1192 + Interview Questions in MATRICES IN GENERAL AWARENESS Page 3 InterviewSolution

101.	The inverse of A = \( \begin{bmatrix}0 & 1& 0 \\[0.3em]1 & 0 &0 \\[0.3em]0 & 0 & 1\end{bmatrix}\)[(0,1,0),(1,0,0),(0,0,1)](a) I (b) A (c) A'(d) -I
Answer» Option : (b) A

Discussion

102.	If A is square matrix such that A2 = A, then (I + A)3 – 7A is equal to (A) A (B) I – A (C) I (D) 3A
Answer» Answer is (C) A² = A (I + A)³ – 7A = (I + A) (I+A)² – 7A = (I + A) {(I + A) (I+A)} -7A = (I + A) (I² + 2A + A²) – 7A [v IA = AI = A] = (I + A) (I+ 2A + A) – 7A (A²= A) = (I + A) (I + 3A) – 7A = I² + AI + 3IA + 3A² – 7A = I + A + 3A + 3A – 7A [∵ A² = A & IA = AI = A] =1

102.

If A is square matrix such that A2 = A, then (I + A)3 – 7A is equal to (A) A (B) I – A (C) I (D) 3A

Answer»

Answer is (C)

A² = A

(I + A)³ – 7A = (I + A) (I+A)² – 7A

= (I + A) {(I + A) (I+A)} -7A

= (I + A) (I² + 2A + A²) – 7A

[v IA = AI = A]

= (I + A) (I+ 2A + A) – 7A (A²= A)

= (I + A) (I + 3A) – 7A

= I² + AI + 3IA + 3A² – 7A

= I + A + 3A + 3A – 7A

[∵ A² = A & IA = AI = A] =1

Discussion

103.	If A and B are two matrices of order 3 × m and 3 × n respectively and m = n, then the order of 5A – 2B is :A. m × 3 B. 3 × 3 C. m × n D. 3 × n
Answer» (D). 3 × n m = n If A and B are two matrices of order 3 × m and 3 × n respectively and m = n Then, A & B have same orders as 3 × n each, So the order of (5A – 2B) should be same as 3 × n. Option (D) is the answer.

103.

If A and B are two matrices of order 3 × m and 3 × n respectively and m = n, then the order of 5A – 2B is :A. m × 3 B. 3 × 3 C. m × n D. 3 × n

Answer»

(D). 3 × n

m = n

If A and B are two matrices of order 3 × m and 3 × n respectively and m = n

Then,

A & B have same orders as 3 × n each,

So the order of (5A – 2B) should be same as 3 × n.

Option (D) is the answer.

Discussion

104.	If \(\begin{bmatrix} 2x+y & 4x \\[0.3em] 5x-7 & 4x \\[0.3em] \end{bmatrix}\) =\(\begin{bmatrix} 7 & 7y-13 \\[0.3em] y & x+6 \\[0.3em] \end{bmatrix}\), then the value of x + y is :A. x = 3, y = 1 B. x = 2, y = 3 C. x = 2, y = 4 D. x = 3, y = 3
Answer» (B). x = 2, y = 3 Comparing the equations, 2x + y = 7 & 4x = x + 6 3x = 6, x = 2 2(2) + y = 7 y = 7 – 4 = 3 x = 2 & y = 3 Option (B) is the answer.

104.

If \(\begin{bmatrix} 2x+y & 4x \\[0.3em] 5x-7 & 4x \\[0.3em] \end{bmatrix}\) =\(\begin{bmatrix} 7 & 7y-13 \\[0.3em] y & x+6 \\[0.3em] \end{bmatrix}\), then the value of x + y is :A. x = 3, y = 1 B. x = 2, y = 3 C. x = 2, y = 4 D. x = 3, y = 3

Answer»

(B). x = 2, y = 3

Comparing the equations,

2x + y = 7 & 4x = x + 6

3x = 6, x = 2

2(2) + y = 7

y = 7 – 4 = 3

x = 2 & y = 3

Option (B) is the answer.

Discussion

105.	If A is a square matrix such that A2 = A, then (I + A)3 – 7A is equal to :A. A B. I – A C. I D. 3A
Answer» C. I (I + A)³ = I³ + A³ + 3A²I + 3AI² (Using the identity of (a + b)³ = a³ + b³ + 3ab (a + b)) (I + A)³ = I + A²(A) + 3AI + 3A [I stands for Identity Matrix] (I + A)³ = I + A² + 3A + 3A (I + A)³ = 7A + I (I + A)³ – 7A 7A + I – 7A = I Option (C) is the answer.

105.

If A is a square matrix such that A2 = A, then (I + A)3 – 7A is equal to :A. A B. I – A C. I D. 3A

Answer»

C. I

(I + A)³ = I³ + A³ + 3A²I + 3AI²

(Using the identity of (a + b)³ = a³ + b³ + 3ab (a + b))

(I + A)³ = I + A²(A) + 3AI + 3A

[I stands for Identity Matrix]

(I + A)³ = I + A² + 3A + 3A

(I + A)³ = 7A + I (I + A)³ – 7A

7A + I – 7A

= I

Option (C) is the answer.

Discussion

106.	If I is the identity matrix and A is a square matrix such A2 = A, then what is the value of (I + A)2 – 3A?
Answer» We are given that, I is the identity matrix. A is a square matrix such that A² = A. We need to find the value of (I + A)² – 3A. We must understand what an identity matrix is. An identity matrix is a square matrix in which all the elements of the principal diagonal are ones and all other elements are zeroes. Take, (I+A)² – 3A = (I)²+ (A)² + 2(I)(A) – 3A [∵, by algebraic identity, (x+y)² = x² + y² + 2xy] ⇒(I+A)² – 3A = (I)(I) + A² + 2(IA) – 3A By property of matrix, (I)(I) = I IA = A ⇒(I+A)² – 3A = I + A² + 2A – 3A ⇒(I+A)² – 3A = I + A + 2A – 3A [∵, given in question, A² = A] ⇒ (I+A)² – 3A = I + 3A – 3A ⇒ (I+A)² – 3A = I + 0 ⇒ (I+A)² – 3A = I Thus, The value of (I+A)²– 3A = I.

106.

If I is the identity matrix and A is a square matrix such A2 = A, then what is the value of (I + A)2 – 3A?

Answer»

We are given that,

I is the identity matrix.

A is a square matrix such that A² = A.

We need to find the value of (I + A)² – 3A.

We must understand what an identity matrix is.

An identity matrix is a square matrix in which all the elements of the principal diagonal are ones and all other elements are zeroes.

Take,

(I+A)² – 3A = (I)²+ (A)² + 2(I)(A) – 3A

[∵, by algebraic identity,

(x+y)² = x² + y² + 2xy]

⇒(I+A)² – 3A = (I)(I) + A² + 2(IA) – 3A

By property of matrix,

(I)(I) = I

IA = A

⇒(I+A)² – 3A = I + A² + 2A – 3A

⇒(I+A)² – 3A = I + A + 2A – 3A

[∵, given in question, A² = A]

⇒ (I+A)² – 3A = I + 3A – 3A

⇒ (I+A)² – 3A = I + 0

⇒ (I+A)² – 3A = I

Thus,

The value of (I+A)²– 3A = I.

Discussion

107.	If A is a square matrix such that A2 = I, then (A – I)3 + (A + I)3 – 7A is equal to : A. A B. I – A C. I + A D. 3A
Answer» (A). A Expansion of the given expression, A³ – I³ + 3AI² – 3A²I + A³ + I³ + 3AI² + 3A²I – 7A 2A³ – 7A + 6AI² 2A²A – 7A + 6A 2AI – A {A² = I} 2A – A A Option (A) is the answer.

107.

If A is a square matrix such that A2 = I, then (A – I)3 + (A + I)3 – 7A is equal to : A. A B. I – A C. I + A D. 3A

Answer»

(A). A

Expansion of the given expression,

A³ – I³ + 3AI² – 3A²I + A³ + I³ + 3AI² + 3A²I – 7A

2A³ – 7A + 6AI²

2A²A – 7A + 6A

2AI – A {A² = I}

2A – A

A

Option (A) is the answer.

Discussion

108.	For what values of `k` the set of equations `2x- 3y + 6z - 5t = 3, y -4z + t=1`, `4x-5y+8z-9t = k` hasA.B.C.D.
Answer» Correct Answer - `(i) kne 7 (ii) k=7` Given equation can be written as, `2x- 3y+ 6z= 5t +3` ` y-4z=1-t` `4x-5y+8z=9t+k` which is of the form `AX = B` Let C be the augmented matrix, then `C=[A:B][[2,-3,6,vdots,5t+3],[0,1,-4,vdots,1-t],[4,-5,8,vdots,9t+k]]` Applying `R_(3) rarr R_(3) - 2R_(1)`, then `C=[[2,-3,6,vdots,5t+3],[0,1,-4,vdots,1-t],[0,-1,-4,vdots,-t+k-6]]` `C=[[2,-3,6,vdots,5t+3],[0,1,-4,vdots,1-t],[0,0,0,vdots,k-7]]` (i) Fpr no solution `R_(A)neR_(C)` `therefore kne7` (ii) For infinite number of solutions `R_(A)=R_(C)` `therefore k=7`

Discussion

109.	If `D=diag [2, 3, 4]`, then `D^(-1)=`A. OB. IC. DD. diag `[(1)/(2), (1)/(3), (1)/(4)]`
Answer» Correct Answer - D

Discussion

110.	If `[{:(x,6),(-1,2w):}]+[{:(" "4,x+y),(z+w," "3):}]=3[{:(x,y),(z,w):}]`, find the values of x, y, z, w.
Answer» Correct Answer - `x=2,y=4,z=1,w=3`

Discussion

111.	Given 3[(x, y), (z, w)] = [(x, 6), (-1, 2w)] + [(4, x + y), (z + w, 3)], find the values of x, y, z and w.
Answer» By definition of equality of matrix as the given matrices are equal, their corresponding elements are equal. Comparing the corresponding elements, we get 3x = x + 4⇒ 2x = 4 ⇒ x = 2 and 3y = 6 + x + y ⇒ 2y = 6 + x ⇒ y = (6 + x)/2 Putting the value of x, we get y = (6 + 2)/2 = 8/2 = 4 Now, 3z = −1 + z + w, 2z = −1 + w z = (-1 + w)/2 .....(ii) Now, 3w = 2w + 3 ⇒ w = 3 Putting the value of w in Eq. (i), we get z = (-1 + 3)/2 = 2/2 = 1 Hence, the values of x, y, z and w are 2, 4, 1 and 3.

111.

Given 3[(x, y), (z, w)] = [(x, 6), (-1, 2w)] + [(4, x + y), (z + w, 3)], find the values of x, y, z and w.

Answer»

By definition of equality of matrix as the given matrices are equal, their corresponding elements are equal. Comparing the corresponding elements, we get

3x = x + 4⇒ 2x = 4 ⇒ x = 2

and 3y = 6 + x + y ⇒ 2y = 6 + x

⇒ y = (6 + x)/2

Putting the value of x, we get

y = (6 + 2)/2 = 8/2 = 4

Now, 3z = −1 + z + w, 2z = −1 + w

z = (-1 + w)/2 .....(ii)

Now, 3w = 2w + 3 ⇒ w = 3

Putting the value of w in Eq. (i), we get

z = (-1 + 3)/2 = 2/2 = 1

Hence, the values of x, y, z and w are 2, 4, 1 and 3.

Discussion

112.	Let P, Q and R be invertible matrices of order 3 such `A=PQ^(-1), B=QR^(-1)` and `C=RP^(-1)`. Then the value of det. `(ABC+BCA+CAB)` is equal to _______.
Answer» Correct Answer - 27 We have `A=PQ^(-1), B=QR^(-1), C=RP^(-1)` `:. ABC=I, BCA=I, CAB=I` `:.` det. `(ABC+BCA+CAB)=` det. `(3I)=3^(3)xx` det. `(I)=27`

Discussion

113.	Find the value of a, b and c from the following equations;\(\begin{bmatrix}{a-b} & {2a+c} \\{2a - b} & {3c+d} \\\end{bmatrix}\) = \(\begin{bmatrix}-1 & 5 \\0 & 13 \\\end{bmatrix}\)
Answer» \(\begin{bmatrix} {a-b} & {2a+c} \\ {2a - b} & {3c+d} \\ \end{bmatrix} \) = \(\begin{bmatrix} -1 & 5 \\ 0 & 13 \\ \end{bmatrix} \) ⇒ a – b = -1, 2a + c = 5, 2a – b = 0, 3c + d = 13 ⇒ a – b = -1 2a – b = 0 – a = -1 ⇒ a = 1 We have, a – b = -1 ⇒ 1 – b = -1 ⇒ b = 2 ⇒ 2a + c = 5 ⇒ 2 + c = 5 ⇒ c = 3 ⇒ 3c + d = 13 ⇒ 9 + d = 13 ⇒ d = 4.

113.

Find the value of a, b and c from the following equations;\(\begin{bmatrix}{a-b} & {2a+c} \\{2a - b} & {3c+d} \\\end{bmatrix}\) = \(\begin{bmatrix}-1 & 5 \\0 & 13 \\\end{bmatrix}\)

Answer»

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

⇒ a – b = -1, 2a + c = 5, 2a – b = 0, 3c + d = 13

⇒ a – b = -1

2a – b = 0

– a = -1

⇒ a = 1

We have, a – b = -1 ⇒ 1 – b = -1 ⇒ b = 2

⇒ 2a + c = 5 ⇒ 2 + c = 5 ⇒ c = 3

⇒ 3c + d = 13 ⇒ 9 + d = 13 ⇒ d = 4.

Discussion

114.	1. Consider a 2 × 2 matrix A = [aij], where aij = \(\frac{(i + j)^2}{2}\)2. Write the transpose of A. 3. Show that A is symmetric.
Answer» 1. A = \(\begin{bmatrix}2 & \frac{9}{8} \\\frac{9}{2} & 8 \\\end{bmatrix}\) 2. A^T = \(\begin{bmatrix}2 & \frac{9}{8} \\\frac{9}{2} & 8 \\\end{bmatrix}\) 3. A^T = A therefore symmetric matrix.

114.

1. Consider a 2 × 2 matrix A = [aij], where aij = \(\frac{(i + j)^2}{2}\)2. Write the transpose of A. 3. Show that A is symmetric.

Answer»

1. A = \(\begin{bmatrix}2 & \frac{9}{8} \\\frac{9}{2} & 8 \\\end{bmatrix}\)

2. A^T = \(\begin{bmatrix}2 & \frac{9}{8} \\\frac{9}{2} & 8 \\\end{bmatrix}\)

3. A^T = A therefore symmetric matrix.

Discussion

115.	Apply the given elementary transformation of the following matrices :A = \(\begin{bmatrix} 1 & 0 \\[0.3em] -1 & 3 \\[0.3em] \end{bmatrix}\), R1 ↔ R2A = [1,0,-1,3], R1 ↔ R2
Answer» Given, A = [1,0,-1,3], R₁ ↔ R₂ A = \(\begin{bmatrix} 1 & 0 \\[0.3em] -1 & 3 \\[0.3em] \end{bmatrix}\) By R₁ ↔ R₂, we get, A ~ \(\begin{bmatrix} -1 & 3 \\[0.3em] 1 & 0 \\[0.3em] \end{bmatrix}\)

115.

Apply the given elementary transformation of the following matrices :A = \(\begin{bmatrix} 1 & 0 \\[0.3em] -1 & 3 \\[0.3em] \end{bmatrix}\), R1 ↔ R2A = [1,0,-1,3], R1 ↔ R2

Answer»

Given,

A = [1,0,-1,3], R₁ ↔ R₂

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

By R₁ ↔ R₂, we get,

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

Discussion

116.	A = [5,4,1,3],C1 ↔ C2; B = [3,1,4,5],R1 ↔ R2.A = \( \begin{bmatrix}5& 4 \\[0.3em]1 & 3 \\[0.3em]\end{bmatrix}\),C1 ↔ C2; B = \( \begin{bmatrix}3& 1 \\[0.3em]4 & 5 \\[0.3em]\end{bmatrix}\), R1 ↔ R2.What do you observe?
Answer» Given, A = [5,4,1,3],C₁ ↔ C₂; B = [3,1,4,5],R₁ ↔ R₂. A = \( \begin{bmatrix} 5& 4 \\[0.3em] 1 & 3 \\[0.3em] \end{bmatrix} \) By C₁ ↔ C₂, we get, A ~ \( \begin{bmatrix} 4&5 \\[0.3em] 3 & 1 \\[0.3em] \end{bmatrix} \)....(1) B = \( \begin{bmatrix} 3& 1 \\[0.3em] 4 & 5 \\[0.3em] \end{bmatrix} \) By R₁ ↔ R₂, we get B ~ \( \begin{bmatrix} 4&5 \\[0.3em] 3 & 1 \\[0.3em] \end{bmatrix} \)....(2) From (1) and (2), We observe that the new matrices are equal.

116.

A = [5,4,1,3],C1 ↔ C2; B = [3,1,4,5],R1 ↔ R2.A = \( \begin{bmatrix}5& 4 \\[0.3em]1 & 3 \\[0.3em]\end{bmatrix}\),C1 ↔ C2; B = \( \begin{bmatrix}3& 1 \\[0.3em]4 & 5 \\[0.3em]\end{bmatrix}\), R1 ↔ R2.What do you observe?

Answer»

Given,

A = [5,4,1,3],C₁ ↔ C₂;

B = [3,1,4,5],R₁ ↔ R₂.

A = \( \begin{bmatrix} 5& 4 \\[0.3em] 1 & 3 \\[0.3em] \end{bmatrix} \)

By C₁ ↔ C₂, we get,

A ~ \( \begin{bmatrix} 4&5 \\[0.3em] 3 & 1 \\[0.3em] \end{bmatrix} \)....(1)

B = \( \begin{bmatrix} 3& 1 \\[0.3em] 4 & 5 \\[0.3em] \end{bmatrix} \)

By R₁ ↔ R₂, we get

B ~ \( \begin{bmatrix} 4&5 \\[0.3em] 3 & 1 \\[0.3em] \end{bmatrix} \)....(2)

From (1) and (2),

We observe that the new matrices are equal.

Discussion

117.	Apply the given elementary transformation of the following matrices :B = \( \begin{bmatrix}1&-1& 3 \\[0.3em]2& 5 & 4 \\[0.3em]\end{bmatrix}\), R1 → R1 → R2B = [1,-1,3,2,5,4],R1 → R1 → R2
Answer» Given, B = [1,-1,3,2,5,4],R₁ → R₁ → R₂ B = \( \begin{bmatrix} 1&-1& 3 \\[0.3em] 2& 5 & 4 \\[0.3em] \end{bmatrix} \), R₁ → R₁ → R₂ gives, B ~ \( \begin{bmatrix} -1&-6& -1 \\[0.3em] 2& 5 & 4 \\[0.3em] \end{bmatrix} \)

117.

Apply the given elementary transformation of the following matrices :B = \( \begin{bmatrix}1&-1& 3 \\[0.3em]2& 5 & 4 \\[0.3em]\end{bmatrix}\), R1 → R1 → R2B = [1,-1,3,2,5,4],R1 → R1 → R2

Answer»

Given,

B = [1,-1,3,2,5,4],R₁ → R₁ → R₂

B = \( \begin{bmatrix} 1&-1& 3 \\[0.3em] 2& 5 & 4 \\[0.3em] \end{bmatrix} \),

R₁ → R₁ → R₂ gives,

B ~ \( \begin{bmatrix} -1&-6& -1 \\[0.3em] 2& 5 & 4 \\[0.3em] \end{bmatrix} \)

Discussion

118.	If A is a skew-symmetric and n ∈ N such that (An)T = λAn, write the value of λ.
Answer» Let A is a skew – symmetric matrix, then A^T= - A …(i) Consider, ⇒ (An)^T = λAⁿ [given] ⇒ (A^T)ⁿ = λAⁿ ⇒ (-A)ⁿ = λAⁿ[from (i)] ⇒ (-1)ⁿ(A)ⁿ = λAⁿ Comparing both the sides, we get λ = (-1)ⁿ

118.

If A is a skew-symmetric and n ∈ N such that (An)T = λAn, write the value of λ.

Answer»

Let A is a skew – symmetric matrix, then

A^T= - A …(i)

Consider,

⇒ (An)^T = λAⁿ [given]

⇒ (A^T)ⁿ = λAⁿ

⇒ (-A)ⁿ = λAⁿ[from (i)]

⇒ (-1)ⁿ(A)ⁿ = λAⁿ

Comparing both the sides, we get

λ = (-1)ⁿ

Discussion

119.	If A =\( \begin{pmatrix} λ& 1 \\[0.3em] -1 & -λ \\[0.3em] \end{pmatrix}\), then A-1 does not exist if λ = ………..((λ,1)(-1,-λ))(a) 0 (b) ± 1 (c) 2 (d) 3
Answer» Option : (b) ± 1

Discussion

120.	Find the value of k so that `A^2=8A+kI` where `A=[(1,0),(-1,7)].`
Answer» We have `A^(2)=[{:(1,0),(-1,7):}][{:(1,0),(-1,7):}]=[{:(1-0,0+0),(-1-7,0+49):}]=[{:(1,0),(-8,49):}],` `(8A+kI)=8.[{:(1,0),(-1,7):}]+k.[{:(1,0),(0,1):}]` `=[{:(8,0),(-8,56):}]+[{:(k,0),(0,k):}]=[{:(8+k," "0),(-8,56+k):}]` `:." "A^(2)=8A+kIimplies[{:(1,0),(-8,49):}]=[{:(8+k," "0),(-8,56+k):}]` `implies" "8+k=1" and "56+k=49impliesimpliesk=-7.` Hence, `k=-7`.

Discussion

121.	What is the maximum number of different elements required to form a symmetric matrix of order 12 ?
Answer» Matrix of order 12 require `12xx12=144` elments. 12 elements for `i=j` `(144-12)/2=66` elements for `I lt j` `:.` Maximum number of different elements required are `66+12=78`

Discussion

122.	the maximum number of different elements requried to from a symmetric matrx of order 6 isA. 15B. 17C. 19D. 21
Answer» Correct Answer - D

Discussion

123.	Construct a `3xx2` matrix whose elements are given by `a_(ij)=(2i-j).`
Answer» `A=[{:(1,9),(3,2),(5,4):}]`

Discussion

124.	Construct a `2xx2`matrix `A=[a_(i j)]`whose elements aregiven by `a_(i j)=((i+2j)^2)/2`.
Answer» `[{:((9)/(2),(25)/(2)),(8,16):}]`

Discussion

125.	If `A=diag [a,b,c]` then show that `A^n=diag[a^n,b^n,c^n]` for all `n in N.`
Answer» We shall prove the result by mathematical induction. When n=1, we have `A^(1)=" diag "[a^(1),b^(1),c^(1)]=" diag "[a, b, c]=A.` So, the result is true for n=1. Let it be true for n=m, so that `A^(m)=" diag "[a^(m),b^(m),c^(m)]" "...(i)` `:." "A^(m+1)=A.A^(m)` `=" diag "[a,b,c]." diag "[a^(m),b^(m),c^(m)]`[using (i)] `=[{:(a,0,0),(0,b,0),(0,0,c):}].[{:(a^(m),0,0),(0,b^(m),0),(0,0,c^(m)):}]` `=[{:(a^(m+1),0,0),(0,b^(m+1),0),(0,0,c^(m+1)):}]=" diag "[a^(m+1),b^(m+1),c^(m+1)].`This shows that the result is true for `n=(m+1)`, whenever it is true for n=m, Hence, by the principle of mathematical iduction, the result is true for all `n inN.`

Discussion

126.	If `A=[{:(2,1),(0,4):}]" and B=[{:(3,4,5),(1,2,3):}]` then A and B are matrices of order `2xx2` and `2xx3` respectively.
Answer» So, A and B are not comparable. Hence, `A+B` is not defined.

Discussion

127.	The order of matrix `A+B^(T)` is `4xx3`, then the order of matrix B is _________ .
Answer» Correct Answer - `3xx4`

Discussion

128.	Construct a `4xx3` matrix whose elements are `a_(ij)=(i)/(j).`
Answer» `[{:(1,(1)/(2),(1)/(3)),(2,1,(2)/(3)),(3,(3)/(2),1),(4,2,(4)/(3)):}]`

Discussion

129.	If order of matrix A is `4xx3` and AB is `4xx5`, then the order of matrix B is _________ .
Answer» Correct Answer - `3xx5`

Discussion

130.	If `A`is any `mxxn`such that `A B`and `B A`are both defined showthat `B`is an `nxxm`matrix.
Answer» Since AB is defined, we have number of rows in B=numbers of columns in A=n. Again, since BA is defined, we have number of columns in B= number of rows in A=m. Hence, the order of B is `nxxm`.

Discussion

131.	If A and B are matrices such that AB and A + B both are defined, thenA. A and B can be any two matricesB. A and B are square matrices not necessarily of the same orderC. A, B are square matries of the same orderD. number of columns of A is same as the number of rows of B
Answer» Correct Answer - C

Discussion

132.	`A , B`are two matrices suchthat `A B`and `A+B`are both defined; showthat `A , B`are square matrices ofthe same order.
Answer» Since `(A+B)` is defined, it follows that both A and B are of the same order, say `(mxxn)`. Thus, order of A is `mxxn`. But, AB is defined. So, number of columns in A=number of rows in B. Consequently, n=m. `:.` A and B are square matrices of the same order.

Discussion

133.	If `A_(2xx3), B_(4xx3)` and `C_(2xx4)` are three matrices, then which of the following is/are defined?A. `AC^(T)B`B. `B^(T)C^(T)A`C. `AB^(T)C`D. `A^(T)BC`
Answer» Correct Answer - B Given A is `2 xx 3` matrix B is `4 xx 3` matrix C is ` 2 xx 4` matrix. `C^(T) " is " 4 xx 2 ` matrix `therefore C^(T)B` is not possible `C^(T) = 4 xx 2 A = 2 xx 3` `therefore C^(T)A` is a matrix of order `4 xx3` `B^(T)` is of order ` 3 xx 4, C^(T)A` is of order ` 4 xx 3` `B^(T)(C^(T)A)` is a matrix of order ` 3 xx 3` Option (b) follows.

Discussion

134.	`A , B`are two matrices suchthat `A B`and `A+B`are both defined; showthat `A , B`are square matrices ofthe same order.A. the order of A and B are not same necessarily,B. No. of columns in `A=No.` of rows in B.C. A and B are square matrices of same order.D. No. of rows in A=No. Of columns in B.
Answer» Correct Answer - C N/a

Discussion

135.	If [2 1 3] `[{:(-1,0,-1),(-1,1,0),(0,1,1):}][{:(1),(0),(-1):}]`=A, then find the value of A.
Answer» We have `[2 1 3] [ {:(-1,0,-1),(-1, 1,0),(0,1,1):}][{:(1),(0),(-1):}]=A` `therefore [2 1 3] [{:(-1,0,-1),(-1,1,0),(0,1,1):}]=[-2-1+0 0+1+3-2+0+3]` Now, `[-341] [{:(1),(0),(-1):}]=A` `A=[ -341 ] [{:(1),(0),(-1):}]` `=[-3+0-1]=[-4]`

Discussion

136.	If `A=[{:(1,2),(-2,1):}],B=[{:(2,3),(3,-4):}]` and `C=[{:(1,0),(-1,0):}]`, verfity (i) A(B+C)=AB+AC.
Answer» We have, `A=[{:(1,2),(-2,1):}]B=[{:(2,3),(3,-4):}]` and `C=[{:(1,0),(-1,0):}]` (i) `(AB)=[{:(1,2),(-2,1):}][{:(2,3),(3,-4):}]=[{:(2+6,3-8),(-4+3,-6-4):}]=[{:(8,-5) ,(-1-,-10):}]` and `(AB)C=[{:(8,-5),(-1,-10),(-1 ,- 10):}][{:(1,0),(-1,0):}]` `=[{:( 8+5,0),(-1,+10,0):}]=[{:(13,0),(9,0):}]` Again, `(BC)=[{:(2,3),(3,-4):}][ {:(-1,0),(7,0):}]` and `A(BC)=[{:(1,2),(-2,1):}][{:(-1,0),(7,0):}]` `=[{:(-1+14,0),(+2+7,0):}]=[{:(13,0),(9,0):}]` `therefore (AB)C=A(BC)` (ii) `(B+C)=[{:(2,3),(3,-4):}]+[{:(1,0),(-1,0):}]=[{:(3,3),(2,-4):}]` and `AB=[{:( 1,2),(-2,1):}][{:(3,3),(2,-4):}]` `=[{:(3+4,3-8),(-6+2,-6-4):}][{:(2,3),(3,-4):}]` `=[{:(2+6 ,3-8),(-4+3,-6-4):}]=[{:(8,-6),(-1,-10):}] ` and `AC=[{:( 1,2),(-2,1):}][{:(1, 0),(-1,0):}]=[{:(1-2,0),(-2-1,0):}]=[{:(-1,0),(-3,0):}]` `therefore AB+AC=[{:( 8,-5),(- 1,-10):}]+[{:(-1,0),(-3,0):}]` `rArr AB+AC=[{:(7,-5),(-4,-10):}]` From Eqs. iii and iv `A(B+C)=AB+AC`

Discussion

137.	If `P=[{:(x,0,0),(0,y,0),(0,0,z):}]` and `Q=[{:(a,0,0),(0,b,0),(0,0 ,c):}]` then prove that `PQ=[{:(xa,0,0),(0,yb,0),(0,0,zc):}]=QP`.
Answer» `PQ=[{:( x,0,0),(0,y,0),(0,0,z):}][{:(a,0,0),(0,b,0),(0,0, c):}]=[{:(xa,0,0),(0,yb,0),(0,0,zc):}]` and `QP=[{:(a,0,0),(0,b,0),(0,0,c):}] [{:(x,0,0),(0,y,0),(0,0,y):}]=[{:(ax ,0,0),(0,by,0),(0,0,zc):}]` Thus, we see that `PQ=QP`

Discussion

138.	If `[1" "x" "1][{:(1,2,3),(4,5,6),(3,2,5):}][{:(" "1),(-2),(" "3):}]=O,` find x.
Answer» Correct Answer - `x=(-5)/(3)`

Discussion

139.	If `S=[((sqrt(3)-1)/(2sqrt(2)),(sqrt(3)+1)/(2sqrt(2))),(-((sqrt(3)+1)/(2sqrt(2))),(sqrt(3)-1)/(2sqrt(2)))], A=[(1,0),(-1,1)]` and `P=S ("adj.A") S^(T)`, then find matrix `S^(T) P^(10) S`.
Answer» `S=[((sqrt(3)-1)/(2sqrt(2)),(sqrt(3)+1)/(2 sqrt(2))),(-((sqrt(3)+1)/(2sqrt(2))),(sqrt(3)-1)/(2sqrt(2)))]` `=[("sin "15^(@),cos 15^(@)),(-"cos "15^(@),sin 15^(@))]` `:. SS^(T)=S^(T)S=I` Now, `S^(T) P^(10) S=S^(T)(S ("adj. A")S^(T))^(10)S` `=S^(T)S("adj. A") S^(T) (S("adj. A")S^(T))^(9)S` `=I ("adj. A")S^(T) (S("adj. A")S^(T))^(9)S` `=("adj. A")S^(T)S("adj. A") S^(T) (S("adj. A")S^(T))^(8) S` `=("adj. A")^(2)S^(T) (S("adj. A")S^(T))^(8)S` ... ... `=("adj. A")^(10)` `A=[(1,0),(-1,1)]` `:.` adj. `A=[(1,0),(1,1)]` `:. ("adj. A")^(2)=[(1,0),(1,1)][(1,0),(1,1)]=[(1,0),(2,1)]` `:. ("adj. A")^(3)=[(1,0),(3,1)]` And so on. `:. ("adj. A")^(10)=[(1,0),(10,1)]=S^(T) P^(10) S`

Discussion

140.	If `A=[{:(1,-1),(2,-1):}],B=[{:(a,-1),(b,-1):}]" and "(A+B)^(2)=(A^(2)+B^(2))` then find the values of a and b.
Answer» Correct Answer - a=1, b=4.

Discussion

141.	Let `A` be a square matrix of order 3 such that adj. (adj. (adj. A)) `=[(16,0,-24),(0,4,0),(0,12,4)]`. Then find (i) `\|A\|` (ii) adj. A
Answer» We know that adj. (adj. A) `=\|A\|^(n-2)A`, where n is order of matrix. `:.` adj. (adj. (adj. A))`=\|"adj. A"\|^(n-2)` adj. A `=(\|A\|^(n-1))^((n-2))` adj. A For `n=3`, adj. (adj. (adj. A))`=\|A\|^(2)` adj. `A=[(16,0,-24),(0,4,0),(0,12,4)]` `:. \|A\|^(6)\|"adj. A"\|=256` `implies \|A\|^(6)\|A\|^(2)=2^(8)` `implies \|A\|=2` `implies` adj. `A=1/4 [(16,0,-24),(0,4,0),(0,12,4)]=[(4,0,-6),(0,1,0),(0,3,1)]`

Discussion

142.	Find the matrix A such that `A.[{:(2,3),(4,5):}]=[{:(0,-4),(10," "3):}].`
Answer» `A=[{:(-8,4),(-19,12):}]`

Discussion

143.	If `A=[(1,2,3),(2,1,2),(2,2,3)]B=[(1,2,2),(-2,-1,-2),(2,2,3)]` and `C=[(-1,-2,-2),(2,1,2),(2,2,3)]` then find the value of tr. `(A+B^(T)+3C)`.
Answer» Correct Answer - 17 tr. `(A+B^(T)+3C)=`tr. `(A)+` tr. `(B^(T))+` tr. `(3C)` `=(1+1+3)+(1-1+3)+3(-1+1+3)` `=17`

Discussion

144.	If `A=[[3,-5],[-4,2]]`, find `A^2-5A-14I`.
Answer» `A^2=A*A=[[3,-5],[-4,2]][[3,-5],[-4,2]]` `=[[29,-25],[-30,24]]` `A^2-5A-14I` `=[[29,-25],[-2-,24]]-[[15,-25],[-20,10]]-[[14,0],[0,14]]` `=[[0,0],[0,0]]` Zero matrix.

Discussion

145.	Find the matrix A such that `[[5-7,],[-2,3]]A=[[-16,-6],[7,2]]`.
Answer» `A=[{:(1,-4),(3,-2):}]`

Discussion

146.	The value of for which the system of equations `ax +y+z=0,x+ay+z=0,x+y+z=0` possess a non-null solution isA. 1B. 2C. `-1`D. `-2`
Answer» Correct Answer - A

Discussion

147.	There are two book shops owned by Suresh and Ganesh. Their sales (in Rupees) for books in three subjects – Physics, Chemistry and Mathematics for two months, July and August 2017 are given by two matrices A and B. i. Find the increase in sales in Rupees from July to August 2017. ii. If both book shops got 10% profit in the month of August 2017, find the profit for each bookseller in each subject in that month.
Answer» i. Increase in sales in rupees from July to August 2017 For Suresh: Increase in sales for Physics books = 6650 – 5600= ₹ 1050 Increase in sales for Chemistry books = 7055 – 6750 = ₹ 305 Increase in sales for Mathematics books = 8905 – 8500 = ₹ 405 For Ganesh: Increase in sales for Physics books = 7000 – 6650 = ₹ 350 Increase in sales for Chemistry books = 7500 – 7055 = ₹ 445 Increase in sales for Mathematics books = 10200 – 8905 = ₹ 1295 ii. Both book shops got 10% profit in the month of August 2017. For Suresh: Profit for Physics books = 6650x10/ 100 = ₹ 665 Profit for Chemistry books = 7055 x 10/100 = ₹ 705.50 Profit for Mathematics books = 8905x10/100 = ₹ 890.50 For Ganesh: Profit for Physics books = 7000x10/ 100 = ₹ 700 Profit for Chemistry books = 7500x10/ 100 = ₹ 750 Profit for Mathematics books = 10200x10/ 100 = ₹ 1020

147.

There are two book shops owned by Suresh and Ganesh. Their sales (in Rupees) for books in three subjects – Physics, Chemistry and Mathematics for two months, July and August 2017 are given by two matrices A and B. i. Find the increase in sales in Rupees from July to August 2017. ii. If both book shops got 10% profit in the month of August 2017, find the profit for each bookseller in each subject in that month.

Answer»

i. Increase in sales in rupees from July to August 2017

For Suresh:

Increase in sales for Physics books = 6650 – 5600= ₹ 1050 Increase in sales for Chemistry books = 7055 – 6750 = ₹ 305 Increase in sales for Mathematics books = 8905 – 8500 = ₹ 405

For Ganesh:

Increase in sales for Physics books = 7000 – 6650 = ₹ 350 Increase in sales for Chemistry books = 7500 – 7055 = ₹ 445 Increase in sales for Mathematics books = 10200 – 8905 = ₹ 1295

ii. Both book shops got 10% profit in the month of August 2017.

For Suresh:

Profit for Physics books = 6650x10/ 100 = ₹ 665

Profit for Chemistry books = 7055 x 10/100 = ₹ 705.50

Profit for Mathematics books = 8905x10/100 = ₹ 890.50

For Ganesh:

Profit for Physics books = 7000x10/ 100 = ₹ 700

Profit for Chemistry books = 7500x10/ 100 = ₹ 750

Profit for Mathematics books = 10200x10/ 100 = ₹ 1020

Discussion

148.	If `A=[{:(1,5),(7,12):}]` and `B=[{:(9,1),( 7,8):}]` then find a matrix C such that `3A+5B+2C` is a null matrix.
Answer» We have, `A=[{:(1, 5),(7,12):}]` and `B=[{:(9,1),(7,8):}]` Let `C=[{:(a,b),(c,d):}] therefore 3A+5B+2C =0` `rArr [{:(3,15),(21,36):}]+[{:(45 ,5),(35,40):}]+[{:(2a,2b),(2c,2d):}]=[{:(0,0),(0,0):}]` `rArr [{:(48,2a,20+2b) ,(56+2c, 76+2d):}]=[{:(0,0),(0,0):}]` `rArr 2a+4B=0rArra=-24` Also, `20+2b=0rArrb=-10` `56 +2c=0rArrc=-28` and `76+2d=0rArrd=-38` `therefore C=[{:(-24,-10),(-28,-38):}]`

Discussion

149.	If \(\begin{bmatrix} a+b&2 \\[0.3em] 5 & b \\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} 6&5 \\[0.3em] 2 & 2 \\[0.3em] \end{bmatrix}\), then find a.
Answer» We are given that, \(\begin{bmatrix} a+b&2 \\[0.3em] 5 & b \\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} 6&5 \\[0.3em] 2 & 2 \\[0.3em] \end{bmatrix}\) 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 \(\begin{bmatrix} a+b&2 \\[0.3em] 5 & b \\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} 6&5 \\[0.3em] 2 & 2 \\[0.3em] \end{bmatrix}\) Corresponding elements of two elements are equal. That is, a + b = 6 …(i) b = 4 …(ii) To solve for a, We have equations (i) and (ii). We can’t solve for a using only equation (i) as equation (i) contains a as well as b. We need to find the value of b from equation (ii) first. From equation (ii), b = 4 Substituting the value of b = 4 in equation (i), a + b = 6 ⇒ a + 4 = 6 ⇒ a = 6 – 4 ⇒ a = 2 Thus, We get a = 2.

149.

If \(\begin{bmatrix} a+b&2 \\[0.3em] 5 & b \\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} 6&5 \\[0.3em] 2 & 2 \\[0.3em] \end{bmatrix}\), then find a.

Answer»

We are given that,

\(\begin{bmatrix} a+b&2 \\[0.3em] 5 & b \\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} 6&5 \\[0.3em] 2 & 2 \\[0.3em] \end{bmatrix}\)

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

\(\begin{bmatrix} a+b&2 \\[0.3em] 5 & b \\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} 6&5 \\[0.3em] 2 & 2 \\[0.3em] \end{bmatrix}\)

Corresponding elements of two elements are equal.

That is,

a + b = 6 …(i)

b = 4 …(ii)

To solve for a,

We have equations (i) and (ii).

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

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

From equation (ii),

b = 4

Substituting the value of b = 4 in equation (i),

a + b = 6

⇒ a + 4 = 6

⇒ a = 6 – 4

⇒ a = 2

Thus,

We get a = 2.

Discussion

150.	if \(\begin{vmatrix}2a+b&3a-b\\c+2d&2c-d\end{vmatrix}=\begin{vmatrix}2&3\\4&-1\end{vmatrix}\), find a, b, c and d.[(2a+b, 3a-b) (c+2d) (2c-d)] = [(2, 3) (4, -1)]
Answer» \(\begin{vmatrix}2a+b&3a-b\\c+2d&2c-d\end{vmatrix}=\begin{vmatrix}2&3\\4&-1\end{vmatrix}\) ∴ By equality of matrices, we get 2a + b = 2 ….(i) 3a – b = 3 ….(ii) c + 2d = 4 ….(iii) 2c – d = -1 ….(iv) Adding (i) and (ii), we get 5a = 5 ∴ a = 1 Substituting a = 1 in (i), we get 2(1) + b = 2 ∴ b = 0 By (iii) + (iv) x = 2, we get 5c = 2 ∴ c = 2/5 Substituting c = 2/5 in (iii), we get 2/5 + 2d =4 ∴ 2d = 4 - 2/5 ∴ 2d = 18/5 ∴ d = 9/5

150.

if \(\begin{vmatrix}2a+b&3a-b\\c+2d&2c-d\end{vmatrix}=\begin{vmatrix}2&3\\4&-1\end{vmatrix}\), find a, b, c and d.[(2a+b, 3a-b) (c+2d) (2c-d)] = [(2, 3) (4, -1)]

Answer»

\(\begin{vmatrix}2a+b&3a-b\\c+2d&2c-d\end{vmatrix}=\begin{vmatrix}2&3\\4&-1\end{vmatrix}\)

∴ By equality of matrices, we get 2a + b = 2 ….(i)

3a – b = 3 ….(ii)

c + 2d = 4 ….(iii)

2c – d = -1 ….(iv)

Adding (i) and (ii), we get 5a = 5

∴ a = 1 Substituting a = 1 in (i), we get 2(1) + b = 2

∴ b = 0

By (iii) + (iv) x = 2, we get 5c = 2

∴ c = 2/5

Substituting c = 2/5 in (iii), we get 2/5 + 2d =4

∴ 2d = 4 - 2/5

∴ 2d = 18/5

∴ d = 9/5

Discussion

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If `[1" "x" "1][{:(1,2,3),(4,5,6),(3,2,5):}][{:(" "1),(-2),(" "3):}]=O,` find x.

If `S=[((sqrt(3)-1)/(2sqrt(2)),(sqrt(3)+1)/(2sqrt(2))),(-((sqrt(3)+1)/(2sqrt(2))),(sqrt(3)-1)/(2sqrt(2)))], A=[(1,0),(-1,1)]` and `P=S ("adj.A") S^(T)`, then find matrix `S^(T) P^(10) S`.

If `A=[{:(1,-1),(2,-1):}],B=[{:(a,-1),(b,-1):}]" and "(A+B)^(2)=(A^(2)+B^(2))` then find the values of a and b.

Let `A` be a square matrix of order 3 such that adj. (adj. (adj. A)) `=[(16,0,-24),(0,4,0),(0,12,4)]`. Then find (i) `|A|` (ii) adj. A

Find the matrix A such that `A.[{:(2,3),(4,5):}]=[{:(0,-4),(10," "3):}].`

If `A=[(1,2,3),(2,1,2),(2,2,3)]B=[(1,2,2),(-2,-1,-2),(2,2,3)]` and `C=[(-1,-2,-2),(2,1,2),(2,2,3)]` then find the value of tr. `(A+B^(T)+3C)`.

If `A=[[3,-5],[-4,2]]`, find `A^2-5A-14I`.

Find the matrix A such that `[[5-7,],[-2,3]]A=[[-16,-6],[7,2]]`.

The value of for which the system of equations `ax +y+z=0,x+ay+z=0,x+y+z=0` possess a non-null solution isA. 1B. 2C. `-1`D. `-2`

If `A=[{:(1,5),(7,12):}]` and `B=[{:(9,1),( 7,8):}]` then find a matrix C such that `3A+5B+2C` is a null matrix.

If \(\begin{bmatrix} a+b&amp;2 \\[0.3em] 5 &amp; b \\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} 6&amp;5 \\[0.3em] 2 &amp; 2 \\[0.3em] \end{bmatrix}\), then find a.

if \(\begin{vmatrix}2a+b&amp;3a-b\\c+2d&amp;2c-d\end{vmatrix}=\begin{vmatrix}2&amp;3\\4&amp;-1\end{vmatrix}\), find a, b, c and d.[(2a+b, 3a-b) (c+2d) (2c-d)] = [(2, 3) (4, -1)]

The inverse of A = \( \begin{bmatrix}0 & 1& 0 \\[0.3em]1 & 0 &0 \\[0.3em]0 & 0 & 1\end{bmatrix}\)[(0,1,0),(1,0,0),(0,0,1)](a) I (b) A (c) A'(d) -I

If \(\begin{bmatrix} 2x+y & 4x \\[0.3em] 5x-7 & 4x \\[0.3em] \end{bmatrix}\) =\(\begin{bmatrix} 7 & 7y-13 \\[0.3em] y & x+6 \\[0.3em] \end{bmatrix}\), then the value of x + y is :A. x = 3, y = 1 B. x = 2, y = 3 C. x = 2, y = 4 D. x = 3, y = 3

Find the value of a, b and c from the following equations;\(\begin{bmatrix}{a-b} & {2a+c} \\{2a - b} & {3c+d} \\\end{bmatrix}\) = \(\begin{bmatrix}-1 & 5 \\0 & 13 \\\end{bmatrix}\)

Apply the given elementary transformation of the following matrices :A = \(\begin{bmatrix} 1 & 0 \\[0.3em] -1 & 3 \\[0.3em] \end{bmatrix}\), R1 ↔ R2A = [1,0,-1,3], R1 ↔ R2

A = [5,4,1,3],C1 ↔ C2; B = [3,1,4,5],R1 ↔ R2.A = \( \begin{bmatrix}5& 4 \\[0.3em]1 & 3 \\[0.3em]\end{bmatrix}\),C1 ↔ C2; B = \( \begin{bmatrix}3& 1 \\[0.3em]4 & 5 \\[0.3em]\end{bmatrix}\), R1 ↔ R2.What do you observe?

Apply the given elementary transformation of the following matrices :B = \( \begin{bmatrix}1&-1& 3 \\[0.3em]2& 5 & 4 \\[0.3em]\end{bmatrix}\), R1 → R1 → R2B = [1,-1,3,2,5,4],R1 → R1 → R2

If A =\( \begin{pmatrix} λ& 1 \\[0.3em] -1 & -λ \\[0.3em] \end{pmatrix}\), then A-1 does not exist if λ = ………..((λ,1)(-1,-λ))(a) 0 (b) ± 1 (c) 2 (d) 3

If \(\begin{bmatrix} a+b&2 \\[0.3em] 5 & b \\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} 6&5 \\[0.3em] 2 & 2 \\[0.3em] \end{bmatrix}\), then find a.

if \(\begin{vmatrix}2a+b&3a-b\\c+2d&2c-d\end{vmatrix}=\begin{vmatrix}2&3\\4&-1\end{vmatrix}\), find a, b, c and d.[(2a+b, 3a-b) (c+2d) (2c-d)] = [(2, 3) (4, -1)]