1192 + Interview Questions in MATRICES IN GENERAL AWARENESS Page 20 InterviewSolution

951.	Three students Anil, Nikhil and Sunil went to a stationary shop and purchased some items. Anil purchased 3 books, 2 erasers and 5 scales. Sunil purchased 5 books, 6 scales and 3 erasers whereas Nikhil purchased 2 books, 3 scales and 4 erasers. Represent the given data in the matrix form.
Answer» Correct Answer - `[{:(3,2,5),(5,3,6),(2,4,3):}]`

Discussion

952.	In applying one or more row operations while finding A–1 by elementary row operations, we obtain all zeros in one or more, then A–1 _________.
Answer» Clearly A^-1 does not exist.

Discussion

953.	If `A=[[1 ,2,-3],[ 5, 0, 2],[ 1,-1, 1]], B=[[3,-1, 2],[ 4, 2 ,5],[ 2, 0, 3]]`and `C=[[4, 1, 2],[ 0, 3, 2],[ 1,-2, 3]]`, then compute `(A+B)` and `(B - C)`. Also, verify that `A + (B - C) = (A + B) - C`.
Answer» `A = [[1,2,-3],[5,0,2],[1,-1,1]]` `B = [[3,-1,2],[4,2,5],[2,0,3]]` `C = [[4,1,2],[0,3,2],[1,-2,3]]` `=> A+B = [[1,2,-3],[5,0,2],[1,-1,1]]+ [[3,-1,2],[4,2,5],[2,0,3]]= [[4,1,-1],[9,2,7],[3,-1,4]]` `=> B-C = [[3,-1,2],[4,2,5],[2,0,3]] - [[4,1,2],[0,3,2],[1,-2,3]] = [[-1,-2,0],[4,-1,3],[1,2,0]]` Now, `A+(B-C) = [[1,2,-3],[5,0,2],[1,-1,1]] + [[-1,-2,0],[4,-1,3],[1,2,0]] = [[0,0,-3],[9,-1,5],[2,1,1]]` `(A+B)-C = [[4,1,-1],[9,2,7],[3,-1,4]] - [[4,1,2],[0,3,2],[1,-2,3]] = [[0,0,-3],[9,-1,5],[2,1,1]]` `:. A+(B-C) = (A+B)-C`

Discussion

954.	Compute the products AB and BA whichever exists in each of the following cases :[a,b] \(\begin{bmatrix} c \\[0.3em] d \\[0.3em] \end{bmatrix}\) + [a b c d] \(\begin{bmatrix} a \\[0.3em] b \\[0.3em] c \\[0.3em] d \end{bmatrix}\)
Answer» [a,b] \(\begin{bmatrix} c \\[0.3em] d \\[0.3em] \end{bmatrix}\) + [a b c d] \(\begin{bmatrix} a \\[0.3em] b \\[0.3em] c \\[0.3em] d \end{bmatrix}\) = [ ac + bd] + [a² + b² + c² + d²] = [ ac + bd = a²+ b² + c² + d²] Hence, [a b] \(\begin{bmatrix} c \\[0.3em] d \end{bmatrix}\) + [ a b c d]\(\begin{bmatrix} a \\[0.3em] b \\[0.3em] c \\[0.3em] d \end{bmatrix}\) = [ ac + bd = a²+ b² + c² + d²]

954.

Compute the products AB and BA whichever exists in each of the following cases :[a,b] \(\begin{bmatrix} c \\[0.3em] d \\[0.3em] \end{bmatrix}\) + [a b c d] \(\begin{bmatrix} a \\[0.3em] b \\[0.3em] c \\[0.3em] d \end{bmatrix}\)

Answer»

[a,b] \(\begin{bmatrix} c \\[0.3em] d \\[0.3em] \end{bmatrix}\) + [a b c d] \(\begin{bmatrix} a \\[0.3em] b \\[0.3em] c \\[0.3em] d \end{bmatrix}\)

= [ ac + bd] + [a² + b² + c² + d²]

= [ ac + bd = a²+ b² + c² + d²]

Hence,

[a b] \(\begin{bmatrix} c \\[0.3em] d \end{bmatrix}\) + [ a b c d]\(\begin{bmatrix} a \\[0.3em] b \\[0.3em] c \\[0.3em] d \end{bmatrix}\)

= [ ac + bd = a²+ b² + c² + d²]

Discussion

955.	`"If A" = [{:(x^(2), y), (5, -4):}] " and "A=A^(-1), " then find " [{:(x^(3), y+x), (1, 2x^(2) + y):}]^(-1)` (a) `(1)/(41)[{:(8, 1), (-1, 5):}]` (b) `(1)/(41)[{:(5, 8), (1, -1):}]` (c) `[{:(5, 1), (-1, 8):}]` (d) `(1)/(41)[{:(5, 1), (-1, 8):}]`
Answer» (i) `"Use A"= [{:(a, b), (c, d):}] " and "A=A^(-1)`, then a =d and ad-bc=-1. (ii) Substitute of values of x and y. Find the inverse of that matrix.

Discussion

956.	Compute the following sums :\(\begin{bmatrix}2 & 1 & 3 \\[0.3em]0 & 3 & 5 \\[0.3em]-1 & 2 & 5\end{bmatrix}+\)\(\begin{bmatrix}1 & -2 & 3 \\[0.3em]2 & 6 & 1 \\[0.3em]0 & -3 & 1\end{bmatrix}\)
Answer» \(\begin{bmatrix}2+1 & 1-2 & 3+3 \\[0.3em]0+2 & 3+6 & 5+1 \\[0.3em]-1+0 & 2-3 & 5+1\end{bmatrix}\) \(=\begin{bmatrix}3 & -1 & 6 \\[0.3em]2 & 9 & 6 \\[0.3em]-1 & -1 & 6\end{bmatrix}\) Hence,\(\begin{bmatrix}2 & 1 & 3 \\[0.3em]0 & 3 & 5 \\[0.3em]-1 & 2 & 5\end{bmatrix}+\) \(\begin{bmatrix}1 & -2 & 3 \\[0.3em]2 & 6 & 1 \\[0.3em]0 & -3 & 1\end{bmatrix}\) =\(\begin{bmatrix}3 & -1 & 6 \\[0.3em]2 & 9 & 6 \\[0.3em]-1 & -1 & 6\end{bmatrix}\)

956.

Compute the following sums :\(\begin{bmatrix}2 & 1 & 3 \\[0.3em]0 & 3 & 5 \\[0.3em]-1 & 2 & 5\end{bmatrix}+\)\(\begin{bmatrix}1 & -2 & 3 \\[0.3em]2 & 6 & 1 \\[0.3em]0 & -3 & 1\end{bmatrix}\)

Answer»

\(\begin{bmatrix}2+1 & 1-2 & 3+3 \\[0.3em]0+2 & 3+6 & 5+1 \\[0.3em]-1+0 & 2-3 & 5+1\end{bmatrix}\)

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

Hence,\(\begin{bmatrix}2 & 1 & 3 \\[0.3em]0 & 3 & 5 \\[0.3em]-1 & 2 & 5\end{bmatrix}+\) \(\begin{bmatrix}1 & -2 & 3 \\[0.3em]2 & 6 & 1 \\[0.3em]0 & -3 & 1\end{bmatrix}\) =\(\begin{bmatrix}3 & -1 & 6 \\[0.3em]2 & 9 & 6 \\[0.3em]-1 & -1 & 6\end{bmatrix}\)

Discussion

957.	Compute : `2{:[(2,3,4,1),(-5,11,6,7)]:}+5{:[(11,13,-3,4),(5,-3,4,-7)]:}-6{:[(6,12,4,-2),(-3,5,11,7)]:}`.
Answer» Correct Answer - `{:[(23,-1,-31,34),(33,-23,-34,-63)]:}`

Discussion

958.	If A = [(2,3),(1,2)], B=[(1,3,2),(4,3,1)], C=[(1),(2)], D=[(4, 6, 8), (5, 7, 9)], then which of the sums A + B, B + C, C + D and B + D is defined?
Answer» Only B + D is defined since matrices of the same order can only be added.

Discussion

959.	Solve the simulteneous linear equations 2x-5y =1, 5x + 3y = 18.
Answer» Given system of linear equations can be written in matrix form as shown below: `[{:(2x-5y), (5x + 3y):}] = [(1),(18)]` The LHS matrix can be further written as product of two matrices as shown below: `[{:(2,-5), (5, 3):}][(x),(y)] = [(1),(18)]` or AX =B. `"Here", A = [{:(2,-5), (5, 3):}]` is called coefficient matrix, X = `[(x),(y)]` is called variable matrix and B = `[(1),(18)]` is called constant matrix. Now, we need to find value of x and y, i.e., the matrix X. To find X, pre-multiplying both the sides of Eq. (1) with `A^(-1)` `rArr A^(-1) (AX) = A^(-1) B` `"or "(A^(-1)A) X = A^(-1) B ["since"A (BC) = (AB)C]` `"or " I X = A^(-1) B, [because A^(-1) A= I] "or" X = A^(-1) B, [IX = X]` `X = A^(-1) B` So, to find X we have to find inverse of coefficient matrix (i.e., A) and multiply it with B. `therefore X = [(x), (y)] = [{:(2, -5), (5, 3):}]^(-1)[(1), (18)]` `= (1)/(2 xx 3-(-5) xx 5)[{:(3, 5), (-5, 2):}][(1), (18)]` `= (1)/(31)[{:(3 xx 1 + 5 xx 18), (-5 xx 1 + 2 xx 18):}]` `= (1)/(31)[(93), (31)] = [(3), (1)]` Thus, X = `[(x), (y)] = [(3), (1)] rArr x = 3, y = 1.` Thus, in general any system of linear equations px + qy =a, and rx + sy =b can be represented in matrix form (i.e., AX = B) as `[{:(p, q), (r, s):}] [(x), (y)] = [(a), (b)]`. Here, A is coefficient matrix = `[{:(p, q), (r, s):}]m X " is variables matrix" = [(x), (y)] " and B" = [(a), (b)]` is constant matrix. `X = A^(-1) B = [{:(p, q), (r, s):}]^(-1) [(a), (b)]`

Discussion

960.	If `I_3` is the identily matrix of order 3, then `(I_3)^(-1)=`
Answer» Correct Answer - c

Discussion

961.	Compute the following sums : \(\begin{bmatrix} 3& -2 \\[0.3em] 1 &4 \\[0.3em] \end{bmatrix}\)\(+\begin{bmatrix} -2&4 \\[0.3em]1 & 3 \\[0.3em] \end{bmatrix} \)
Answer» \(=\begin{bmatrix} 3-2& -2+4 \\[0.3em] 1+1 &4+3 \\[0.3em] \end{bmatrix}\) \(=\begin{bmatrix} 1& 2 \\[0.3em] 2 &7 \\[0.3em] \end{bmatrix}\) Hence, \(\begin{bmatrix} 3& -2 \\[0.3em] 1 &4 \\[0.3em] \end{bmatrix}\)\(+\begin{bmatrix} -2&4 \\[0.3em]1 & 3 \\[0.3em] \end{bmatrix} \) = \(\begin{bmatrix} 1& 2 \\[0.3em] 2 &7 \\[0.3em] \end{bmatrix}\)

961.

Compute the following sums : \(\begin{bmatrix} 3& -2 \\[0.3em] 1 &4 \\[0.3em] \end{bmatrix}\)\(+\begin{bmatrix} -2&4 \\[0.3em]1 & 3 \\[0.3em] \end{bmatrix} \)

Answer»

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

\(=\begin{bmatrix} 1& 2 \\[0.3em] 2 &7 \\[0.3em] \end{bmatrix}\)

Hence, \(\begin{bmatrix} 3& -2 \\[0.3em] 1 &4 \\[0.3em] \end{bmatrix}\)\(+\begin{bmatrix} -2&4 \\[0.3em]1 & 3 \\[0.3em] \end{bmatrix} \) = \(\begin{bmatrix} 1& 2 \\[0.3em] 2 &7 \\[0.3em] \end{bmatrix}\)

Discussion

962.	the matrix S is rotation through an angle `45^(@)` and G is th reflection about the line `y=2x`, then `(SG)^(2)` is equal toA. 7IB. 5IC. 3ID. I
Answer» Correct Answer - D

Discussion

963.	If `{:[(4,5)]:}{:[(1,6),(x,3)]:}{:[(-2),(3)]:}=[9]`, then the value of x isA. 9B. -9C. 10D. -10
Answer» Correct Answer - C (i) Evaluate the product in LIIS and equate it to RHS. (ii) Find the product of first two matrices and then multiply product matrix with third matrix of LHS. (iii) Equate the above resultant matrix with [9] and find the value of x.

Discussion

964.	If `A={:[(8),(-1),(-7)]:}andB={:[(2,5,-6)]:}," then find "3A+B^(T)`.A. `{:[(26),(2),(-27)]:}`B. `{:[(36),(4),(-1)]:}`C. `{:[(18),(2),(-9)]:}`D. `{:[(2),(26),(-27)]:}`
Answer» Correct Answer - A Given that `A={:[(8),(-1),(-7)]:}` `B=[2" "5" "-6]` `B^(T)={:[(2),(5),(-6)]:},3A={:[(24),(-3),(-21)]:}` `3A+B^(T)={:[(24+2),(-3+5),(-6-22)]:}={:[(26),(2),(-27)]:}`

Discussion

965.	If `A={:[(2,3,5,6),(-1,2,-3,4)]:}andB={:[(4,-5,3,7),(8,-3,-1,2)]:}`, then find `A^(T)+B^(T)and(A+B)^(T)`. What do you notice ?
Answer» Correct Answer - `(A+B)^(T)=(A^(T)+B^(T))`

Discussion

966.	If `A={:[(6,x),(7,-6)]:}andA^(2)=I`, then x= _______ .A. 5B. `(-5)/(2)`C. -5D. 1
Answer» Correct Answer - C `A={:[(6,x),(7,-6)]:}` `A^(2)={:[(6,x),(7,-7)]:}{:[(6,x),(7,-6)]:}` `={:[(36+7x,6x-6x),(42-42,7x+36)]:}`, `={:[(36+7x,0),(0,36+7x)]:}`. Given, `A^(2)=IrArr{:[(36+7x,0),(0,36+7x)]:}` `={:[(1,0),(0,1)]:}` `36+7x=1` `x=-5`.

Discussion

967.	If `A=[(-2,3),(-1,1)]` then `A^(3) ` is equal toA. 2AB. AC. 2I.D. I
Answer» Correct Answer - D

Discussion

968.	Find the values of a and b if A = B, where \(A=\begin{bmatrix} a+4& 3b \\[0.3em] 8 & -6 \\[0.3em] \end{bmatrix},\)\(B=\begin{bmatrix} 2a+2&b^2+2 \\[0.3em] 8 & b^2-10 \\[0.3em] \end{bmatrix} \)
Answer» Given two matrices are equal. i.e, A = B. \(\begin{bmatrix} a+4& 3b \\[0.3em] 8 & -6 \\[0.3em] \end{bmatrix}\)\(=\begin{bmatrix} 2a+2&b^2+2 \\[0.3em] 8 & b^2-10 \\[0.3em] \end{bmatrix} \) We know that if two matrices are equal then the elements of each matrices are also equal. ∴ a + 4 = 2a + 2 ⇒ a – 2a = 2 – 4 ⇒ – a = – 2 ⇒ a = 2 … (1) And 3b = b² + 2 ⇒ b² – 3b + 2 = 0 ⇒ b² – 2b – b + 2 = 0 ⇒ b(b – 2) – 1(b – 2) = 0 ⇒ (b – 2)(b – 1) = 0 ⇒ b = 2 or 1 … (2) And – 6 = b² – 10 ⇒ b² = – 10 + 6 ⇒ b² = – 4 ⇒ b = ±2i (No real solution) … (3) ∴ a = 2, b = 2 or 1

968.

Find the values of a and b if A = B, where \(A=\begin{bmatrix} a+4& 3b \\[0.3em] 8 & -6 \\[0.3em] \end{bmatrix},\)\(B=\begin{bmatrix} 2a+2&b^2+2 \\[0.3em] 8 & b^2-10 \\[0.3em] \end{bmatrix} \)

Answer»

Given two matrices are equal.

i.e, A = B.

\(\begin{bmatrix} a+4& 3b \\[0.3em] 8 & -6 \\[0.3em] \end{bmatrix}\)\(=\begin{bmatrix} 2a+2&b^2+2 \\[0.3em] 8 & b^2-10 \\[0.3em] \end{bmatrix} \)

We know that if two matrices are equal then the elements of each matrices are also equal.

∴ a + 4 = 2a + 2

⇒ a – 2a = 2 – 4

⇒ – a = – 2

⇒ a = 2 … (1)

And 3b = b² + 2

⇒ b² – 3b + 2 = 0

⇒ b² – 2b – b + 2 = 0

⇒ b(b – 2) – 1(b – 2) = 0

⇒ (b – 2)(b – 1) = 0

⇒ b = 2 or 1 … (2)

And – 6 = b² – 10

⇒ b² = – 10 + 6

⇒ b² = – 4

⇒ b = ±2i

(No real solution) … (3)

∴ a = 2, b = 2 or 1

Discussion

969.	if `\|[4,x+2],[2x-3,x+1]\|` is a symmetric then `x=`A. 3B. 5C. 2D. 4
Answer» Correct Answer - b

Discussion

970.	`{:[(-6,5),(-7,6)]^(-1)=:}`A. `{:[(-6,5),(-7,6)]:}`B. `{:[(6,-5),(-7,6)]:}`C. `{:[(6,5),(7,6)]:}`D. `{:[(6,-5),(7,-6)]:}`
Answer» Correct Answer - a

Discussion

971.	`A_(x)={:[(0,0,x),(0,0,0)]:}` then find the maximum number of possibilities of matrix `B_(x)`, in which x can be placed in `a_(11),a_(21),or a_(31)`, position such `A_(x)B_(x)=O_(2xx1)`.A. 1B. 2C. 3D. Cannot be determined
Answer» Correct Answer - C Check for the possibilities of `B_(x)` such that `A_(x)B_(x)=O`.

Discussion

972.	If `{:A+B=[(1,0),(1,1)]andA-2B=[(-1,1),(0,-1)]:}," then "A=`A. `{:[(1,1),(2,1)]:}`B. `{:[(2//3,1//3),(1//3,2//3)]:}`C. `{:[(1//3,1//3),(2//3,1//3)]:}`D. none of these
Answer» Correct Answer - c

Discussion

973.	If `A=[(2,2,1),(1,3,1),(1,2,2)]` and the sum of eigen values of A is m anda product of eigen values of A is n, then m+n is equal toA. 10B. 12C. 14D. 16
Answer» Correct Answer - B

Discussion

974.	Find the values of x and y if : \(\begin{bmatrix} x+10& y^2+2y \\[0.3em] 0 & -4 \\[0.3em] \end{bmatrix}\)\(=\begin{bmatrix} 3x+4&3 \\[0.3em] 0 & y^2-5y \\[0.3em] \end{bmatrix} \)
Answer» Given two matrices are equal. i.e, A = B. \(\begin{bmatrix} x+10& y^2+2y \\[0.3em] 0 & -4 \\[0.3em] \end{bmatrix}\)\(=\begin{bmatrix} 3x+4&3 \\[0.3em] 0 & y^2-5y \\[0.3em] \end{bmatrix} \) We know that if two matrices are equal then the elements of each matrices are also equal. ∴ x + 10 = 3x + 4 ⇒ x – 3x = 4 – 10 ⇒ – 2x = – 6 ⇒ x = 3 …(1) And y²+ 2y = 3 ⇒ y² + 2y – 3 = 0 ⇒ y² + 3y – y – 3 = 0 ⇒ y(y + 3) – 1(y + 3) = 0 ⇒ (y + 3)(y – 1) = 0 ⇒ y = – 3 or 1 …(2) And y² – 5y = – 4 ⇒ y² – 5y + 4 = 0 ⇒ y² – 4y – y + 4 = 0 ⇒ y(y – 4) – 1(y – 4) = 0 ⇒ (y – 4)(y – 1) = 0 ⇒ y = 4 or 1 …(3) ∴ The common value is x = 3 and y = 1

974.

Find the values of x and y if : \(\begin{bmatrix} x+10& y^2+2y \\[0.3em] 0 & -4 \\[0.3em] \end{bmatrix}\)\(=\begin{bmatrix} 3x+4&3 \\[0.3em] 0 & y^2-5y \\[0.3em] \end{bmatrix} \)

Answer»

Given two matrices are equal.

i.e, A = B.

\(\begin{bmatrix} x+10& y^2+2y \\[0.3em] 0 & -4 \\[0.3em] \end{bmatrix}\)\(=\begin{bmatrix} 3x+4&3 \\[0.3em] 0 & y^2-5y \\[0.3em] \end{bmatrix} \)

We know that if two matrices are equal then the elements of each matrices are also equal.

∴ x + 10 = 3x + 4

⇒ x – 3x = 4 – 10

⇒ – 2x = – 6 ⇒ x = 3 …(1)

And y²+ 2y = 3

⇒ y² + 2y – 3 = 0

⇒ y² + 3y – y – 3 = 0

⇒ y(y + 3) – 1(y + 3) = 0

⇒ (y + 3)(y – 1) = 0

⇒ y = – 3 or 1 …(2)

And y² – 5y = – 4

⇒ y² – 5y + 4 = 0

⇒ y² – 4y – y + 4 = 0

⇒ y(y – 4) – 1(y – 4) = 0

⇒ (y – 4)(y – 1) = 0

⇒ y = 4 or 1 …(3)

∴ The common value is x = 3 and y = 1

Discussion

975.	If `A-2B={:[(3,6),(7,6)]:}andA-3B={:[(2,6),(7,5)]:}`, then the matrix A isA. `{:[(5,6),(7,8)]:}`B. `{:[(5,8),(7,6)]:}`C. `{:[(6,5),(7,8)]:}`D. `{:[(5,8),(2,6)]:}`
Answer» Correct Answer - A (i) Multiply A-2B by 3. (ii) Multiply A-3B by 2, then subtract to get A.

Discussion

976.	For what value of x and y are the following matrices equal?\(A=\begin{bmatrix}2x+1& 2y \\[0.3em]0 & y^2-2y \\[0.3em]\end{bmatrix},\)\(B=\begin{bmatrix}x+3& y^2+2 \\[0.3em]0 & -6 \\[0.3em]\end{bmatrix}\)
Answer» Given two matrices are equal. i.e, A = B. \(\begin{bmatrix}2x+1& 2y \\[0.3em]0 & y^2-2y \\[0.3em]\end{bmatrix}\)\(=\begin{bmatrix}x+3& y^2+2 \\[0.3em]0 & -6 \\[0.3em]\end{bmatrix}\) We know that if two matrices are equal, then the elements of each matrices are also equal. ∴2x + 1 = x + 3 ⇒2x – x = 3 – 1 ⇒x = 2 …(1) And 2y = y² + 2 ⇒ y² – 2y + 2 = 0 ⇒ y = \(\frac{-2±\sqrt{4-8}}{2}\) ⇒ y = \(\frac{-2±2i}{2}\) ⇒ y = \(\frac{2(-1±i)}{2}\) ⇒ y = -1 ± i (No real solutions) … (2) And y² – 5y = – 6 ⇒ y² – 5y + 6 = 0 ⇒ y² – 3y – 2y + 6 = 0 ⇒ y(y – 3) – 2(y – 3) = 0 ⇒ (y – 3)(y – 2) = 0 ⇒ y = 3 or 2 … (3) ∴From the above equations we can say that A and B can’t be equal for any value of y.

976.

For what value of x and y are the following matrices equal?\(A=\begin{bmatrix}2x+1& 2y \\[0.3em]0 & y^2-2y \\[0.3em]\end{bmatrix},\)\(B=\begin{bmatrix}x+3& y^2+2 \\[0.3em]0 & -6 \\[0.3em]\end{bmatrix}\)

Answer»

Given two matrices are equal.

i.e, A = B.

\(\begin{bmatrix}2x+1& 2y \\[0.3em]0 & y^2-2y \\[0.3em]\end{bmatrix}\)\(=\begin{bmatrix}x+3& y^2+2 \\[0.3em]0 & -6 \\[0.3em]\end{bmatrix}\)

We know that if two matrices are equal, then the elements of each matrices are also equal.

∴2x + 1 = x + 3

⇒2x – x = 3 – 1

⇒x = 2 …(1)

And 2y = y² + 2

⇒ y² – 2y + 2 = 0

⇒ y = \(\frac{-2±\sqrt{4-8}}{2}\)

⇒ y = \(\frac{-2±2i}{2}\)

⇒ y = \(\frac{2(-1±i)}{2}\)

⇒ y = -1 ± i

(No real solutions) … (2)

And y² – 5y = – 6

⇒ y² – 5y + 6 = 0

⇒ y² – 3y – 2y + 6 = 0

⇒ y(y – 3) – 2(y – 3) = 0

⇒ (y – 3)(y – 2) = 0

⇒ y = 3

or 2 … (3)

∴From the above equations we can say that A and B can’t be equal for any value of y.

Discussion

977.	Match the following lists :
Answer» `a rarr r, b rarr s, c rarr p, r, d rarr p, q, r, s`. a. Since A is idempotent, `A^(2)=A^(3)=A^(4)= =A`. Now, `(A+I)^(n)=I+ .^(n)C_(1)A+ .^(n)C_(2)A^(2)+ +.^(n)C_(n)A^(n)` `=I+.^(n)C_(1)A+.^(n)C_(2)A+ + .^(n)C_(n)A` `=I+(.^(n)C_(1)+.^(n)C_(1)+ +.^(n)C_(n))A` `=I+(2^(n)-1)A` `implies 2^(n)-1=127` `implies n=7` b. We have, `(I-A) (I+A+A^(2)+ + A^(7))` `=I+A+A^(2)+ +A^(7)+ (-A-A^(2)-A^(3)-A^(4)-A^(8))` `=I-A^(8)` `=I ("if "A^(8)=O)` c. Here matrix A is skew-symmetric and since `\|A\|=\|-A^(T)\|=(-1)^(n)\|A\|`, so `\|A\| (1-(-1)^(n))=0`. As n is odd, hence `\|A\|=0`, Hence, A is singular. d. If A is symmetric, `A^(-1)` is also symmetric for matrix of any order.

Discussion

978.	Match the following lists :
Answer» `a rarr s, b rarr p, c rarr q, d rarr r`. a. Since A is idempotent, we have `A^(2)=A` `implies A^(3)=A A^(2)=A A=A^(2)=A, A^(4)=A A^(3)=A A=A^(2)=A` `implies A^(n)=A` `implies (I-A)^(n)=^(n)C_(0)I-.^(n)C_(1)A+.^(n)C_(2)A^(2)- .^(n)C_(3)A^(3)+` `= I+(-^(n)C_(1)+ .^(n)C_(2)-^(n)C_(3)+)A` `=I+[(.^(n)C_(0)-^(n)C_(1)+^(n)C_(3)+)-^(n)C_(0)]A` `=I-A` b. A is involuntary. Hence, `A^(2)=I` `implies A^(3)=A^(5)= = A` and `A^(2)=A^(4)=A^(6)= =I` `implies (I-A)^(n)=^(n)C_(0)I-^(n)C_(1)A+ .^(n)C_(2)A^(2)-^(n)C_(3)A^(3)+` `=^(n)C_(0)I-^(n)C_(1)A+.^(n)C_(1)I-.^(n)C_(3)A+^(n)C_(4)I-` `=(.^(n)C_(0)+.^(n)C_(2)+.^(n)C_(4)+)I-(.^(n)C_(1)A+.^(n)C_(3)+.^(n)C_(5)+)A` `=2^(n-1) (I-A)` `implies [(I-A)^(n)]A^(-1)=2^(n-1) (I-a)A^(-1)=2^(n-1) (A^(-1)-I)` c. If A is nilpotent of index 2, then `A^(2)=A^(3)=A^(4)=A^(n)=O` `implies (I-A)^(n)=^(n)C_(0)I- .^(n)C_(1)A+.^(n)C_(2)A^(2)- .^(n)C_(3)A^(3)+` `=I-nA+O+O+` `=I-nA` d. A is orthogonal. Hence, `A A^(T)=I` `implies (A^(T))^(-1) =A`

Discussion

979.	If `A{:[(1,2,3,5,6),(4,-1,2,-3,1)]:}andB={:[(5,3),(2,1),(-3,5),(4,1),(1,2)]:}`, then find AB and BA.
Answer» Correct Answer - `X={:[(26,37),(1,20)]:},{:[(17,7,21,16,33),(6,3,8,7,13),(17,-11,1,-30,-13),(8,7,14,17,25),(9,0,7,-1,8)]:}`

Discussion

980.	If `A={:[(3,3),(3,3)]:}andB={:[(1,-1),(-1,1)]:}`, then find AB and BA. What do you observe ?
Answer» Correct Answer - `A={:[(0,0),(0,0)]:}` We can observe that AB=BA=O and none of the two matrices A or B is null matrix.

Discussion

981.	If `A={:[(5,-6),(2,4)]:}andB=A^(T)," then "A^(T)-B^(T)` is ________ .A. `{:[(0,6),(6,0)]:}`B. `{:[(0,8),(-8,0)]:}`C. `{:[(0,-8),(8,0)]:}`D. `{:[(8,0),(0,8)]:}`
Answer» Correct Answer - B `A={:[(5,-6),(2,4)]:}` `B=A^(T)={:[(5,2),(-6,4)]:}` `A^(T){:[(5,2),(-6,4)]:}:B^(T)={:[(5,-6),(2,4)]:}` `A^(T)-B^(T)={:[(5-5,2+6),(-6-2,4-4)]:}={:[(0,8),(-8,0)]:}`

Discussion

982.	Find the matrix X such that – A + 3B + X = 0, where \(\begin{bmatrix}-2&6 \\[0.3em] 5 &8 \end{bmatrix}\) and B = \(\begin{bmatrix}1&2 \\[0.3em] -2 &3 \end{bmatrix}\)
Answer» Given , –A + 3B + X = 0 ⇒ X = A – 3B ⇒ X = \(\begin{bmatrix}-2&6 \\[0.3em] 5 &8 \end{bmatrix}\) -3\(\begin{bmatrix}1&2 \\[0.3em] -2 &3 \end{bmatrix}\) = \(\begin{bmatrix}-2&6 \\[0.3em] 5 &8 \end{bmatrix}\) - \(\begin{bmatrix} 3 & 6 \\[0.3em] -6 & 9 \end{bmatrix}\) = \(\begin{bmatrix}-2-3 &6-6 \\[0.3em] 5-(-6)&8-9 \end{bmatrix}\) = \(\begin{bmatrix}-5&0 \\[0.3em] 11 &-1 \end{bmatrix}\)

982.

Find the matrix X such that – A + 3B + X = 0, where \(\begin{bmatrix}-2&6 \\[0.3em] 5 &8 \end{bmatrix}\) and B = \(\begin{bmatrix}1&2 \\[0.3em] -2 &3 \end{bmatrix}\)

Answer»

Given , –A + 3B + X = 0

⇒ X = A – 3B

⇒ X = \(\begin{bmatrix}-2&6 \\[0.3em] 5 &8 \end{bmatrix}\) -3\(\begin{bmatrix}1&2 \\[0.3em] -2 &3 \end{bmatrix}\)

= \(\begin{bmatrix}-2&6 \\[0.3em] 5 &8 \end{bmatrix}\) - \(\begin{bmatrix} 3 & 6 \\[0.3em] -6 & 9 \end{bmatrix}\) = \(\begin{bmatrix}-2-3 &6-6 \\[0.3em] 5-(-6)&8-9 \end{bmatrix}\) = \(\begin{bmatrix}-5&0 \\[0.3em] 11 &-1 \end{bmatrix}\)

Discussion

983.	For what values of x and y are the following matrices equal`A=[(2x+1,3y),(0,y^2-5y)] B= [(x+3,y^2+2),(0,-6)] `
Answer» With the given matrices we have, `2x+1 = x+3 => x = 2` `3y = y^2+2 =>y^2-3y+2 = 0` `=>y^2-2y-y+2 = 0` `=>y(y-2)-1(y-2) = 0` `=> (y- 2)(y-1) = 0` `=>y = 2 and y = 1` `y^2-5y = -6` `=>y^2-5y+6 = 0` `=>y2-3y-2y+6 = 0` `=>(y-3)(y-2) = 0` `=>y=3 and y = 2` So, `y=2` is the common solution that satisfies both equation. `:. y = 2` So, `x = 2, y =2`

Discussion

984.	Match the following lists : A. `{:(a,b,c,d),(s,r,q,p):}`B. `{:(a,b,c,d),(s,p,q,r):}`C. `{:(a,b,c,d),(q,p,s,r):}`D. `{:(a,b,c,d),(s,q,r,p):}`
Answer» Correct Answer - C a. `\|M_(r)\|=1/(r-1)-1/r` `lim_(n rarr oo) sum_(r=2)^(n) (1/(r-1)-1/r)` `=lim_(n rarr oo) [(1/1-1/2)+(1/2-1/3)+...+(1/(n-1)-1/n)]` `=1-0=1` b. `(A+B)^(2)=A^(2)+B^(2)` `implies AB+BA=O` `implies AB=-BA` `implies \|AB\|=\|-BA\|` `implies \|A\|\|B\|=-\|B\|\|A\|" "` [A and B are odd orderes matrices] `implies \|B\|=-\|B\| (\|A\|=2)` `implies \|B\|=0` c. `k^(2)=\|C\|=("det. A")^(2)=^(2) implies k=4` d. `A^(4)=-4I implies lambda =4`

Discussion

985.	If `A={:[(1,4),(-2,5)]:}andB={:[(-6,-1),(-8,2)]:}`, then AB + BA = ________ .A. `{:[(-42,22),(40,10)]:}`B. `{:[(-42,-22),(-40,-10)]:}`C. `{:[(-42,22),(-40,10)]:}`D. `{:[(42,22),(40,10)]:}`
Answer» Correct Answer - B Given `A={:[(1,4),(-2,5)]:}andB={:[(-6,-1),(-8,2)]:}` `AB={:[(1,4),(-2,5)]:}{:[(-6,-1),(-8,2)]:}` `={:[(1xx-6+4xx-8,1xx-1+4xx2),(-2x-6+5x-8,-2xx-1+5xx2)]:}` `={:[(-6-32,-1+8),(+12-40,2+10)]:}={:[(-38,7),(-28,12)]:}`. `BA={:[(-6,-1),(-8,2)]:}{:[(1,4),(-2,5)]:}` `={:[(-6+2,-24-5),(-8-4,-32+10)]:}` `={:[(-4,-29),(-12,-22)]:}`. `AB+BA={:[(-38-4,7-29),(-28-12,12-22)]:}` `={:[(-42,-22),(-40,-10)]:}`.

Discussion

986.	The order of the matrix \(\begin{bmatrix} -1 \\[0.3em] 3 \\[0.3em] 4 \end{bmatrix}\) is :(a) 1 × 3 (b) 3 × 1 (c) 1 × 1 (d) 3 × 3
Answer» (b) Since there are 3 rows and 1 column, the order is 3 × 1.

Discussion

987.	Consider a matrix `A=[a_("ij")]` of order `3xx3` such that `a_("ij")=(k)^(i+j)` where `k in I`. Match List I with List II and select the correct answer using the codes given below the lists. A. `{:(a,b,c,d),(r,p,s,q):}`B. `{:(a,b,c,d),(s,p,q,r):}`C. `{:(a,b,c,d),(r,p,q,s):}`D. `{:(a,b,c,d),(q,p,r,s):}`
Answer» Correct Answer - C Let `A=[A_("ij")]_(3xx3)`, where `a_("ij")=(k)^(i+j)` So, `A=[(k^(2),k^(3),k^(4)),(k^(3),k^(4),k^(5)),(k^(4),k^(5),k^(6))]` a. If A is singular, then `\|A\|=0` `implies k^(2).k^(3).k^(4) \|(1,1,1),(k,k,k),(k^(2),k^(2),k^(2))\|=0`, `implies k in I` b. If A is null matrix, then `k in {0}` c. There is no value of `k` for `A` to be skew-symmetric matrix which is not null-matrix. `:. k in phi` d. If `A^(2)=3A`, then `[(k^(2),k^(3),k^(4)),(k^(3),k^(4),k^(5)),(k^(4),k^(5),k^(6))][(k^(2),k^(3),k^(4)),(k^(3),k^(4),k^(5)),(k^(4),k^(5),k^(6))]=[(3k^(2),3k^(3),3k^(4)),(3k^(3),3k^(4),3k^(5)),(3k^(4),3k^(5),3k^(6))]` `implies [(k^(4)+k^(6)+k^(8),k^(5)+k^(7)+k^(9),k^(6)+k^(8)+k^(10)),(k^(5)+k^(7)+k^(9),k^(6)+k^(8)+k^(10),k^(7)+k^(9)+k^(11)),(k^(6)+k^(8)+k^(10),k^(7)+k^(9)+k^(11),k^(8)+k^(10)+k^(12))]` `=[(3k^(2),3k^(3),3k^(4)),(3k^(3),3k^(4),3k^(5)),(3k^(4),3k^(5),3k^(6))]` `implies k in {-1, 0, 1}`

Discussion

988.	If `A={:[(2,1),(0,3)]:}andf(x)=x^(2)-4x+3`, then find f(A).A. `{:[(-1,1),(0,0)]:}`B. `{:[(0,0),(1,-1)]:}`C. `{:[(0,1),(2,3)]:}`D. `{:[(0,0),(1,2)]:}`
Answer» Correct Answer - A Given `A={:[(2,1),(0,3)]:}` `f(x)=x^(2)-4x+3` `f(A)=A^(2)-4A+3I` `A^(2)={:[(2,1),(0,3)]:}{:[(2,1),(0,3)]:}` `{:[(4+0,2+3),(0+0,0+9)]:}` `{:[(4,5),(0,9)]:}` `4A={:[(8,4),(0,12)]:}` `3I={:[(3,0),(0,3)]:}` `A^(2)-4A+3I={:[(4-8+3,5-4+0),(0-0+0,9-12+3)]:}` `={:[(-1,1),(0,0)]:}`.

Discussion

989.	If `y={:((4,-2),(3,0)):}andf(y)=y^(2)-y-6`, find f(y).A. `{:((0,6),(9,12)):}`B. `{:((0,6),(9,-12)):}`C. `{:((0,-6),(9,-12)):}`D. `{:((0,6),(9,-12)):}`
Answer» Correct Answer - C

Discussion

990.	If `{:[(1,-1,2),(3,a,1)]:}{:[(-1),(3),(2)]:}={:[(b),(14)]:}`, then find (a+b).
Answer» Correct Answer - 5

Discussion

991.	Which of the following pair of matrices are equal ?
Answer» (c) For equality of matrices, the order should be same and corresponding elements should be equal.

Discussion

992.	If matrix A is given by `A=[[6,11] , [2,4]]` then determinant of `A^(2005)-6A^(2004)` isA. `2^(2006)`B. `(-11)2^(2005)`C. `-2^(2005).7`D. `(-9) 2^(2004)`
Answer» Correct Answer - B `\|A^(2005)-6A^(2004)\|=\|A\|^(2004)\|A-6I\|` `=2^(2004)\|(0,11),(2,-2)\|=(-22) 2^(2004)=(-11) (2)^(2005)`

Discussion

993.	If `P={:[(-1,0),(2,-1)]:}andf(x)=x^(2)-2x+2`, then find f(P).
Answer» Correct Answer - `f(P)={:[(5,0),(-8,5)]:}`

Discussion

994.	A square matrix A has 9 elements. What is the possible order of A ? (a) 1 × 9 (b) 9 × 9 (c) 3 × 3 (d) 2 × 7
Answer» (c) The factors of 9 are 1, 3 and 9 so, the possible orders of a matrix containing 9 elements is 1 ×9, 9 × 1, 3 ×3. In a square matrix the number of rows is equal to the number of columns so the required order is 3 × 3.

Discussion

995.	If `B-A^(T)={:[(3,4,-2),(5,-3,7)]:}andB^(T)+A={:[(2,0),(5,-1),(3,4)]:}`, then find Matrix A.A. `{:[(1//2,-3//2),(1//2,1),(-5//2,3//2)]:}`B. `{:[(1//2,3//2),(1//2,1),(-5//2,3//2)]:}`C. `{:[(-1//2,-5//2),(1//2,1),(5//2,-3//2)]:}`D. `{:[(1//2,3//2),(1//2,1),(-5//2,3//2)]:}`
Answer» Correct Answer - C (i) Transpose `B^(T)+A={:[(2,0),(5,-1),(3,4)]:}` and add it to the other equation. (ii) `(B-A^(T))^(T)=B^(T)-A`. (iii) Subtract `(B^(T)-A)" from "B^(T)+A`.

Discussion

996.	If `A={:[(-10,11),(6,-3)]:}={:[(-32,46)]:}`, then find A.
Answer» Correct Answer - [5 3]

Discussion

997.	If `P={:[(2005,2004),(2004,2005)]:}`, then find X such that PX=XP=P.
Answer» Correct Answer - `{:[(1,0),(0,1)]:}`

Discussion

998.	The matrix \(\begin{bmatrix}-12\\[0.3em]10\\[0.3em]13\\[0.3em]4\end{bmatrix}\) is a :(a) square matrix (b) row matrix (c) column matrix (d) null matrix
Answer» (c) column matrix

Discussion

999.	If `A={:[(-5,3),(2,1),(1,5)]:}andB={:[(14,-13),(1,3),(14,-3)]:}`, then find the matrix X such that AX = B.
Answer» Correct Answer - `{:[(-1,2),(3,-1)]:}`

Discussion

1000.	If `A=((p,q),(0,1))`, then show that `A^(8)=((p^(8),q((p^(8)-1)/(p-1))),(0,1))`
Answer» `A^(2)=((p,q),(0,1))((p,q),(0,1))` `=((p^(2),pq+q),(0,1))=((p^(2), q(p+1)),(0,1))` `A^(3)=A.A^(2)` `=((p,q),(0,1))((p^(2), pq+q),(0,1))=((p^(3),p^(2)q+pq+q),(0,1))=((p^(3), q(p^(2)+p+1)),(0,1))` Similarly, `A^(4)=((p^(4), q(p^(3)+p^(2)+p+1)),(0,1))` and so on. `:. A^(8)=((p^(8), q(p^(7)+p^(8)+...+1)),(0,1))=((p^(8),q((p^(8)-1)/(p-1))),(0,1))`

Discussion

Explore topic-wise InterviewSolutions in .

Three students Anil, Nikhil and Sunil went to a stationary shop and purchased some items. Anil purchased 3 books, 2 erasers and 5 scales. Sunil purchased 5 books, 6 scales and 3 erasers whereas Nikhil purchased 2 books, 3 scales and 4 erasers. Represent the given data in the matrix form.

In applying one or more row operations while finding A–1 by elementary row operations, we obtain all zeros in one or more, then A–1 _________.

If `A=[[1 ,2,-3],[ 5, 0, 2],[ 1,-1, 1]], B=[[3,-1, 2],[ 4, 2 ,5],[ 2, 0, 3]]`and `C=[[4, 1, 2],[ 0, 3, 2],[ 1,-2, 3]]`, then compute `(A+B)` and `(B - C)`. Also, verify that `A + (B - C) = (A + B) - C`.

Compute the products AB and BA whichever exists in each of the following cases :[a,b] \(\begin{bmatrix} c \\[0.3em] d \\[0.3em] \end{bmatrix}\) + [a b c d] \(\begin{bmatrix} a \\[0.3em] b \\[0.3em] c \\[0.3em] d \end{bmatrix}\)

`"If A" = [{:(x^(2), y), (5, -4):}] " and "A=A^(-1), " then find " [{:(x^(3), y+x), (1, 2x^(2) + y):}]^(-1)` (a) `(1)/(41)[{:(8, 1), (-1, 5):}]` (b) `(1)/(41)[{:(5, 8), (1, -1):}]` (c) `[{:(5, 1), (-1, 8):}]` (d) `(1)/(41)[{:(5, 1), (-1, 8):}]`

Compute the following sums :\(\begin{bmatrix}2 &amp; 1 &amp; 3 \\[0.3em]0 &amp; 3 &amp; 5 \\[0.3em]-1 &amp; 2 &amp; 5\end{bmatrix}+​​\)\(\begin{bmatrix}1 &amp; -2 &amp; 3 \\[0.3em]2 &amp; 6 &amp; 1 \\[0.3em]0 &amp; -3 &amp; 1\end{bmatrix}​​\)

Compute : `2{:[(2,3,4,1),(-5,11,6,7)]:}+5{:[(11,13,-3,4),(5,-3,4,-7)]:}-6{:[(6,12,4,-2),(-3,5,11,7)]:}`.

If A = [(2,3),(1,2)], B=[(1,3,2),(4,3,1)], C=[(1),(2)], D=[(4, 6, 8), (5, 7, 9)], then which of the sums A + B, B + C, C + D and B + D is defined?

Solve the simulteneous linear equations 2x-5y =1, 5x + 3y = 18.

If `I_3` is the identily matrix of order 3, then `(I_3)^(-1)=`

Compute the following sums : \(\begin{bmatrix} 3&amp; -2 \\[0.3em] 1 &amp;4 \\[0.3em] \end{bmatrix}\)\(+\begin{bmatrix} -2&amp;4 \\[0.3em]1 &amp; 3 \\[0.3em] \end{bmatrix} ​​\)

the matrix S is rotation through an angle `45^(@)` and G is th reflection about the line `y=2x`, then `(SG)^(2)` is equal toA. 7IB. 5IC. 3ID. I

If `{:[(4,5)]:}{:[(1,6),(x,3)]:}{:[(-2),(3)]:}=[9]`, then the value of x isA. 9B. -9C. 10D. -10

If `A={:[(8),(-1),(-7)]:}andB={:[(2,5,-6)]:}," then find "3A+B^(T)`.A. `{:[(26),(2),(-27)]:}`B. `{:[(36),(4),(-1)]:}`C. `{:[(18),(2),(-9)]:}`D. `{:[(2),(26),(-27)]:}`

If `A={:[(2,3,5,6),(-1,2,-3,4)]:}andB={:[(4,-5,3,7),(8,-3,-1,2)]:}`, then find `A^(T)+B^(T)and(A+B)^(T)`. What do you notice ?

If `A={:[(6,x),(7,-6)]:}andA^(2)=I`, then x= _______ .A. 5B. `(-5)/(2)`C. -5D. 1

If `A=[(-2,3),(-1,1)]` then `A^(3) ` is equal toA. 2AB. AC. 2I.D. I

Find the values of a and b if A = B, where \(A=\begin{bmatrix} a+4&amp; 3b \\[0.3em] 8 &amp; -6 \\[0.3em] \end{bmatrix},\)\(B=\begin{bmatrix} 2a+2&amp;b^2+2 \\[0.3em] 8 &amp; b^2-10 \\[0.3em] \end{bmatrix} ​​\)

if `|[4,x+2],[2x-3,x+1]|` is a symmetric then `x=`A. 3B. 5C. 2D. 4

`{:[(-6,5),(-7,6)]^(-1)=:}`A. `{:[(-6,5),(-7,6)]:}`B. `{:[(6,-5),(-7,6)]:}`C. `{:[(6,5),(7,6)]:}`D. `{:[(6,-5),(7,-6)]:}`

`A_(x)={:[(0,0,x),(0,0,0)]:}` then find the maximum number of possibilities of matrix `B_(x)`, in which x can be placed in `a_(11),a_(21),or a_(31)`, position such `A_(x)B_(x)=O_(2xx1)`.A. 1B. 2C. 3D. Cannot be determined

If `{:A+B=[(1,0),(1,1)]andA-2B=[(-1,1),(0,-1)]:}," then "A=`A. `{:[(1,1),(2,1)]:}`B. `{:[(2//3,1//3),(1//3,2//3)]:}`C. `{:[(1//3,1//3),(2//3,1//3)]:}`D. none of these

If `A=[(2,2,1),(1,3,1),(1,2,2)]` and the sum of eigen values of A is m anda product of eigen values of A is n, then m+n is equal toA. 10B. 12C. 14D. 16

Find the values of x and y if : \(\begin{bmatrix} x+10&amp; y^2+2y \\[0.3em] 0 &amp; -4 \\[0.3em] \end{bmatrix}\)\(=\begin{bmatrix} 3x+4&amp;3 \\[0.3em] 0 &amp; y^2-5y \\[0.3em] \end{bmatrix} ​​\)

If `A-2B={:[(3,6),(7,6)]:}andA-3B={:[(2,6),(7,5)]:}`, then the matrix A isA. `{:[(5,6),(7,8)]:}`B. `{:[(5,8),(7,6)]:}`C. `{:[(6,5),(7,8)]:}`D. `{:[(5,8),(2,6)]:}`

For what value of x and y are the following matrices equal?\(A=\begin{bmatrix}2x+1&amp; 2y \\[0.3em]0 &amp; y^2-2y \\[0.3em]\end{bmatrix},​​\)\(B=\begin{bmatrix}x+3&amp; y^2+2 \\[0.3em]0 &amp; -6 \\[0.3em]\end{bmatrix}​​\)

Match the following lists :

Match the following lists :

If `A{:[(1,2,3,5,6),(4,-1,2,-3,1)]:}andB={:[(5,3),(2,1),(-3,5),(4,1),(1,2)]:}`, then find AB and BA.

If `A={:[(3,3),(3,3)]:}andB={:[(1,-1),(-1,1)]:}`, then find AB and BA. What do you observe ?

If `A={:[(5,-6),(2,4)]:}andB=A^(T)," then "A^(T)-B^(T)` is ________ .A. `{:[(0,6),(6,0)]:}`B. `{:[(0,8),(-8,0)]:}`C. `{:[(0,-8),(8,0)]:}`D. `{:[(8,0),(0,8)]:}`

Find the matrix X such that – A + 3B + X = 0, where \(\begin{bmatrix}-2&amp;6 \\[0.3em] 5 &amp;8 \end{bmatrix}\) and B = \(\begin{bmatrix}1&amp;2 \\[0.3em] -2 &amp;3 \end{bmatrix}\)

For what values of x and y are the following matrices equal`A=[(2x+1,3y),(0,y^2-5y)] B= [(x+3,y^2+2),(0,-6)] `

Match the following lists : A. `{:(a,b,c,d),(s,r,q,p):}`B. `{:(a,b,c,d),(s,p,q,r):}`C. `{:(a,b,c,d),(q,p,s,r):}`D. `{:(a,b,c,d),(s,q,r,p):}`

If `A={:[(1,4),(-2,5)]:}andB={:[(-6,-1),(-8,2)]:}`, then AB + BA = ________ .A. `{:[(-42,22),(40,10)]:}`B. `{:[(-42,-22),(-40,-10)]:}`C. `{:[(-42,22),(-40,10)]:}`D. `{:[(42,22),(40,10)]:}`

The order of the matrix \(\begin{bmatrix} -1 \\[0.3em] 3 \\[0.3em] 4 \end{bmatrix}\) is :(a) 1 × 3 (b) 3 × 1 (c) 1 × 1 (d) 3 × 3

If `A={:[(2,1),(0,3)]:}andf(x)=x^(2)-4x+3`, then find f(A).A. `{:[(-1,1),(0,0)]:}`B. `{:[(0,0),(1,-1)]:}`C. `{:[(0,1),(2,3)]:}`D. `{:[(0,0),(1,2)]:}`

If `y={:((4,-2),(3,0)):}andf(y)=y^(2)-y-6`, find f(y).A. `{:((0,6),(9,12)):}`B. `{:((0,6),(9,-12)):}`C. `{:((0,-6),(9,-12)):}`D. `{:((0,6),(9,-12)):}`

If `{:[(1,-1,2),(3,a,1)]:}{:[(-1),(3),(2)]:}={:[(b),(14)]:}`, then find (a+b).

Which of the following pair of matrices are equal ?

If matrix A is given by `A=[[6,11] , [2,4]]` then determinant of `A^(2005)-6A^(2004)` isA. `2^(2006)`B. `(-11)2^(2005)`C. `-2^(2005).7`D. `(-9) 2^(2004)`

If `P={:[(-1,0),(2,-1)]:}andf(x)=x^(2)-2x+2`, then find f(P).

A square matrix A has 9 elements. What is the possible order of A ? (a) 1 × 9 (b) 9 × 9 (c) 3 × 3 (d) 2 × 7

If `B-A^(T)={:[(3,4,-2),(5,-3,7)]:}andB^(T)+A={:[(2,0),(5,-1),(3,4)]:}`, then find Matrix A.A. `{:[(1//2,-3//2),(1//2,1),(-5//2,3//2)]:}`B. `{:[(1//2,3//2),(1//2,1),(-5//2,3//2)]:}`C. `{:[(-1//2,-5//2),(1//2,1),(5//2,-3//2)]:}`D. `{:[(1//2,3//2),(1//2,1),(-5//2,3//2)]:}`

If `A={:[(-10,11),(6,-3)]:}={:[(-32,46)]:}`, then find A.

If `P={:[(2005,2004),(2004,2005)]:}`, then find X such that PX=XP=P.

The matrix \(\begin{bmatrix}-12\\[0.3em]10\\[0.3em]13\\[0.3em]4\end{bmatrix}\) is a :(a) square matrix (b) row matrix (c) column matrix (d) null matrix

If `A={:[(-5,3),(2,1),(1,5)]:}andB={:[(14,-13),(1,3),(14,-3)]:}`, then find the matrix X such that AX = B.

If `A=((p,q),(0,1))`, then show that `A^(8)=((p^(8),q((p^(8)-1)/(p-1))),(0,1))`

Compute the following sums :\(\begin{bmatrix}2 & 1 & 3 \\[0.3em]0 & 3 & 5 \\[0.3em]-1 & 2 & 5\end{bmatrix}+\)\(\begin{bmatrix}1 & -2 & 3 \\[0.3em]2 & 6 & 1 \\[0.3em]0 & -3 & 1\end{bmatrix}\)

Compute the following sums : \(\begin{bmatrix} 3& -2 \\[0.3em] 1 &4 \\[0.3em] \end{bmatrix}\)\(+\begin{bmatrix} -2&4 \\[0.3em]1 & 3 \\[0.3em] \end{bmatrix} \)

Find the values of a and b if A = B, where \(A=\begin{bmatrix} a+4& 3b \\[0.3em] 8 & -6 \\[0.3em] \end{bmatrix},\)\(B=\begin{bmatrix} 2a+2&b^2+2 \\[0.3em] 8 & b^2-10 \\[0.3em] \end{bmatrix} \)

Find the values of x and y if : \(\begin{bmatrix} x+10& y^2+2y \\[0.3em] 0 & -4 \\[0.3em] \end{bmatrix}\)\(=\begin{bmatrix} 3x+4&3 \\[0.3em] 0 & y^2-5y \\[0.3em] \end{bmatrix} \)

For what value of x and y are the following matrices equal?\(A=\begin{bmatrix}2x+1& 2y \\[0.3em]0 & y^2-2y \\[0.3em]\end{bmatrix},\)\(B=\begin{bmatrix}x+3& y^2+2 \\[0.3em]0 & -6 \\[0.3em]\end{bmatrix}\)

Find the matrix X such that – A + 3B + X = 0, where \(\begin{bmatrix}-2&6 \\[0.3em] 5 &8 \end{bmatrix}\) and B = \(\begin{bmatrix}1&2 \\[0.3em] -2 &3 \end{bmatrix}\)