1192 + Interview Questions in MATRICES IN GENERAL AWARENESS Page 17 InterviewSolution

801.	Find the inverse of matrix`[{:(1," "1,-2),(1," "0," "3),(5," "0," "1):}]`
Answer» `(1)/(2).[{:(1,-1,1),(-8," "6,-2),(5,-3,1):}]`

Discussion

802.	Find the inverse of matrix`[{:(6,7),(8,9):}]`
Answer» `[{:((-9)/(2),(7)/(3)),(4,-3):}]`

Discussion

803.	Find the inverse of matrix `[{:(4,0),(2,5):}]`
Answer» `[{:((1)/(4),0),((-1)/(10),(1)/(5)):}]`

Discussion

804.	Find the determinant of a matrix `[{:(2,-3),(4," "5):}]`
Answer» `[{:((5)/(22),(3)/(22)),((-2)/(11),(1)/(11)):}]`

Discussion

805.	Find the inverse of matrix `[{:(2,5),(-3,1):}]`
Answer» `(1)/(17).[{:(1,-5),(3," "2):}]`

Discussion

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

Discussion

807.	Find the adjoint of matrix`[{:(1,2),(3,7):}]`
Answer» `[{:(7,-2),(-3," "1):}]`

Discussion

808.	If matrix `A=[1 2 3]`, write `AA^T`.
Answer» Correct Answer - [14]

Discussion

809.	Let `A=[3 2 5 4 1 3 0 6 7]`Express `A`as sum oftwo matrices such that one is symmetric and the other is skew symmetric.
Answer» `A=[{:(3,3,(5)/(2)),(3,1,(9)/(2)),((5)/(2),(9)/(2),7):}]+[{:(0,-1,(5)/(2)),(1,0,(-3)/(2)),((-5)/(2),(3)/(2),0):}]`

Discussion

810.	If `[2 3 5 7] [1-3-2 4]=[-4 6-9x],`write the value of `x`
Answer» Correct Answer - x=13

Discussion

811.	Expess the matrix A as the sum of a symmetric and a skew-symmetric matrix, where `A=[{:(3,-1,0),(2," "0,3),(1,-1,2):}].`
Answer» `A=[{:(3,(1)/(2),(1)/(2)),((1)/(2),0,1),((1)/(2),1,2):}]+[{:(0,(-3)/(2),(-1)/(2)),((3)/(2),0," "2),((1)/(2),-2," "0):}]`

Discussion

812.	Express the matrix `A=[{:(-1,5,1),(2,3,4),(7,0,9):}]` as the sum of a symmetric and a skew-symmetric matrix.
Answer» `A=[{:(-1,(7)/(2),4),((7)/(2),3,2),(4,2,9):}]+[{:(0,(3)/(2),-3),((-3)/(2),0," "2),(3,-2," "0):}]`

Discussion

813.	Express the matrix `A=[{:(2,3),(-1,4):}]` as the sum of a symmetric matrix and a skew-symmetric matrix.
Answer» `A=[{:(2,1),(1,4):}]+[{:(0,2),(-2,0):}]`

Discussion

814.	Express the matrix `A=[{:(3,-4),(1,-1):}]` as the sum of a symmetric matrix and a skew-symmetric matrix.
Answer» `A=[{:(3,(-3)/(2)),((-3)/(2),-1):}]+[{:(0,(-5)/(2)),((5)/(2),0):}]`

Discussion

815.	Total number of possible matrices of order 3 × 3 with each entry 2 or 0 is A. 9 B. 27 C. 81 D. 512
Answer» As matrix has total 3× 3 = 9 elements. As each element can take 2 values (0 or 2) ∴ By simple counting principle we can say that total number of possible matrices = total number of ways in which 9 elements can take possible values = 2⁹ = 512 Clearly it matches with option D. ∴ Option (D) is the only correct answer.

815.

Total number of possible matrices of order 3 × 3 with each entry 2 or 0 is A. 9 B. 27 C. 81 D. 512

Answer»

As matrix has total 3× 3 = 9 elements.

As each element can take 2 values (0 or 2)

∴ By simple counting principle we can say that total number of possible matrices = total number of ways in which 9 elements can take possible values = 2⁹ = 512

Clearly it matches with option D.

∴ Option (D) is the only correct answer.

Discussion

816.	The matrix P = \(\begin{bmatrix} 0 & 0 & 4 \\[0.3em] 0 & 4 & 0 \\[0.3em] 4 & 0 & 0 \end{bmatrix}\) is a A. square matrixB. diagonal matrixC. unit matrixD. none
Answer» As P has equal number of rows and columns and thus it matches with the definition of square matrix. The given matrix does not satisfy the definition of unit and diagonal matrices. ∴ Option (A) is the only correct answer.

816.

The matrix P = \(\begin{bmatrix} 0 & 0 & 4 \\[0.3em] 0 & 4 & 0 \\[0.3em] 4 & 0 & 0 \end{bmatrix}\) is a A. square matrixB. diagonal matrixC. unit matrixD. none

Answer»

As P has equal number of rows and columns and thus it matches with the definition of square matrix.

The given matrix does not satisfy the definition of unit and diagonal matrices.

∴ Option (A) is the only correct answer.

Discussion

817.	If the matrix A = \(\begin{bmatrix}5 &2 & x \\[0.3em]y & z &-3 \\[0.3em]4 & t & -7\end{bmatrix}\) is a symmetric matrix, find x, y, z and t.
Answer» Given, A = \(\begin{bmatrix}5 &2 & x \\[0.3em]y & z &-3 \\[0.3em]4 & t & -7\end{bmatrix}\) is a symmetric matrix. We know that, A = [a_ij]_{m x n}is a symmetric matrix if a_{ij =}a_ji So, x = a₁₃ = a₃₁ = 4 y = a₂₁ = a₁₂ = 2 z = a₂₂= _a22 = z t = a₃₂ = a₂₃ = - 3 Hence, X = 4, y = 2, t = - 3 and z can have any value.

817.

If the matrix A = \(\begin{bmatrix}5 &2 & x \\[0.3em]y & z &-3 \\[0.3em]4 & t & -7\end{bmatrix}\) is a symmetric matrix, find x, y, z and t.

Answer»

Given,

A = \(\begin{bmatrix}5 &2 & x \\[0.3em]y & z &-3 \\[0.3em]4 & t & -7\end{bmatrix}\) is a symmetric matrix.

We know that,

A = [a_ij]_{m x n}is a symmetric matrix if a_{ij =}a_ji

So,

x = a₁₃ = a₃₁ = 4

y = a₂₁ = a₁₂ = 2

z = a₂₂= _a22 = z

t = a₃₂ = a₂₃ = - 3

Hence,

X = 4, y = 2, t = - 3 and z can have any value.

Discussion

818.	If A `=[{:(3,-3,4),(2,-3,4),(0,-1,1):}]`and B is the adjoint of A, find the value of`\|AB+2I\|`,where l is the identity matrix of order 3.
Answer» `because " "A = [{:(3,-3,4),(2,-3,4),(0,-1,1):}]` `therefore" " \|A\| = [{:(3,-3,4),(2,-3,4),(0,-1,1):}]` `=3(-3+4) + 3(2-0) + 4(-2-0) = 1 ne 0` `therefore \|AB + 2I\| = (\|A\| + 2)^(3)` ` =(1+2)^(3) = 3^(3) =27`

Discussion

819.	If a matrix has 8 elements, what are the possible orders it can have? What if it has 5 elements?
Answer» If a matrix is of order m x n elements, it has mn elements. So, if the matrix has 8 elements, we will find the ordered pairs m and n. mn = 8 Then, ordered pairs m and n can be m x n be (8 x 1),(1 x 8),(4 x 2),(2 x 4) Now, if it has 5 elements Possible orders are (5 x 1), (1 x 5).

819.

If a matrix has 8 elements, what are the possible orders it can have? What if it has 5 elements?

Answer»

If a matrix is of order m x n elements, it has mn elements. So, if the matrix has 8 elements, we will find the ordered pairs m and n.

mn = 8

Then, ordered pairs m and n can be

m x n be (8 x 1),(1 x 8),(4 x 2),(2 x 4)

Now, if it has 5 elements

Possible orders are (5 x 1), (1 x 5).

Discussion

820.	Choose the correct answer If A, B are symmetric matrices of same order, then AB – BA is a (A) Skew symmetric matrix (B) Symmetric matrix (C) Zero matrix (D) Identity matrix
Answer» Given:` A_(nXn)=A_(nXn)^T` `B_(nXn)=B_(nXn)^T` proof `(AB-BA)^T=(AB)^T-(BA)^T` =`B^TA^T-A^TB^T` =BA-AB =-(AB-BA) So, it is a skew symmetric matrix

Discussion

821.	If A and B are symmetric matrices, prove that AB – BA is a skew symmetric matrix.
Answer» Given `A^T=A`(1) `B^T=B`(2) Proof `(AB-BA)^T=(AB)^T-(BA)^T` =`B^TA^T-A^TB^T` =`-(A^TB^T-B^TA^T)` =-(AB-BA) From equation (1) and (2) so, (AB-BA) is a skew symmetric matrix

Discussion

822.	If A and B are matrices of same order, then (AB’ – BA’) is a A. skew symmetric matrix B. null matrix C. symmetric matrix D. unit matrix
Answer» Let C = (AB’ – BA’) C’ = (AB’ – BA’)’ ⇒ C’ = (AB’)’ – (BA’)’ ⇒ C’ = (B’)’A’ – (A’)’B’ ⇒ C’ = BA’ – AB’ ⇒ C’ = -C ∴ C is a skew-symmetric matrix. Clearly Option (A) matches with our deduction. ∴ Option (A) is the correct.

822.

If A and B are matrices of same order, then (AB’ – BA’) is a A. skew symmetric matrix B. null matrix C. symmetric matrix D. unit matrix

Answer»

Let C = (AB’ – BA’)

C’ = (AB’ – BA’)’

⇒ C’ = (AB’)’ – (BA’)’

⇒ C’ = (B’)’A’ – (A’)’B’

⇒ C’ = BA’ – AB’

⇒ C’ = -C

∴ C is a skew-symmetric matrix.

Clearly Option (A) matches with our deduction.

∴ Option (A) is the correct.

Discussion

823.	If A and B are symmetric matrices of the same order, show that (AB – BA) is a skew symmetric matrix.
Answer» We are given that A and B are symmetric matrices of the same order then, we need to show that (AB – BA) is a skew symmetric matrix. Let us consider P is a matrix of the same order as A and B And let P = (AB – BA), we have A = A’ and B = B’ then, P’ = (AB – BA)’ P’ = ((AB)’ – (BA)’) …….using reversal law we have (CD)’=D’C’ P’ = (B’A’ – A’B’) P’ = (BA – AB) P’ = -P Hence, P is a skew symmetric matrix.

823.

If A and B are symmetric matrices of the same order, show that (AB – BA) is a skew symmetric matrix.

Answer»

We are given that A and B are symmetric matrices of the same order then,

we need to show that (AB – BA) is a skew symmetric matrix.

Let us consider P is a matrix of the same order as A and B

And let P = (AB – BA),

we have A = A’ and B = B’

then, P’ = (AB – BA)’

P’ = ((AB)’ – (BA)’) …….using reversal law we have (CD)’=D’C’

P’ = (B’A’ – A’B’)

P’ = (BA – AB)

P’ = -P

Hence, P is a skew symmetric matrix.

Discussion

824.	For `3xx3`matrices `Ma n dN ,`which of the following statement (s) is (are) NOT correct ?`N^T M N`is symmetricor skew-symmetric,according as `m`is symmetric or skew-symmetric.`M N-N M`is skew-symmetric for allsymmetric matrices `Ma n dNdot``M N`is symmetric for all symmetricmatrices `M a n dN``(a d jM)(a d jN)=a d j(M N)`for all invertible matrices `Ma n dNdot`A. `N^(T) MN` is symmetric or skew-symmetric, according as M is symmetric of skew-symmetricB. `MN-NM` is skew-symmetric for all symmetric matrices M and NC. MN is symmetric for all symmetric matrices M and ND. (adj M) (adj N) = adj (MN) for all invertible matrices M and N
Answer» Correct Answer - C::D (a) `(N^(T) MN)^(T) = N^(T) M^(T) (N^(T))^(T)=N^(T)M^(T)N=N^(T) MN` or `-N^(T) MN` According as M is symmetric ro skew-symmetric. `therefore` Correct. (b) `(MN-NM)^(T) = (MN)^(T)- (NM)^(T) =N^(T) M^(T) - M^(T)N^(T) ` `= NM=MN ` [`because`+ M, N are symmetric] `=-(MN-NM)` `therefore ` correct (c) `(MN)^(T) = N^(T) M^(T) = NMneMN` [`because` M, N are symmetric] `therefore` Incorrect. (d) `(adj M) (adjN) = adj (NM) ne adj(MN)` `therefore ` Incorrect.

Discussion

825.	If A, B are square matrices of same order and B is a skew-symmetric matrix, show that A’ BA is skew symmetric.
Answer» A matrix is said to be skew-symmetric if A = -A’ Given, B is a skew-symmetric matrix. ∴ B = -B’ Let C = A’ BA …(1) We have to prove C is skew-symmetric. To prove: C = -C’ As C’ = (A’BA)’ We know that: (AB)’ = B’A’ ⇒ C’ = (A’BA)’ = A’B’(A’)’ ⇒ C’ = A’B’A {∵ (A’)’ = A} ⇒ C’ = A’(-B)A ⇒ C’ = -A’BA …(2) From equation 1 and 2: We have, C’ = -C Thus we say that C = A’ BA is a skew-symmetric matrix.

825.

If A, B are square matrices of same order and B is a skew-symmetric matrix, show that A’ BA is skew symmetric.

Answer»

A matrix is said to be skew-symmetric if A = -A’

Given, B is a skew-symmetric matrix.

∴ B = -B’

Let C = A’ BA …(1)

We have to prove C is skew-symmetric.

To prove: C = -C’

As C’ = (A’BA)’

We know that: (AB)’ = B’A’

⇒ C’ = (A’BA)’ = A’B’(A’)’

⇒ C’ = A’B’A {∵ (A’)’ = A}

⇒ C’ = A’(-B)A

⇒ C’ = -A’BA …(2)

From equation 1 and 2:

We have,

C’ = -C

Thus we say that C = A’ BA is a skew-symmetric matrix.

Discussion

826.	For `3xx3`matrices `Ma n dN ,`which of the following statement (s) is (are) NOT correct ?`N^T M N`is symmetricor skew-symmetric,according as `m`is symmetric or skew-symmetric.`M N-N M`is skew-symmetric for allsymmetric matrices `Ma n dNdot``M N`is symmetric for all symmetricmatrices `M a n dN``(a d jM)(a d jN)=a d j(M N)`for all invertible matrices `Ma n dNdot`A. `N^(T)MN` is symmetric or skew-symmetric, according as M is symmetric or skew-symmetricB. `MN-NM` is skew0symmetric for all symmetric matrices M and NC. MN is symmetric for all symmetric matrices M and ND. (adj M ) (adj N) = adj (MN) for all inveriblr matrices M and N.
Answer» Correct Answer - C::D (1) `(N^(T)MN)^(T)=N^(T)M^(T)N=N^(T)MN` if `M` is symmetric and is `-N^(T)MN` if `M` is skew symmetric (2) `(MN-NM)^(T)=N^(T) M^(T)-M^(T)N^(T)=NM-MN=-(MN-NM)`. So, `(MN-NM)` is skew symmetric (3) `(MN)^(T)=N^(T)M^(T)=NM ne MN` if `M` and `N` are symmetric. So, MN is not symmetric (4) (adj. M) (adj. N) = adj `(NM) ne` adj `(MN)`

Discussion

827.	A and B are square matrices of order `3xx3` , A is an orthogonal matrix and B is a skew symmetric matrix. Which of the following statement is not trueA. `abs(AB)=1`B. `abs(AB)=0`C. `abs(AB)=-1`D. none of these
Answer» Correct Answer - B We have, A an orthogonal matrix `rArr absA=pm1` B a skew-symmetric matrix of odd order `rArr absB=0` `:. abs(AB)=absAabsBrArr abs(AB)=(pm1)xx0=0`.

Discussion

828.	If \(\begin{bmatrix}-2&-3&1\\[0.3em]-y-2&-1&4\end{bmatrix} = \begin{bmatrix}x + 2&-3&1\\[0.3em]5&-1&4\end{bmatrix}\), then find the value of x and y.
Answer» \(\begin{bmatrix}-2&-3&1\\[0.3em]-y-2&-1&4\end{bmatrix} = \begin{bmatrix}x + 2&-3&1\\[0.3em]5&-1&4\end{bmatrix}\) On comparing, x + 2 = -2 ∴ x = -4 – y – 2 = 5 ⇒ y = -7 Hence, x = -4, y = -7

828.

If \(\begin{bmatrix}-2&-3&1\\[0.3em]-y-2&-1&4\end{bmatrix} = \begin{bmatrix}x + 2&-3&1\\[0.3em]5&-1&4\end{bmatrix}\), then find the value of x and y.

Answer»

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

On comparing,

x + 2 = -2 ∴ x = -4

– y – 2 = 5 ⇒ y = -7

Hence, x = -4, y = -7

Discussion

829.	If A and B are symmetric matrices, then ABA isA. symmetric matrixB. skew-symmetric matrixC. diagonal matrixD. scalar matrix
Answer» Correct Answer - A

Discussion

830.	If `A=[5a-b3 2]`and A adj `A=AA^T`, then `5a+b`is equal to:(1) `-1`(2) 5(3) 4 (4)13A. -1B. 5C. 4D. 13
Answer» Correct Answer - B It is given that matrix `{:A=[(5a,-b),(3,2)]:}` satisfies the equation `A(adj A) =A A^T` `rArr absAI_2=A A^T` `rArr (10a+3b){:[(1,0),(0,1)]=[(5a,-b),(3,2)][(5a,3),(-b,2)]:}` `rArr {:[(10a+3b,0),(0,10a+3b)]=[(25a^2+b^2,15a-2b),(15a-2b,13)]:}` `rArr 10a+3b=13,15a-2b=0and 25a^2+b^2=10a+3b` `a=2/5and b=3` `:. 5a+b=2+3=5`

Discussion

831.	If `A = [(5a,-b),(3,2)]`and `A adj A=AA^T`, then `5a+b`is equal to:A. 5B. 4C. 13D. `-1`
Answer» Correct Answer - A A adj `A=A A^(T)` `implies \|A\|I=A A^(T)` `implies (10a+3b) [(1,0),(0,1)]=[(5a,-b),(3,2)][(5a,3),(-b,2)]` `implies 25a^(2)+b^(2)=10a+3b, 15a-2b=0, 10a+3b=13` `implies 10a+(3.15a)/2=13` `implies 65a=2xx13` `implies a=2/5` `implies 5a=2` `implies 2b=6` `implies b=3` `:. 5a+b=5`

Discussion

832.	If `A=[(2,3),(4,6)]` , then `A^(-1)=`A. `[(1,2),(-(3)/(2),2)]`B. `[(2,-3),(4,6)]`C. `[(-2,4),(-3,6)]`D. Does not exist
Answer» Correct Answer - D

Discussion

833.	If `A=[5a-b3 2]`and A adj `A=AA^T`, then `5a+b`is equal to:(1) `-1`(2) 5(3) 4 (4)13A. 5B. 13C. 4D. -1
Answer» Correct Answer - A `because A" adj "A = A A^(T)` `rArr A^(-1) (A" adj " A) = A^(-1) (A A^(T))` `rArr (A^(-1)A)" adj " A = (A^(-1) A) A^(T)` `rArr I ( " ADj " A ) = IA^(T)` or ` " adj " A = A^(T)` or `[[2,b],[-3,5a]]=[[5a,3],[-b,2]]` `rArr 5 a = 2 and b = 3` `therefore 5a + b = 5`

Discussion

834.	If A = \(\begin{bmatrix} 0 & 2 \\[0.3em] 3& -4 \\[0.3em] \end{bmatrix}\) and kA = \(\begin{bmatrix} 0 & 3a \\[0.3em] 2b & 24 \\[0.3em] \end{bmatrix}\), then the values of k, a, b, are respectivelyA. –6, –12, –18 B. –6, 4, 9 C. –6, –4, –9 D. –6, 12, 18
Answer» A = \(\begin{bmatrix} 0 & 2 \\[0.3em] 3& -4 \\[0.3em] \end{bmatrix}\) and kA = \(\begin{bmatrix} 0 & 3a \\[0.3em] 2b & 24 \\[0.3em] \end{bmatrix}\) = \(\begin{bmatrix} 0 & 2k \\[0.3em] 3k & -4k \\[0.3em] \end{bmatrix}\) Comparing the equations, -4k = 24 k = -6 3k = 2b 3(-6) = 2b 2b = -18 b= - 9 3a = 2k 3a = 2(-6) 3a = -12 a = -4 Values are, k = -6, a = -4 & b = -9 Option (C) is the answer.

834.

If A = \(\begin{bmatrix} 0 & 2 \\[0.3em] 3& -4 \\[0.3em] \end{bmatrix}\) and kA = \(\begin{bmatrix} 0 & 3a \\[0.3em] 2b & 24 \\[0.3em] \end{bmatrix}\), then the values of k, a, b, are respectivelyA. –6, –12, –18 B. –6, 4, 9 C. –6, –4, –9 D. –6, 12, 18

Answer»

A = \(\begin{bmatrix} 0 & 2 \\[0.3em] 3& -4 \\[0.3em] \end{bmatrix}\) and kA = \(\begin{bmatrix} 0 & 3a \\[0.3em] 2b & 24 \\[0.3em] \end{bmatrix}\)

= \(\begin{bmatrix} 0 & 2k \\[0.3em] 3k & -4k \\[0.3em] \end{bmatrix}\)

Comparing the equations,

-4k = 24

k = -6

3k = 2b

3(-6) = 2b

2b = -18

b= - 9

3a = 2k

3a = 2(-6)

3a = -12

a = -4

Values are,

k = -6, a = -4 & b = -9

Option (C) is the answer.

Discussion

835.	If the matrix A is both symmetric and skew-symmetric, then A is a(a) diagonal matrix(b) zero matrix(c) square matrix(d) scalar matrix
Answer» (b) Zero matrix

Discussion

836.	let `z= (-1+sqrt(3i))/2, where i=sqrt(-1) and r,s epsilon P1,2,3}. Let P= [((-z)^r, z^(2s)),(z^(2s), z^r)]` and I be the idenfity matrix or order 2. Then the total number of ordered pairs (r,s) or which `P^2=-I` isA. `1/2 abs(a-b)`B. `1/2 abs(a+b)`C. ` abs(a-b)`D. ` abs(a+b)`
Answer» Correct Answer - A `because Z = (-1+ sqrt(3)i)/2 = omega` ` rArr omega = 1 and 1 + omega + omega ^(2) = 0 ` Now, `P = [[(-omega)^(r),omega^(2s)],[omega^(2s), omega^(r)]]` `therefore P^(2) = [[(-omega)^(r),omega^(2s)],[omega^(2s), omega^(r)]] [[(-omega)^(r),omega^(2s)],[omega^(2s), omega^(r)]]` ` =[[omega^(2r)+omega^(4s),omega^(2s)((-omega)^(r)+omega^(r))],[omega^(2s)((-omega)^(r)+omega^(r)), omega^(4s)+omega^(2r)]]` ` =[[omega^(2r)+omega^(4s),omega^(2s)((-omega)^(r)+omega^(r))],[omega^(2s)((-omega)^(r)+omega^(s)), omega^(s)+omega^(2r)]] " " (because omega^(3) = r)` `because P^(2) = -I= [[-1,0],[0,-1]]` ...(ii) Form Eqs. (i) and (ii), we get `omega^(2r) +omega^(s)=-1` and ` omega^(2s) ((-omega)^(r)+omega^(r))=0` `rArr r` is odd and `s = r` but not a multiple of 3, Which is possible `therefore ` only one pair is there.

Discussion

837.	If `A=[{:(1, 1, 1),(1, 2, "-3"),(2,"-1",3):}]` , then adj A isA. `[{:(3, "-9", "-5"),("-4", 1, 3),("-5",4,1):}]`B. `[{:(3, "-4", "-5"),("-9", 1, 4),("-5",3,1):}]`C. `[{:("-3", 4, 5),(9, "-1", "-4"),(5,"-3","-1"):}]`D. `[{:(3, "-9","-5"),(4, "-1", 3),("-5",4,1):}]`
Answer» Correct Answer - B

Discussion

838.	If `A={:[(a,0,0),(0,a,0),(0,0,a)],:}" then: "\|A\|*\|adj.A\|=`A. `a^(3)`B. `a^(6)`C. `a^(9)`D. `a^(27)`
Answer» Correct Answer - C

Discussion

839.	Let `p=[(3,-1,-2),(2,0,alpha),(3,-5,0)],` where `alpha in RR.` Suppose `Q=[q_(ij)]` is a matrix such that `PQ=kl,` where `k in RR, k != 0 and l` is the identity matrix of order 3. If `q_23=-k/8 and det(Q)=k^2/2,` thenA. `alpha0,k=8`B. `4alpha-k+8=0`C. `det(PadjQ)=2^9`D. `det(Q adjP)2^13`
Answer» Correct Answer - B::C We have, `PQ =kl and det (Q) =k^2/2` `rArr det(PQ)=det(kI) and det (Q)=k^2/2` `rArr det (P) det(Q) =k^3and det (Q) =k^2/2` `rArr k^2/2det(P) =k^3and det(Q) =k^2/2` `rArr det (P) =2k and det (Q) =k^2/2` Again, PQ = kI `rArr Q=P^(-1) (kI) =kP^(-1)` `rArr Q=k(1/absPadjP)` `rArr Q=k(1/(2k) adjP)` `rArr Q=1/2adjP` `rArr q_23 =1/2(adjP) _23 =1/2" cofactor " P_32 =-1/2{:abs((3,-2),(2,alpha)):}` `rArr =-k/8=-1/2(3alpha +4)` `rArr k=12alpha +16 ...(i)` Now, `det (P) =2K` `rArr {:abs((3,-1,-2),(2,0,alpha),(3,-5,0))=2k:}` `rArr 15 alpha-3alpha+20=2k` `rArr k=6alpha+10` Solving (i) and (ii), we get : `alpha=-1 k=4`. `:. 4alpha-k+8=-4-4+8=0` So, option (b) is correct. Now, `det(P(adjQ)) =abs(PadjQ)=absPabs(adjQ)` `=2kabs(Q)^2=2k(k^2/2)^2=1/2k^5=2^9` and, `det (Q(adjP))= abs(QadjP)=absQ abs(adjP)` `=absQabs(P)^2=k^2/2xx(2k)^2=2k^4=2xx(4)^45=2xx(4)^4=2^9` So, option (c) is correct.

Discussion

840.	Let `p=[(3,-1,-2),(2,0,alpha),(3,-5,0)],` where `alpha in RR.` Suppose `Q=[q_(ij)]` is a matrix such that `PQ=kl,` where `k in RR, k != 0 and l` is the identity matrix of order 3. If `q_23=-k/8 and det(Q)=k^2/2,` thenA. `alpha = 0, k=8`B. `4alpha -k + 8 =0`C. `det (padj(Q) ) = 2^(9)`D. `det (Qadj(P) ) = 2^(13)`
Answer» Correct Answer - B::C `because PQ = kI rArr (P.Q)/k = I rArr P^(-1) = Q/k" " ...(i)` Also `abs(P) = 12 alpha +20 " " (ii)` and `" "`given ` q_(23) = (-k)/8` Comoaring the third element of `2^(nd)` row no both sides, we get `1/((12alpha + 20))(-(3alpha + 4)) = 1/k xx (-k)/8` `rArr 24 alpha + 32= 12 alpha + 20` ` alpha = -1` ...(iii) From (ii), `abs(P)=8` ...(iv) Also `PQ= kI` `rArrabs(PQ) = abs(kI)` `rArr abs(P) abs(Q) = k^(3)` `rArr 8xx (k^(2))/2 = k^(3) " " (because abs(P) = 8, abs(Q) = k^(2)/2)` `therefore k= 4 " " ...(v)` (b) `4 alpha - k + 8 = -4 -4 + 8 = 0` (c) `det(P " adj" (Q)) = abs(P) abs("adj" Q) = abs(P) abs(Q)^(2) = 8xx8^(2) = 2^(9)` (d) `det(Q " adj" (P)) = abs(Q) abs("adj" P) = abs(Q) abs(P)^(2) = 8xx8^(2) = 2^(9)`

Discussion

841.	Fill in the blanks:If A is skew symmetric, then kA is a ______. (k is any scalar)
Answer» A skew symmetric matrix. Given A’=-A ⇒ (kA)’=k(A)’=k(-A) ⇒ (kA)’=-(kA)

841.

Fill in the blanks:If A is skew symmetric, then kA is a ______. (k is any scalar)

Answer»

A skew symmetric matrix.

Given A’=-A

⇒ (kA)’=k(A)’=k(-A)

⇒ (kA)’=-(kA)

Discussion

842.	Matrices a and B satisfy `AB=B^(-1)`, where `B=[(2,-1),(2,0)]`. Find (i) without finding `B^(-1)`, the value of K for which `KA-2B^(-1)+I=O`. (ii) without finding `A^(-1)`, the matrix X satifying `A^(-1) XA=B`.
Answer» (i) `AB=B^(-1)` or `AB B=B^(-1)B` or `AB^(2)=I` Now, `KA-2B^(-1)+I=O` or `KAB-2B^(-1)B+IB=O` or `KAB-2I+B=O` or `KAB^(2)-2B+B^(2)=O` or `KI-2B+B^(2)=O` or `K[(1,0),(0,1)]-2[(2,-1),(2,0)]+[(2,-1),(2,0)][(2,-1),(2,0)] =[(0,0),(0,0)]` or `[(K,0),(0,K)]-[(4,-2),(4,0)]+[(2,-2),(4,-1)]=[(0,0),(0,0)]` or `[(K-2,0),(0,K-2)]=[(0,0),(0,0)]` ot `K=2` (ii) `A^(-1)XA=B` or `A A^(-1)XA=AB` or `IXA=AB` or `XAB=AB^(2)` or `XAB=I` or `XAB^(2)=B` or `XI=B` or `X=B`

Discussion

843.	A=`[{:(0,3),(2,0):}] and A^(-1)=lambda` (adj, A) then `lambda` is equal toA. `(-1)/(6)`B. `(1)/(3)`C. `(-1)/(3)`D. `(1)/(6)`
Answer» Correct Answer - a

Discussion

844.	If A is an invertible square matrix; then `adj A^T = (adjA)^T`A. `2absA`B. `2absAI`C. null matrixD. unit matrix
Answer» Correct Answer - C

Discussion

845.	If `A={:[(4,2),(5,3)]:}" then :"\|adj(adjA)\|=`A. 1B. 2C. 3D. 4
Answer» Correct Answer - B

Discussion

846.	If `k` is a scalar and `l` is a unit matrix of order 3 then `adj(kl)` is equal toA. `k^(3)I`B. `k^(2)I`C. `-k^(3)I`D. `-k^(2)I`
Answer» Correct Answer - b

Discussion

847.	If x is square materix of order `3xx3` and `lambda` is a scalar, then adj `(lambda X)` is equal toA. `lambda adj X`B. `lambda ^(3) adj X`C. `lambda ^(2)adj X`D. `lambda ^(4) adj X`
Answer» Correct Answer - c

Discussion

848.	If A and B are square matrices of the same order, then (i) (AB)=......... (ii) (KA)=.......... (where, k is any scalar) (iii) [k(A-B)]=......
Answer» (AB)=BA (ii) (KA)=kA [k(A-B)]=k(A-B)

Discussion

849.	If `{:A=[(1,x),(x^7,4y)]a,B=[(-3,1),(1,0)]and adjA+B=[(1,0),(0,1)]:}`, then the values of x and y are respectivelyA. (1,1)B. (-1,1)C. (1,0)D. none of these
Answer» Correct Answer - A

Discussion

850.	If `A=[(a,b),(c,d)]`, then `adj(adjA)` is equal toA. adj AB. AC. `A^(T)`D. `-A`
Answer» Correct Answer - B

Discussion

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If \(\begin{bmatrix}-2&-3&1\\[0.3em]-y-2&-1&4\end{bmatrix} = \begin{bmatrix}x + 2&-3&1\\[0.3em]5&-1&4\end{bmatrix}\), then find the value of x and y.

If A = \(\begin{bmatrix} 0 & 2 \\[0.3em] 3& -4 \\[0.3em] \end{bmatrix}\) and kA = \(\begin{bmatrix} 0 & 3a \\[0.3em] 2b & 24 \\[0.3em] \end{bmatrix}\), then the values of k, a, b, are respectivelyA. –6, –12, –18 B. –6, 4, 9 C. –6, –4, –9 D. –6, 12, 18