760.	If A is a matrix of order `mxx m` such that `A^(2) +A + 2I = O`, thenA. A is non-singularB. A is symmetricC. `abs(A)ne 0 `D. `A^(-1) = 1/2(A+I)`
Answer» Correct Answer - A::C::D Here, `A(A + I) = -2I` ...(i) ` rArr abs(A(A +I))=abs(-2I) = (-2)^(m)ne 0 ` Thus, `abs(A) ne 0,` also, `I= -1/2 A(A+I)` [from Eq. (i)] `therefore A^(-1) = -1/2 (A + I)`

Discussion

761.	If `A=[{:(1,2),(-1,3):}]B=[{:(4,0),(1,5):}],C=[{:(2,0),(1,-2):}]` a=4 and b=-2, then show that (i) (a+b)B=aB+bB (ii) a(C-A)=aC-aA (iii) `(bA)^(T)=bA^(T)`
Answer» we have, `A=[{:(1,2),(-1,3):}],B=[{:(4,0),(1,6):}]` `C=[{:(2,0),(1,-2):}]` and a=4,b=-2 (i) ` A+(B+C)=[{:( 1,2) ,(-1,3):}]+[{(:(6,0),(2,3):}]=[{:(7,2),(1, 6):}]` and `(A+B)+C= [{:(5,2),( 0,8):}]+[{:(2,0),(1,-2):}]` `=[{:(7,2),(1,6):}]=A+(B+C)` (ii) (BC)`=[{:(4,0) ,(1,5):}][{:(2,0),(1,-2):}]=[{:(8,0),(7,-10):}]` and ` A(BC)=[{:(1,2),(-1, 3):}][{:(8,0),(7,-10):}]` `=[{:(8+14, 0 -20),(-8+21,0-30):}]=[{:(22,-20),(13,3 0):}]` and Also `(AB)=[{:(1,2),(-1,3):}][{:(4,0),(1,5):}]=[{:(6,10),(-1,15):}]` `(AB)C=[{:(6 ,1 0) ,( -1,15):}][{:(2,0),(1,-2):}]` `=[{:(22,-20),(13,-30):}]=A(BC)` (iii) `(a+ b)B=(4-2)[{:(4, 0),(1,5):}]` `=[{:(8,0),(2,10):}]` and `aB=bB=4B-2B` `=[{:(16,0),(4,20):}]-[{:(8,0),(2,10):}]` `=[{:(8, 0),(2,10):}]` `=(a+b)B` (iv) `(C-A)=[{:(2-1,0-2),(1+1,-2-3):}]=[{:(1,-2),(2,-5):}]` and `a(C-A)=[{:(4,-6),(8,-20):}]` Also, `aC=aA=[{:(8,0),(4-8)-[{:(4,8),(-4,12):}]=[{:( 4,-6),(8,-20):}]` (v) `A^(T)=[{:(1,2) ,( -1,3):}]=[{:(1,-1),(2,3):}]` Now, `(A^(T))^(T)= [{:(1,2),( -1,3):}]^(T)` =A (vi) `(bA)^(T)= [{:(-2,-4),(2,-6) :}]^(T)` `=[{:(-2 ,2),(-4,-6):}]` and `A^(T)=[{:(1,-1),(2,3):}]` `bA^(T)=[{:(-2,2),(-4,-6):}]=(bA)^(T)` (vii) `AB=[{:( 1,2),(-1,3):}][{:(4,0),(1,5):}]=[{:(4+2, 0+12),(-4+3,0+15):}]=[{:(6,10),(-1,15):}]` `therefore (AB)^(T)=[{:( 6,-1),( 10,15):}]` Now, `B^(T)A^(T)=[{:(4,1),(0,5):}][{:(1,-1),(2,3):}]=[{:(6,-1),(10,15):}]` `= (AB)^(T)` (viii) `(A-B) =[{:( 1-4,2-0),(-1-1, 3-5):}]=[{:(-3,2),(-2,-2):}]` `(A-B)C=[{:( -3,2),(-2,-2):}][{:(2,0),(1,-2):}]=[{:(4,-4),(1,- 6):}]` and `BC=[{:(4,0),(1,5):}][{:(2,0),(1 ,-2):}]=[{:(8,0),(7,-10):}]` `therefore AC- BC=[{:(4-8,-4 ,-0) ,(1-7,-6+10):}]` `=(A-B)C` (ix) `(A-B)^(T)=[{:(1-4, 2-0),(-1-1,3-5):}]^(T)` `=[{:(-3,-2),(2,-2):}]=(A-B)^(T)`

Discussion

762.	If `A^(2) - 3 A + 2I = 0,` then A is equal toA. `I`B. `2I`C. `[[3,-2],[1,0]]`D. `[[3,1],[-2,0]]`
Answer» Correct Answer - A::B::C::D `because A^(2) - 3 A + 2I = 0` …(i) `rArr A^(2) - 2 AI + 2I^(2) = 0 ` `rArr (A-I) (A -2I) = 0` `therefore A = I or A = 2I` Characteristic Eq. (i) is `lambda^(2) - 3lambda + 2 = 0 rArr lambda = 1,2` It is clear that alternate (c) and (d) have the characteristic equation `lambda^(2) - 3lambda + 2 = 0 `.

Discussion

763.	If A and B are two matrices such that their product AB is a null matrix, thenA. det `A ne 0 rArr B` must be a null matrixB. det `B ne 0 rArr A ` must be a null matrixC. alteast one of the two matrices must be singularD. if neither det A nor det B is zero, then the given statement is not possible
Answer» Correct Answer - C::D `because AB = 0 ` ` rArr abs(AB) = 0 rArr abs(A) abs(B) = 0` or (det A) (det B) = 0 rArr Either det ` A = 0 or det B - 0` Hence, atleast one of the two matrices must be singular otherwise this statement is not possible.

Discussion

764.	If the matrix `{:[(a,b),(c,d)]:}` is commutative with matrix `{:[(1,1),(0,1)]:}`, thenA. `a=0,b=c`B. `b=0,c=d`C. `c=0,d=a`D. `d=0,a=b`
Answer» Correct Answer - C

Discussion

765.	Matrix m ultiplication is commutative.
Answer» False Since, `AB neBA` is possible when AB and BA are both defined.

Discussion

766.	For what value of x, is the matrix A = \( \begin{bmatrix}0&1 & -2 \\[0.3em]-1 & 0 &3 \\[0.3em]x & -3 &0\end{bmatrix}\) a skew-symmetric matrix?
Answer» We are given that, A =\( \begin{bmatrix}0&1 & -2 \\[0.3em]-1 & 0 &3 \\[0.3em]x & -3 &0\end{bmatrix}\) is a skew-symmetric matrix. We need to find the value of x. Let us understand what skew-symmetric matrix is. A skew-symmetric matrix is a square matrix whose transpose equals its negative, that, it satisfies the condition A skew symmetric matrix ⇔ A^T = - A First, let us find –A. - A =-1 x\( \begin{bmatrix}0&1 & -2 \\[0.3em]-1 & 0 &3 \\[0.3em]x & -3 &0\end{bmatrix}\) ⇒ - A =\( \begin{bmatrix}0&-1 & 2 \\[0.3em]1 & 0 &-3 \\[0.3em]-x & 3 &0\end{bmatrix}\) Let us find the transpose of A. We know that the transpose of a matrix is a new matrix whose rows are the columns of the original. In matrix A, 1^strow of A = (0 ,1 ,-2) 2^nd row of A = (-1 ,0 ,3) 3^rd row of A = (x ,-3,0) In the formation of matrix A^T, 1^st column of A^T = 1^st row of A = (0 ,1 ,-2) 2^ndcolumn of A^T = 2^nd row of A = (-1,0, 3) 3^rd column of A^T = 3^rd row of A = (x, -3 ,0) So, A^T =\( \begin{bmatrix}0&-1 & x \\[0.3em]1 & 0 &-3 \\[0.3em]-2 & 3 &0\end{bmatrix}\) Substituting the matrices –A and A^T , we get - A = A^T ⇒ \( \begin{bmatrix}0&-1 & 2 \\[0.3em]1 & 0 &-3 \\[0.3em]-x & 3 &0\end{bmatrix}\)⁼\( \begin{bmatrix}0&-1 & x \\[0.3em]1 & 0 &-3 \\[0.3em]-2 & 3 &0\end{bmatrix}\) We know by the property of matrices, \(\begin{bmatrix} a_{11}& a_{12} \\[0.3em] a_{21} & a_{22} \\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} b_{11}& b_{12} \\[0.3em] b_{21} & b_{22} \\[0.3em] \end{bmatrix}\) This implies, a₁₁= b₁₁, a₁₂ = b₁₂, a₂₁ = b₂₁ and a₂₂= b₂₂ By comparing the corresponding elements of the two matrices, x = 2 Thus, The value of x = 2.

766.

For what value of x, is the matrix A = \( \begin{bmatrix}0&1 & -2 \\[0.3em]-1 & 0 &3 \\[0.3em]x & -3 &0\end{bmatrix}\) a skew-symmetric matrix?

Answer»

We are given that,

A =\( \begin{bmatrix}0&1 & -2 \\[0.3em]-1 & 0 &3 \\[0.3em]x & -3 &0\end{bmatrix}\) is a skew-symmetric matrix.

We need to find the value of x.

Let us understand what skew-symmetric matrix is.

A skew-symmetric matrix is a square matrix whose transpose equals its negative, that, it satisfies the condition

A skew symmetric matrix ⇔ A^T = - A

First, let us find –A.

- A =-1 x\( \begin{bmatrix}0&1 & -2 \\[0.3em]-1 & 0 &3 \\[0.3em]x & -3 &0\end{bmatrix}\)

⇒ - A =\( \begin{bmatrix}0&-1 & 2 \\[0.3em]1 & 0 &-3 \\[0.3em]-x & 3 &0\end{bmatrix}\)

Let us find the transpose of A.

We know that the transpose of a matrix is a new matrix whose rows are the columns of the original.

In matrix A,

1^strow of A = (0 ,1 ,-2)

2^nd row of A = (-1 ,0 ,3)

3^rd row of A = (x ,-3,0)

In the formation of matrix A^T,

1^st column of A^T = 1^st row of A = (0 ,1 ,-2)

2^ndcolumn of A^T = 2^nd row of A = (-1,0, 3)

3^rd column of A^T = 3^rd row of A = (x, -3 ,0)

So,

A^T =\( \begin{bmatrix}0&-1 & x \\[0.3em]1 & 0 &-3 \\[0.3em]-2 & 3 &0\end{bmatrix}\)

Substituting the matrices –A and A^T , we get

- A = A^T

⇒ \( \begin{bmatrix}0&-1 & 2 \\[0.3em]1 & 0 &-3 \\[0.3em]-x & 3 &0\end{bmatrix}\)⁼\( \begin{bmatrix}0&-1 & x \\[0.3em]1 & 0 &-3 \\[0.3em]-2 & 3 &0\end{bmatrix}\)

We know by the property of matrices,

\(\begin{bmatrix} a_{11}& a_{12} \\[0.3em] a_{21} & a_{22} \\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} b_{11}& b_{12} \\[0.3em] b_{21} & b_{22} \\[0.3em] \end{bmatrix}\)

This implies,

a₁₁= b₁₁,

a₁₂ = b₁₂,

a₂₁ = b₂₁ and

a₂₂= b₂₂

By comparing the corresponding elements of the two matrices,

x = 2

Thus,

The value of x = 2.

Discussion

767.	The statement are True or False:Matrix addition is associative as well as commutative.
Answer» True 1. A+B=B+A (commutative) 2. (A+B)+C= A+(B+C) (associative)

767.

The statement are True or False:Matrix addition is associative as well as commutative.

Answer»

True

1. A+B=B+A (commutative)

2. (A+B)+C= A+(B+C) (associative)

Discussion

768.	State the statement is True or False. If A and B are two square matrices of the same order, then A + B = B + A.
Answer» True Since, matrix addition is commutative i.e., A + B = B +A, where A and B are two square matrices.

768.

State the statement is True or False. If A and B are two square matrices of the same order, then A + B = B + A.

Answer»

True

Since, matrix addition is commutative i.e., A + B = B +A, where A and B are two square matrices.

Discussion

769.	State the statement is True or False. Matrix addition is associative as well as commutative.
Answer» True Matrix addition is associative as well as commutative i.e., (A + B) + C = A + (B + C) and A + B = B + A, where A, B and C are matrices of same order.

769.

State the statement is True or False. Matrix addition is associative as well as commutative.

Answer»

True

Matrix addition is associative as well as commutative i.e., (A + B) + C = A + (B + C) and A + B = B + A, where A, B and C are matrices of same order.

Discussion

770.	State the statement is True or False. Matrix multiplication is commutative.
Answer» False If AB is defined, it is not necessary that BA is defined. Also if AB and BA are defined, it not”necessary that they have same order. Further if AB and BA are defined and have same order, it is not necessary their corresponding elements are equal. So, in general AB^BA

770.

State the statement is True or False. Matrix multiplication is commutative.

Answer»

False

If AB is defined, it is not necessary that BA is defined.

Also if AB and BA are defined, it not”necessary that they have same order. Further if AB and BA are defined and have same order, it is not necessary their corresponding elements are equal.

So, in general AB^BA

Discussion

771.	State the statement is True or False. A square matrix where every element is unity is called an identity matrix.
Answer» False Since, in an identity matrix, the diagonal elements are one and rest are all zero.

771.

State the statement is True or False. A square matrix where every element is unity is called an identity matrix.

Answer»

False

Since, in an identity matrix, the diagonal elements are one and rest are all zero.

Discussion

772.	Let A and B be square matrices of the order 3 × 3. Is (AB)2 = A2B2 ? Give reasons.
Answer» We are given that, A and B are square matrices of the order 3 × 3. We need to check whether (AB)² = A²B² is true or not. Take (AB)². (AB)² = (AB)(AB) [∵ A and B are of order (3 × 3) each, A and B can be multiplied; A and B be any matrices of order (3 × 3)] ⇒ (AB)² = ABAB [∵ (AB)(AB) = ABAB] ⇒ (AB)² = AABB [∵ ABAB = AABB; as A can be multiplied with itself and B can be multiplied by itself] ⇒ (AB)² = A²B² So, note that, (AB)² = A²B² is possible. But this is possible if and only if BA = AB. And BA = AB is always true whenever A and B are square matrices of any order. And for BA = AB, (AB)² = A²B²

772.

Let A and B be square matrices of the order 3 × 3. Is (AB)2 = A2B2 ? Give reasons.

Answer»

We are given that,

A and B are square matrices of the order 3 × 3.

We need to check whether (AB)² = A²B² is true or not.

Take (AB)².

(AB)² = (AB)(AB)

[∵ A and B are of order (3 × 3) each, A and B can be multiplied; A and B be any matrices of order (3 × 3)]

⇒ (AB)² = ABAB

[∵ (AB)(AB) = ABAB]

⇒ (AB)² = AABB

[∵ ABAB = AABB; as A can be multiplied with itself and B can be multiplied by itself]

⇒ (AB)² = A²B²

So, note that, (AB)² = A²B² is possible.

But this is possible if and only if BA = AB.

And BA = AB is always true whenever A and B are square matrices of any order. And for BA = AB,

(AB)² = A²B²

Discussion

773.	If \(\begin{bmatrix}2x+y &3y \\[0.3em]0 & 4 \\[0.3em]\end{bmatrix}\)= \(\begin{bmatrix}6&0 \\[0.3em]6 & 4 \\[0.3em]\end{bmatrix}\), then find x.
Answer» We are given that, \(\begin{bmatrix}2x+y &3y \\[0.3em]0 & 4 \\[0.3em]\end{bmatrix}\)= \(\begin{bmatrix}6&0 \\[0.3em]6 & 4 \\[0.3em]\end{bmatrix}\) We need to find the value of x. We know by the property of matrices, \(\begin{bmatrix} a_{11}& a_{12} \\[0.3em] a_{21} & a_{22} \\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} b_{11}& b_{12} \\[0.3em] b_{21} & b_{22} \\[0.3em] \end{bmatrix}\) This implies, a₁₁= b₁₁, a₁₂ = b₁₂, a₂₁ = b₂₁ and a₂₂= b₂₂ So, if we have \(\begin{bmatrix}2x+y &3y \\[0.3em]0 & 4 \\[0.3em]\end{bmatrix}\)= \(\begin{bmatrix}6&0 \\[0.3em]6 & 4 \\[0.3em]\end{bmatrix}\) Corresponding elements of two elements are equal. That is, 2x + y = 6 …(i) 3y = 0 …(ii) To solve for x, We have equations (i) and (ii). We can’t solve for x using only equation (i) as equation (i) contains x as well as y. We need to find the value of y from equation (ii) first. From equation (ii), 3y = 0 ⇒ y = \(\frac{0}{3}\) ⇒ y = 0 Substituting y = 0 in equation (i), 2x + y = 6 ⇒ 2x + (0) = 6 ⇒ 2x = 6 – 0 ⇒ 2x = 6 ⇒ x = \(\frac{6}{2}\) ⇒ x = 3 Thus, we get x = 3.

773.

If \(\begin{bmatrix}2x+y &3y \\[0.3em]0 & 4 \\[0.3em]\end{bmatrix}\)= \(\begin{bmatrix}6&0 \\[0.3em]6 & 4 \\[0.3em]\end{bmatrix}\), then find x.

Answer»

We are given that,

\(\begin{bmatrix}2x+y &3y \\[0.3em]0 & 4 \\[0.3em]\end{bmatrix}\)= \(\begin{bmatrix}6&0 \\[0.3em]6 & 4 \\[0.3em]\end{bmatrix}\)

We need to find the value of x.

We know by the property of matrices,

\(\begin{bmatrix} a_{11}& a_{12} \\[0.3em] a_{21} & a_{22} \\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} b_{11}& b_{12} \\[0.3em] b_{21} & b_{22} \\[0.3em] \end{bmatrix}\)

This implies,

a₁₁= b₁₁,

a₁₂ = b₁₂,

a₂₁ = b₂₁ and

a₂₂= b₂₂

So, if we have

\(\begin{bmatrix}2x+y &3y \\[0.3em]0 & 4 \\[0.3em]\end{bmatrix}\)= \(\begin{bmatrix}6&0 \\[0.3em]6 & 4 \\[0.3em]\end{bmatrix}\)

Corresponding elements of two elements are equal.

That is,

2x + y = 6 …(i)

3y = 0 …(ii)

To solve for x,

We have equations (i) and (ii).

We can’t solve for x using only equation (i) as equation (i) contains x as well as y.

We need to find the value of y from equation (ii) first.

From equation (ii),

3y = 0

⇒ y = \(\frac{0}{3}\)

⇒ y = 0

Substituting y = 0 in equation (i),

2x + y = 6

⇒ 2x + (0) = 6

⇒ 2x = 6 – 0

⇒ 2x = 6

⇒ x = \(\frac{6}{2}\)

⇒ x = 3

Thus, we get

x = 3.

Discussion

774.	If matrix A = [1 2 3] , write AAT.
Answer» We are given that, A = [1 2 3] We need to compute AA^T . We know that the transpose of a matrix is a new matrix whose rows are the columns of the original. So, Transpose of matrix A will be given as A^T = \( \begin{bmatrix} 1 \\[0.3em] 2 \\[0.3em] 3 \end{bmatrix}\) Multiplying A by A^T, AA^T = [1 2 3]\( \begin{bmatrix} 1 \\[0.3em] 2 \\[0.3em] 3 \end{bmatrix}\) In multiplication of matrices, [ a₁₁ a₁₂ a₁₃]\( \begin{bmatrix} b_{11} \\[0.3em] b_{21} \\[0.3em] b_{31} \end{bmatrix}\) Dot multiply the matching members of 1^strow of first matrix and 1 st column of second matrix and then sum up. (a₁₁ a₁₂ a₁₃)(b₁₁ b₂₁ b₃₁) = a₁₁ × b₁₁ + a₁₂ × b₂₁ + a₁₃× b₃₁ So, (1 2 3)(1 2 3) = 1 × 1 + 2 × 2 + 3 × 3 ⇒ (1 2 3)(1 2 3) = 1 + 4 + 9 ⇒ (1 2 3)(1 2 3) = 14 Thus, [1 2 3]\( \begin{bmatrix} 1 \\[0.3em] 2 \\[0.3em] 3 \end{bmatrix}\)= [14]

774.

If matrix A = [1 2 3] , write AAT.

Answer»

We are given that,

A = [1 2 3]

We need to compute AA^T .

We know that the transpose of a matrix is a new matrix whose rows are the columns of the original.

So,

Transpose of matrix A will be given as

A^T = \( \begin{bmatrix} 1 \\[0.3em] 2 \\[0.3em] 3 \end{bmatrix}\)

Multiplying A by A^T,

AA^T = [1 2 3]\( \begin{bmatrix} 1 \\[0.3em] 2 \\[0.3em] 3 \end{bmatrix}\)

In multiplication of matrices,

[ a₁₁ a₁₂ a₁₃]\( \begin{bmatrix} b_{11} \\[0.3em] b_{21} \\[0.3em] b_{31} \end{bmatrix}\)

Dot multiply the matching members of 1^strow of first matrix and 1 st column of second matrix and then sum up.

(a₁₁ a₁₂ a₁₃)(b₁₁ b₂₁ b₃₁) = a₁₁ × b₁₁ + a₁₂ × b₂₁ + a₁₃× b₃₁

So,

(1 2 3)(1 2 3) = 1 × 1 + 2 × 2 + 3 × 3

⇒ (1 2 3)(1 2 3) = 1 + 4 + 9

⇒ (1 2 3)(1 2 3) = 14

Thus,

[1 2 3]\( \begin{bmatrix} 1 \\[0.3em] 2 \\[0.3em] 3 \end{bmatrix}\)= [14]

Discussion

775.	Find the value of x if \(\begin{bmatrix} 3x+y& -y \\[0.3em] 2y-x &3\\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} 1& 2\\[0.3em] -5 & 3 \\[0.3em] \end{bmatrix}\).
Answer» We are given that, \(\begin{bmatrix} 3x+y& -y \\[0.3em] 2y-x &3\\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} 1& 2\\[0.3em] -5 & 3 \\[0.3em] \end{bmatrix}\) We need to find the values of x and y. We know by the property of matrices, \(\begin{bmatrix} a_{11}& a_{12} \\[0.3em] a_{21} & a_{22} \\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} b_{11}& b_{12} \\[0.3em] b_{21} & b_{22} \\[0.3em] \end{bmatrix}\) This implies, a₁₁= b₁₁, a₁₂ = b₁₂, a₂₁ = b₂₁ and a₂₂= b₂₂ So, if we have \(\begin{bmatrix} 3x+y& -y \\[0.3em] 2y-x &3\\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} 1& 2\\[0.3em] -5 & 3 \\[0.3em] \end{bmatrix}\) Corresponding elements of two matrices are equal. That is, 3x + y = 1 …(i) - y = 2 …(ii) 2y – x = - 5 …(iii) 3 = 3 To solve for x and y, We have equations (i), (ii) and (iii). From equation (ii), -y = 2 Multiplying both sides by -1, -1 × -y = - 1 × 2 ⇒ y = - 2 Substituting y = - 2 in either of the equations (i) or (iii), say (i) 3x + y = 1 ⇒ 3x + (-2) = 1 ⇒ 3x – 2 = 1 ⇒ 3x = 1 + 2 ⇒ 3x = 3 ⇒ x = \(\frac{3}{3}\) ⇒ x = 1 Thus, We get x = 1 and y = - 2.

775.

Find the value of x if \(\begin{bmatrix} 3x+y& -y \\[0.3em] 2y-x &3\\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} 1& 2\\[0.3em] -5 & 3 \\[0.3em] \end{bmatrix}\).

Answer»

We are given that,

\(\begin{bmatrix} 3x+y& -y \\[0.3em] 2y-x &3\\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} 1& 2\\[0.3em] -5 & 3 \\[0.3em] \end{bmatrix}\)

We need to find the values of x and y.

We know by the property of matrices,

\(\begin{bmatrix} a_{11}& a_{12} \\[0.3em] a_{21} & a_{22} \\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} b_{11}& b_{12} \\[0.3em] b_{21} & b_{22} \\[0.3em] \end{bmatrix}\)

This implies,

a₁₁= b₁₁,

a₁₂ = b₁₂,

a₂₁ = b₂₁ and

a₂₂= b₂₂

So, if we have

\(\begin{bmatrix} 3x+y& -y \\[0.3em] 2y-x &3\\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} 1& 2\\[0.3em] -5 & 3 \\[0.3em] \end{bmatrix}\)

Corresponding elements of two matrices are equal.

That is,

3x + y = 1 …(i)

- y = 2 …(ii)

2y – x = - 5 …(iii)

3 = 3

To solve for x and y,

We have equations (i), (ii) and (iii).

From equation (ii),

-y = 2

Multiplying both sides by -1,

-1 × -y = - 1 × 2

⇒ y = - 2

Substituting y = - 2 in either of the equations (i) or (iii), say (i)

3x + y = 1

⇒ 3x + (-2) = 1

⇒ 3x – 2 = 1

⇒ 3x = 1 + 2

⇒ 3x = 3

⇒ x = \(\frac{3}{3}\)

⇒ x = 1

Thus, We get

x = 1 and y = - 2.

Discussion

776.	If A = \(\begin{bmatrix} 4 & 3 \\[0.3em] 1 & 2 \end{bmatrix}\) and B = \(\begin{bmatrix} -4 \\[0.3em] 3 \end{bmatrix},\) write AB.
Answer» Given, A = \(\begin{bmatrix} 4 & 3 \\[0.3em] 1 & 2 \end{bmatrix}\) and B = \(\begin{bmatrix} -4 \\[0.3em] 3 \end{bmatrix}\) So, AB will be, AB = \(\begin{bmatrix} 4 & 3 \\[0.3em] 1 & 2 \end{bmatrix}\) x \(\begin{bmatrix} -4 \\[0.3em] 3 \end{bmatrix}\) = \(\begin{bmatrix} -16+9 \\[0.3em] -4+6 \end{bmatrix}\) = \(\begin{bmatrix} -7 \\[0.3em] 2 \end{bmatrix}\)

776.

If A = \(\begin{bmatrix} 4 & 3 \\[0.3em] 1 & 2 \end{bmatrix}\) and B = \(\begin{bmatrix} -4 \\[0.3em] 3 \end{bmatrix},\) write AB.

Answer»

Given,

A = \(\begin{bmatrix} 4 & 3 \\[0.3em] 1 & 2 \end{bmatrix}\) and B = \(\begin{bmatrix} -4 \\[0.3em] 3 \end{bmatrix}\)

So,

AB will be,

AB = \(\begin{bmatrix} 4 & 3 \\[0.3em] 1 & 2 \end{bmatrix}\) x \(\begin{bmatrix} -4 \\[0.3em] 3 \end{bmatrix}\)

= \(\begin{bmatrix} -16+9 \\[0.3em] -4+6 \end{bmatrix}\)

= \(\begin{bmatrix} -7 \\[0.3em] 2 \end{bmatrix}\)

Discussion

777.	Find the value of y if \(\begin{bmatrix} x-y& 2 \\[0.3em] x &5\\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} 2& 2\\[0.3em] 3 & 5 \\[0.3em] \end{bmatrix}\).
Answer» We are given that, \(\begin{bmatrix} x-y& 2 \\[0.3em] x &5\\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} 2& 2\\[0.3em] 3 & 5 \\[0.3em] \end{bmatrix}\) We need to find the values of x and y. We know by the property of matrices, \(\begin{bmatrix} a_{11}& a_{12} \\[0.3em] a_{21} & a_{22} \\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} b_{11}& b_{12} \\[0.3em] b_{21} & b_{22} \\[0.3em] \end{bmatrix}\) This implies, a₁₁= b₁₁, a₁₂ = b₁₂, a₂₁ = b₂₁ and a₂₂= b₂₂ So, if we have \(\begin{bmatrix} x-y& 2 \\[0.3em] x &5\\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} 2& 2\\[0.3em] 3 & 5 \\[0.3em] \end{bmatrix}\) Corresponding elements of two matrices are equal. That is, x – y = 2 …(i) 2 = 2 x = 3 …(ii) 5 = 5 To solve for x and y, We have equations (i) and (ii). From equation (ii), x = 3 Substituting x = 3 in equation (i), we get 3 – y = 2 ⇒ y = 3 – 2 ⇒ y = 1 Thus, We get x = 3 and y = 1.

777.

Find the value of y if \(\begin{bmatrix} x-y& 2 \\[0.3em] x &5\\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} 2& 2\\[0.3em] 3 & 5 \\[0.3em] \end{bmatrix}\).

Answer»

We are given that,

\(\begin{bmatrix} x-y& 2 \\[0.3em] x &5\\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} 2& 2\\[0.3em] 3 & 5 \\[0.3em] \end{bmatrix}\)

We need to find the values of x and y.

We know by the property of matrices,

\(\begin{bmatrix} a_{11}& a_{12} \\[0.3em] a_{21} & a_{22} \\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} b_{11}& b_{12} \\[0.3em] b_{21} & b_{22} \\[0.3em] \end{bmatrix}\)

This implies,

a₁₁= b₁₁,

a₁₂ = b₁₂,

a₂₁ = b₂₁ and

a₂₂= b₂₂

So, if we have

\(\begin{bmatrix} x-y& 2 \\[0.3em] x &5\\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} 2& 2\\[0.3em] 3 & 5 \\[0.3em] \end{bmatrix}\)

Corresponding elements of two matrices are equal.

That is,

x – y = 2 …(i)

2 = 2

x = 3 …(ii)

5 = 5

To solve for x and y,

We have equations (i) and (ii).

From equation (ii),

x = 3

Substituting x = 3 in equation (i), we get

3 – y = 2

⇒ y = 3 – 2

⇒ y = 1

Thus,

We get x = 3 and y = 1.

Discussion

778.	If A =\(\begin{bmatrix}1 \\[0.3em]2 \\[0.3em]3\end{bmatrix}\), write AAT.
Answer» Given, A =\(\begin{bmatrix}1 \\[0.3em]2 \\[0.3em]3\end{bmatrix}_{3 \times 1}\) Now, Firstly we find the A^T A^T = [1 2 3]_1x3 So, The product AA^T will be, AA^T = \(\begin{bmatrix}1 \\[0.3em]2 \\[0.3em]3\end{bmatrix}\)[1 2 3] = \(\begin{bmatrix}1 & 2 & 3 \\[0.3em]2 & 4 & 6 \\[0.3em]3 & 6 & 9\end{bmatrix}_{3\times3}\)

778.

If A =\(\begin{bmatrix}1 \\[0.3em]2 \\[0.3em]3\end{bmatrix}\), write AAT.

Answer»

Given,

A =\(\begin{bmatrix}1 \\[0.3em]2 \\[0.3em]3\end{bmatrix}_{3 \times 1}\)

Now,

Firstly we find the A^T

A^T = [1 2 3]_1x3

So,

The product AA^T will be,

AA^T = \(\begin{bmatrix}1 \\[0.3em]2 \\[0.3em]3\end{bmatrix}\)[1 2 3]

= \(\begin{bmatrix}1 & 2 & 3 \\[0.3em]2 & 4 & 6 \\[0.3em]3 & 6 & 9\end{bmatrix}_{3\times3}\)

Discussion

779.	Find the value of x from the following : \(\begin{bmatrix} 2x-y& 5 \\[0.3em] 3 &y\\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} 6& 5\\[0.3em] 3 & -2 \\[0.3em] \end{bmatrix}\).
Answer» We are given with matrix equation \(\begin{bmatrix} 2x-y& 5 \\[0.3em] 3 &y\\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} 6& 5\\[0.3em] 3 & -2 \\[0.3em] \end{bmatrix}\) We need to find the values of x and y. We know by the property of matrices, \(\begin{bmatrix} a_{11}& a_{12} \\[0.3em] a_{21} & a_{22} \\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} b_{11}& b_{12} \\[0.3em] b_{21} & b_{22} \\[0.3em] \end{bmatrix}\) This implies, a₁₁= b₁₁, a₁₂ = b₁₂, a₂₁ = b₂₁ and a₂₂ = b₂₂ So, if we have \(\begin{bmatrix} 2x-y& 5 \\[0.3em] 3 &y\\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} 6& 5\\[0.3em] 3 & -2 \\[0.3em] \end{bmatrix}\) Corresponding elements of two matrices are equal. That is, 2x – y = 6 …(i) 5 = 5 3 = 3 y = - 2 …(ii) To solve for x and y, We have equations (i) and (ii). From equation (ii), y = - 2 Substituting y = - 2 in equation (i), we get 2x – y = 6 ⇒ 2x – (- 2) = 6 ⇒ 2x + 2 = 6 ⇒ 2x = 6 – 2 ⇒ 2x = 4 ⇒ x = \(\frac{4}{2}\) ⇒ x = 2 Thus, We get x = 2 and y = - 2.

779.

Find the value of x from the following : \(\begin{bmatrix} 2x-y& 5 \\[0.3em] 3 &y\\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} 6& 5\\[0.3em] 3 & -2 \\[0.3em] \end{bmatrix}\).

Answer»

We are given with matrix equation
\(\begin{bmatrix} 2x-y& 5 \\[0.3em] 3 &y\\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} 6& 5\\[0.3em] 3 & -2 \\[0.3em] \end{bmatrix}\)

We need to find the values of x and y.

We know by the property of matrices,

\(\begin{bmatrix} a_{11}& a_{12} \\[0.3em] a_{21} & a_{22} \\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} b_{11}& b_{12} \\[0.3em] b_{21} & b_{22} \\[0.3em] \end{bmatrix}\)

This implies,

a₁₁= b₁₁,

a₁₂ = b₁₂,

a₂₁ = b₂₁ and

a₂₂ = b₂₂

So, if we have

\(\begin{bmatrix} 2x-y& 5 \\[0.3em] 3 &y\\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} 6& 5\\[0.3em] 3 & -2 \\[0.3em] \end{bmatrix}\)

Corresponding elements of two matrices are equal.

That is,

2x – y = 6 …(i)

5 = 5

3 = 3

y = - 2 …(ii)

To solve for x and y,

We have equations (i) and (ii).

From equation (ii),

y = - 2

Substituting y = - 2 in equation (i), we get

2x – y = 6

⇒ 2x – (- 2) = 6

⇒ 2x + 2 = 6

⇒ 2x = 6 – 2

⇒ 2x = 4

⇒ x = \(\frac{4}{2}\)

⇒ x = 2

Thus,

We get x = 2 and y = - 2.

Discussion

780.	If \(\begin{bmatrix} x+3& 4 \\[0.3em] y-4 & x+y \\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} 5& 4 \\[0.3em] 3 & 9 \\[0.3em] \end{bmatrix}\), find the value of x and y.
Answer» We are given with, \(\begin{bmatrix} x+3& 4 \\[0.3em] y-4 & x+y \\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} 5& 4 \\[0.3em] 3 & 9 \\[0.3em] \end{bmatrix}\) We need to find the values of x and y. We know by the property of matrices, \(\begin{bmatrix} a_{11}& a_{12} \\[0.3em] a_{21} & a_{22} \\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} b_{11}& b_{12} \\[0.3em] b_{21} & b_{22} \\[0.3em] \end{bmatrix}\) This implies, a₁₁= b₁₁, a₁₂ = b₁₂, a₂₁ = b₂₁ and a₂₂ = b₂₂ So, if we have \(\begin{bmatrix} x+3& 4 \\[0.3em] y-4 & x+y \\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} 5& 4 \\[0.3em] 3 & 9 \\[0.3em] \end{bmatrix}\) Corresponding elements of two matrices are equal. That is, x + 3 = 5 …(i) 4 = 4 y – 4 = 3 …(ii) x + y = 9 …(iii) To solve for x and y, We have three equations (i), (ii) and (iii). From equation (i), x + 3 = 5 ⇒ x = 5 – 3 ⇒ x = 2 From equation (ii), y – 4 = 3 ⇒ y = 3 + 4 ⇒ y = 7 We need not solve equation (iii) as we have got the values of x and y, Thus, The values of x = 2 and y = 7.

780.

If \(\begin{bmatrix} x+3& 4 \\[0.3em] y-4 & x+y \\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} 5& 4 \\[0.3em] 3 & 9 \\[0.3em] \end{bmatrix}\), find the value of x and y.

Answer»

We are given with,

\(\begin{bmatrix} x+3& 4 \\[0.3em] y-4 & x+y \\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} 5& 4 \\[0.3em] 3 & 9 \\[0.3em] \end{bmatrix}\)

We need to find the values of x and y.

We know by the property of matrices,

\(\begin{bmatrix} a_{11}& a_{12} \\[0.3em] a_{21} & a_{22} \\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} b_{11}& b_{12} \\[0.3em] b_{21} & b_{22} \\[0.3em] \end{bmatrix}\)

This implies,

a₁₁= b₁₁,

a₁₂ = b₁₂,

a₂₁ = b₂₁ and

a₂₂ = b₂₂

So, if we have

\(\begin{bmatrix} x+3& 4 \\[0.3em] y-4 & x+y \\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} 5& 4 \\[0.3em] 3 & 9 \\[0.3em] \end{bmatrix}\)

Corresponding elements of two matrices are equal.

That is,

x + 3 = 5 …(i)

4 = 4

y – 4 = 3 …(ii)

x + y = 9 …(iii)

To solve for x and y,

We have three equations (i), (ii) and (iii).

From equation (i),

x + 3 = 5

⇒ x = 5 – 3

⇒ x = 2

From equation (ii),

y – 4 = 3

⇒ y = 3 + 4

⇒ y = 7

We need not solve equation (iii) as we have got the values of x and y,

Thus,

The values of x = 2 and y = 7.

Discussion

781.	Write a square matrix which is both symmetric as well as skew-symmetric.
Answer» We must understand what symmetric matrix is. A symmetric matrix is a square matrix that is equal to its transpose. A symmetric matrix ⇔ A = A^T Now, Let us understand what skew-symmetric matrix is. A skew-symmetric matrix is a square matrix whose transpose equals its negative, that, it satisfies the condition A skew symmetric matrix ⇔ A T = - A And, A square matrix is a matrix with the same number of rows and columns. An n-by-n matrix is known as a square matrix of order n. We need to find a square matrix which is both symmetric as well as skew symmetric. Take a 2 × 2 null matrix. Say, A =\( \begin{bmatrix}0 &0 \\[0.3em]0 & 0 \\[0.3em]\end{bmatrix}\) Let us take transpose of the matrix A. We know that, The transpose of a matrix is a new matrix whose rows are the columns of the original. So, A^T=\( \begin{bmatrix}0 &0 \\[0.3em]0 & 0 \\[0.3em]\end{bmatrix}\) Since, A = A^T . ∴ A is symmetric. Take the same matrix and multiply it with -1. - A = -1 x\( \begin{bmatrix}0 &0 \\[0.3em]0 & 0 \\[0.3em]\end{bmatrix}\) ⇒ - A = -\( \begin{bmatrix}0 &0 \\[0.3em]0 & 0 \\[0.3em]\end{bmatrix}\) ⇒ - A = \( \begin{bmatrix}0 &0 \\[0.3em]0 & 0 \\[0.3em]\end{bmatrix}\) Let us take transpose of the matrix –A. So, - A^T=\( \begin{bmatrix}0 &0 \\[0.3em]0 & 0 \\[0.3em]\end{bmatrix}\) Since, AT = -A ∴ A is skew-symmetric. Thus, A (a null matrix) is both symmetric as well as skew-symmetric.

781.

Write a square matrix which is both symmetric as well as skew-symmetric.

Answer»

We must understand what symmetric matrix is.

A symmetric matrix is a square matrix that is equal to its transpose.

A symmetric matrix ⇔ A = A^T

Now,

Let us understand what skew-symmetric matrix is.

A skew-symmetric matrix is a square matrix whose transpose equals its negative, that, it satisfies the condition

A skew symmetric matrix ⇔ A T = - A

And,

A square matrix is a matrix with the same number of rows and columns. An n-by-n matrix is known as a square matrix of order n.

We need to find a square matrix which is both symmetric as well as skew symmetric.

Take a 2 × 2 null matrix.

Say,

A =\( \begin{bmatrix}0 &0 \\[0.3em]0 & 0 \\[0.3em]\end{bmatrix}\)

Let us take transpose of the matrix A.

We know that,

The transpose of a matrix is a new matrix whose rows are the columns of the original.

So,

A^T=\( \begin{bmatrix}0 &0 \\[0.3em]0 & 0 \\[0.3em]\end{bmatrix}\)

Since,

A = A^T .

∴ A is symmetric.

Take the same matrix and multiply it with -1.

- A = -1 x\( \begin{bmatrix}0 &0 \\[0.3em]0 & 0 \\[0.3em]\end{bmatrix}\)

⇒ - A = -\( \begin{bmatrix}0 &0 \\[0.3em]0 & 0 \\[0.3em]\end{bmatrix}\)

⇒ - A = \( \begin{bmatrix}0 &0 \\[0.3em]0 & 0 \\[0.3em]\end{bmatrix}\)

Let us take transpose of the matrix –A.

So,

- A^T=\( \begin{bmatrix}0 &0 \\[0.3em]0 & 0 \\[0.3em]\end{bmatrix}\)

Since,

AT = -A

∴ A is skew-symmetric.

Thus,

A (a null matrix) is both symmetric as well as skew-symmetric.

Discussion

782.	If A and B are two square matrices of the same order, then A+B=B+A.
Answer» True, Since matrix addition is commutative i.e., A+B=B+A where A and B are two square matrices.

Discussion

783.	If A and B are two m atrices of the same order, then A-B=B-A.
Answer» False Since, the addition of two matrices of same order are commutative.

Discussion

784.	If A dn B be `3xx3` matrices the AB=0 implies (A) `A=0 or B=0` (B) `A=0 and B=0` (C) `\|A\|=0 or \|B\|=0` (D) `\|A\|=0 and \|B\|=0`
Answer» False Since, for two non-zero matrices A and B of same order, it can be possible that AB=0= null matrix

Discussion

785.	Let A; B; C be square matrices of the same order n. If A is a non singular matrix; then `AB = AC` then `B = C`
Answer» If AB=AC=0 then it can be possible that B and C are two non-zero matrices such that `BneC` `therefore A.B=0=A.C` Let `A=[{:(1,0),(0,0):} ]B=[{:(0,0),(1,3):}]` and `C=[{:(0,0),(3,1):}]` `therefore AB=[{:(1,0),(0,0):}][{:(0,0),(1,3):}]=[{:(0,0),(0,0):}]` and `AC=[{:(1,0),(0, 0):}][{:(0,0),(1,3):}]=[{:(0,0),(0,0):}]` and `AC=[{:(1,0),(0,0):}][{:(0,0),(3,1):}]=[{:(0,0),(0,0):}]` `rArr AB=AC` but `BneC`

Discussion

786.	The trace of the matrix`A=[1-5 7 0 7 9 11 8 9]`is(a) 17(b) 25 (c) 3 (d) 12A. 17B. 25C. 3D. 12
Answer» Correct Answer - A

Discussion

787.	Transpose of a column matrix is a column matrix.
Answer» False, Transpose of a column matrix is a row matrix.

Discussion

788.	show that the characterstic roots of an idempotent matrix are either zero or unity.
Answer» Let A be an idempotent matrix, then `A^(2)=A` If `lambda` be an eigen value of the matrix A correspondin to eigen vector X,so that `AX=lambdaX` where `X!=0` from Eq. (ii) , `A(AX)=A(lambdaX)` `rArr (A A)X=lambda(AX)` ` rArr A^(2)x=lambda (lambdaX)` `rArr AX = lambda^(2)X` `rArr lambdaX=lambda^(2)X` `rArr (lambda-lambda^(2))X=0` `rArr lambda-lambda^(2)=0` `therefore lambda=0` or lambda=1`

Discussion

789.	If A and P are the square matrices of the same order and if P be invertible, show that the matrices A and `P^(-1)` have the same characteristic roots.
Answer» Let `P^(-1)AP=B` `therefore " " \|B-lambdaI\|=\|P^(-1)AP-lambdaI\|` `=\|P^(-1)AP-P^(-1)lambdaP\|` `=\|P^(-1)(A-lambdaI)p\|` `=P^(-1)\|\|A-lambdaI\|\|P\|` `=(1)/(\|P\|)\|A-lambda\|\|P\|=\|A-lambdaI\|`

Discussion

790.	State with reason whether the statement is true or false. A, B and C are matrices of order 2 × 2.(A – B)2 = A2 – 2A . B + B2
Answer» False. Laws of algebra for factorization and expansion are not applicable to matrices.

790.

State with reason whether the statement is true or false. A, B and C are matrices of order 2 × 2.(A – B)2 = A2 – 2A . B + B2

Answer»

False.

Laws of algebra for factorization and expansion are not applicable to matrices.

Discussion

791.	State with reason whether the statement is true or false. A, B and C are matrices of order 2 × 2.A2 – B2 = (A + B) (A – B)
Answer» False. Laws of algebra for factorization and expansion are not applicable to matrices.

791.

State with reason whether the statement is true or false. A, B and C are matrices of order 2 × 2.A2 – B2 = (A + B) (A – B)

Answer»

False.

Laws of algebra for factorization and expansion are not applicable to matrices.

Discussion

792.	State with reason whether the statement is true or false. A, B and C are matrices of order 2 × 2. (A – B) . C = A . C – B . C
Answer» True. Multiplication of matrices is distributive over subtraction.

792.

State with reason whether the statement is true or false. A, B and C are matrices of order 2 × 2. (A – B) . C = A . C – B . C

Answer»

True.

Multiplication of matrices is distributive over subtraction.

Discussion

793.	State with reason whether the statement is true or false. A, B and C are matrices of order 2 × 2.A. (B – C) = A . B – A . C
Answer» True. Multiplication of matrices is distributive over subtraction.

793.

State with reason whether the statement is true or false. A, B and C are matrices of order 2 × 2.A. (B – C) = A . B – A . C

Answer»

True.

Multiplication of matrices is distributive over subtraction.

Discussion

794.	State with reason whether the statement is true or false. A, B and C are matrices of order 2 × 2.(A + B) . C = A . C + B . C
Answer» True. Multiplication of matrices is distributive over addition.

794.

State with reason whether the statement is true or false. A, B and C are matrices of order 2 × 2.(A + B) . C = A . C + B . C

Answer»

True.

Multiplication of matrices is distributive over addition.

Discussion

795.	State with reason whether the statement is true or false. A, B and C are matrices of order 2 × 2.(B . C). A = B . (C . A)
Answer» True. Multiplication of matrices is associative.

795.

State with reason whether the statement is true or false. A, B and C are matrices of order 2 × 2.(B . C). A = B . (C . A)

Answer»

True.

Multiplication of matrices is associative.

Discussion

796.	State with reason whether the statement is true or false. A, B and C are matrices of order 2 × 2.A + B = B + A
Answer» True. Addition of matrices is commutative.

796.

State with reason whether the statement is true or false. A, B and C are matrices of order 2 × 2.A + B = B + A

Answer»

True.

Addition of matrices is commutative.

Discussion

797.	The rank of a null matrix is
Answer» Correct Answer - C

Discussion

798.	If A and B are any two 2 × 2 matrices such that AB = BA = B and B is not a zero matrix, what can you say about the matrix A?
Answer» AB = BA = B We know that AI = IA = I, where I is the identity matrix. Hence, B is the identity matrix.

798.

If A and B are any two 2 × 2 matrices such that AB = BA = B and B is not a zero matrix, what can you say about the matrix A?

Answer»

AB = BA = B We know that AI = IA = I, where I is the identity matrix. Hence, B is the identity matrix.

Discussion

799.	The cooperative stores of a particular school has 10 dozen physics books, 8 dozen chemistry books and 5 dozen mathematics books. Their selling prices are Rs 8.30, Rs 3.45 and Rs 4.50 each respectively. Find the total amount the store will receive from selling all the items.
Answer» Given, The details of stock of various types of books. Physics : 10 dozen books Chemistry : 8 dozen books Mathematics : 5 dozen books Hence, the number of dozens of books available in the store can be represented in matrix form with each column corresponding to a different subject as - X = [10 8 5] The price of each of the items is also given. Cost of one physics book = Rs 8.30 ⇒ Cost of one dozen physics books = 12 × Rs 8.30 ∴ Cost of one dozen physics books = Rs 99.60 Cost of one chemistry book = Rs 3.45 ⇒ Cost of one dozen chemistry books = 12 × Rs 3.45 ∴ Cost of one dozen chemistry books = Rs 41.40 Cost of one mathematics book = Rs 4.50 ⇒ Cost of one dozen mathematics books = 12 × Rs 4.50 ∴ Cost of one dozen mathematics books = Rs 54 Hence, the cost of purchasing a dozen books of each subject can be represented in matrix form with each row corresponding to a different subject as – Y =\(\begin{bmatrix} 99.60 \\[0.3em] 41.60 \\[0.3em] 54 \end{bmatrix}\) Now, The amount received by the store upon selling all the available books can be found by taking the product of the matrices X and Y. XY = [10 8 5] \(\begin{bmatrix} 99.60 \\[0.3em] 41.60 \\[0.3em] 54 \end{bmatrix}\) ⇒ XY = [10 × 99.60 + 8 × 41.40 + 5 × 54] ⇒ XY = [996 + 331.20 + 270] ∴ XY = [1597.20] Thus, The total amount the store will receive from selling all the items is Rs 1597.20.

799.

The cooperative stores of a particular school has 10 dozen physics books, 8 dozen chemistry books and 5 dozen mathematics books. Their selling prices are Rs 8.30, Rs 3.45 and Rs 4.50 each respectively. Find the total amount the store will receive from selling all the items.

Answer»

Given,

The details of stock of various types of books.

Physics : 10 dozen books

Chemistry : 8 dozen books

Mathematics : 5 dozen books

Hence, the number of dozens of books available in the store can be represented in matrix form with each column corresponding to a different subject as -

X = [10 8 5]

The price of each of the items is also given.

Cost of one physics book = Rs 8.30

⇒ Cost of one dozen physics books = 12 × Rs 8.30

∴ Cost of one dozen physics books = Rs 99.60

Cost of one chemistry book = Rs 3.45

⇒ Cost of one dozen chemistry books = 12 × Rs 3.45

∴ Cost of one dozen chemistry books = Rs 41.40 Cost of one mathematics book = Rs 4.50

⇒ Cost of one dozen mathematics books = 12 × Rs 4.50

∴ Cost of one dozen mathematics books = Rs 54

Hence, the cost of purchasing a dozen books of each subject can be represented in matrix form with each row corresponding to a different subject as –

Y =\(\begin{bmatrix} 99.60 \\[0.3em] 41.60 \\[0.3em] 54 \end{bmatrix}\)

Now,

The amount received by the store upon selling all the available books can be found by taking the product of the matrices X and Y.

XY = [10 8 5] \(\begin{bmatrix} 99.60 \\[0.3em] 41.60 \\[0.3em] 54 \end{bmatrix}\)

⇒ XY = [10 × 99.60 + 8 × 41.40 + 5 × 54]

⇒ XY = [996 + 331.20 + 270]

∴ XY = [1597.20]

Thus,

The total amount the store will receive from selling all the items is Rs 1597.20.

Discussion

800.	Find the inverse of matrix`[{:(1, -3, " "1),(8, " "2 ," " 0),(-3, " " 2, " " 1):}]`
Answer» `(-1)/(5).[{:(2,0,-3),(1,-1," "0),(-2,-1," "2):}]`

Discussion

Explore topic-wise InterviewSolutions in .

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The rank of a null matrix is

If A and B are any two 2 × 2 matrices such that AB = BA = B and B is not a zero matrix, what can you say about the matrix A?

The cooperative stores of a particular school has 10 dozen physics books, 8 dozen chemistry books and 5 dozen mathematics books. Their selling prices are Rs 8.30, Rs 3.45 and Rs 4.50 each respectively. Find the total amount the store will receive from selling all the items.

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