1192 + Interview Questions in MATRICES IN GENERAL AWARENESS Page 14 InterviewSolution

651.	If `[{:(x, 0), (0, y):}] + [{:(-y, 4),(7, x):}] = [{:(4, 4), (7, 6):}]` then x = ____ and y=___.
Answer» Correct Answer - 5 and 1

Discussion

652.	If `A = [{:(8, -7), (9, -8):}], " then " (A^(2007))^(-1)` =___.A. IB. 2AC. ABD. 2007I
Answer» Correct Answer - C Calculate `A^(2), " then find " (A^(2007))^(-1)`

Discussion

653.	If `A=[(x,-2),(3,7)]` and `A^(-1)=[(7/34,1/17),((-3)/34,2/17)]`,then the value of x isA. 2B. 3C. `-4`D. 4
Answer» Correct Answer - d

Discussion

654.	If `A={:[(p,-1),(q,1)]:},B={:[(1,-1),(1,2)]:}and(A+B)^(2)=A^(2)+2AB+B^(2)`, then p-q= _______ .A. 2B. -1C. 3D. 1
Answer» Correct Answer - B

Discussion

655.	The value of `\|{:("cos"theta, -"sin"theta), ("sin" theta, "cos"theta):}\|` =___.
Answer» Correct Answer - 1

Discussion

656.	If `{:((a,3),(4,5)):}{:((3,-2),(b,8)):}={:((30,20),(52,a)):}`, then find 2a+b-c.A. 16B. 15C. 12D. -20
Answer» Correct Answer - D

Discussion

657.	Let `P=[(1,2,1),(0,1,-1),(3,1,1)]`. If the product PQ has inverse `R=[(-1,0,1),(1,1,3),(2,0,2)]` then `Q^(-1)` equalsA. `[(3,2,9),(-1,1,1),(0,1,8)]`B. `[(5,2,9),(-1,1,1),(0,1,7)]`C. `[(2,-1,0),(10,6,3),(8,6,4)]`D. none of these
Answer» Correct Answer - C `(PQ)^(-1)=R` `implies Q^(-1)xxP^(-1) =R` `implies Q^(-1) =RP=[(-1,0,1),(1,1,3),(2,0,2)][(1,2,1),(0,1,-1),(3,1,1)]` `=[(2,-1,0),(10,6,3),(8,6,4)]`

Discussion

658.	If `A=[(2x,0),(x,x)]` and `A^(-1)=[(1,0),(-1,2)]` then x equals toA. 2B. `(1)/(2)`C. 1D. 3
Answer» Correct Answer - b

Discussion

659.	If K is real number, then `(KA)^(-1)` =____.
Answer» Correct Answer - I

Discussion

660.	If `A={:[(-13,24),(-7,13)]:}`, then `A^(8)` = _______ .A. AB. IC. OD. None of these
Answer» Correct Answer - B `A={:[(-13,24),(-8,13)]:}` `A^(2){:[(-13,24),(-7,13)]:}{:[(-13,24),(-7,13)]:}` `{:[(169-168,-312+312),(91-91,-168+169)]:}={:[(1,0),(0,1)]:}` `A^(2)=I` `(A^(2))^(4)=I^(4)` `A^(8)=I`.

Discussion

661.	If `P=[(1,2,4),(3,1,0),(0,0,1)], Q=[(1,-2,-3),(-3,1,9),(0,0,-5)]`then `(PQ)^(-1)` equals toA. zero matrixB. `I_(3)`C. diag `[-5, -5, -5]`D. `-(1)/(5)I_(3)`
Answer» Correct Answer - D

Discussion

662.	If `P={:[(2,-1),(-1,3)]:},Q={:[(1,-2),(-3,0)]:}andR={:[(-2,1),(2,-1)]:}`. then find PQ+PR.A. `{:[(1,2),(2,1)]:}`B. `{:[(1,1),(2,2)]:}`C. `{:[(1,0),(0,1)]:}`D. `{:[(-1,-1),(-2,-2)]:}`
Answer» Correct Answer - D

Discussion

663.	If `A(x)={:[(e^(x),e^(x)),(e^(-x),e^(-x))]:}` then A(x) A(y)=A. A(x+y)A(x-y)B. A(x+y)+A(x-y)C. A(x+y)-A(x-y)D. A(x+y)
Answer» Correct Answer - B `A(x)={:[(e^(x),e^(x)),(e^(-x),e^(-x))]:}` `A(y)={:[(e^(y),e^(y)),(e^-y,e^(-y))]:}` `A(x)A(y)={:[(e^(x+y)+e^(x-y),e^(-x)e^(y)+e^(x)e^(-y)),(e^(-x)e^(y)+e^(-x)e^(-y),e^(-x)e^(y)+e^(-x)e^(-y))]:}` `{:[(e^(x+y)+e^(x-y),e^(x+y)+e^(x-y)),(e^(-x+y)+e^(-x-y),e^(-x+y)+e^(-x-y))]:}` `{:[(e^(x+y)+e^(x-y),e^(x+y)+e^(x-y)),(e^(-(x-y))+e^(-(x+y)),e^(-(x-y))+e^(-(x+y)))]:}` =A(x+y)+A(x-y).

Discussion

664.	If `{:[(2,4),(p,1)]:}{:[(-1,2),(3,1)]:}={:[(10,q),(-2,r)]:}`, then pq= _______ .A. 4(r-1)B. 5rC. 4r+2D. r
Answer» Correct Answer - A Given `{:[(2,4),(p,1)]:}{:[(-1,2),(3,1)]:}={:[(10,q),(-2,r)]:}` `{:[(-2+12,4+4),(-p+3,2p+1)]:}={:[(10,q),(-2,r)]:}` `{:[(10,8),(-p+3,2p+1)]:}={:[(10,q),(-2,r)]:}` `-p+3=-2,q=8,r=2p+1` `-p=-5,r=2(5)+1` `p=5,r=11`. Therefore pq `=pq=5(8)=40=4(r-1)`.

Discussion

665.	The matrix `A = [{:(a, d), (c, a):}]` is singular then a =____.
Answer» Correct Answer - `sqrt(bc)`

Discussion

666.	If A `= [s 2] " and "B = [(x), (y)]` then AB = _____.
Answer» Correct Answer - `(sx + 2y)_(1 xx 1)`

Discussion

667.	If `A={:[(6,-7,1),(3,-2,1)]:}andB={:[(5,1,2),(-4,3,-5)]:}` then find X such that 2A-B+X=O.A. `{:[(7,6,0),(1,3,-5)]:}`B. `{:[(10,1,2),(3,2,1)]:}`C. `{:[(7,7,0),(10,3,5)]:}`D. `{:[(-7,15,0),(-10,7,-7)]:}`
Answer» Correct Answer - D `A={:[(6,-7,1),(3,-2,1)]:}rArr{:[(12,-14,2),(6,-4,2)]:}` Given that 2A-B+X=O 2A-B+X2A+B=I+B-2A (Adding B and subtracting 2A) X=O+B-2A `O^(1)={:[(0,0,0),(0,0,0)]:}` `B={:[(5,1,2),(-4,3,-5)]:}` `X={:[(0+5-12,0+1+14,0+2-2),(0-4-6,0+3+4,0-5-2)]:}` `={:[(7,15,0),(-10,7,-7)]:}`.

Discussion

668.	If A and B commute, then `(A + B)^(2)` =___.
Answer» Correct Answer - `A^(2) + 2AB + B^(2)`

Discussion

669.	If A and B are commute then `(A+B)^(2)` =A. `A^(2)+B^(2)`B. `A^(2)+2AB+B^(2)`C. `A^(2)-B^(2)`D. `A^(2)-2AB+B^(2)`
Answer» Correct Answer - B If A and B are commute, then AB=BA.

Discussion

670.	If `A={:((1,-3,4),(2,1,-2)):},B={:((-2,-4,5),(1,-1,3)):}` and 5A-3B+2X=O, then X=A. `{:((-11,3,-5),(-7,-8,19)):}`B. `(1)/(2){:((-11,3,-5),(-7,-8,19)):}`C. `(1)/(2){:((11,-3,5),(7,8,-19)):}`D. None of these
Answer» Correct Answer - B 2X=3B-5A

Discussion

671.	A and B are two square matrices of same order. If `AB =B^(-1), " then " A^(-1)` =____.A. BAB. `A^(2)`C. `B^(2)`D. B
Answer» Correct Answer - C `AB = B^(-1)` `(AB) B = B^(-1) * B` `AB^(2) =I` `A^(-1) * A * B^(2) = A^(-1) * I` `I * B^(2) = A^(-1)` `therefore A^(-1) = B^(2)`.

Discussion

672.	If `{:[(0,3,-5),(-3,0,6),(5,-6,0)]:}` is aA. scalar matrix.B. symmetric matrix.C. skew symmetric.D. diagonal matrix.
Answer» Correct Answer - C Find `A^(T)and-A`.

Discussion

673.	If `A = [[(2, 0)], [(5, -3)]] , B = [(-2, 1), (3, -1)] ` then find the trace of `(AB^T)^T`A. 10B. 14C. `-4`D. `-18`
Answer» Correct Answer - B `A = ({:(2, 0), (5, -3):}) B =({:(-2, 1), (3, -1):})` `B^(T) = ({:(-2, 3), (1, -1):})` `A xx B^(T) = ({:(2, 0), (5, -3):}) ({:(-2, 3), (1, -1):})` ` = ({:(-4+10, 6-0), (-10-3, 15+3):}) = ({:(-4, 6), (-13, 18):})` `(AB^(T))^(T) = ({:(-4, -13), (6, 18):})` Trace of `(AB^(T))^(T) = -4 + 18 =14.`

Discussion

674.	If `A{:[(1,4,-3),(5,7,9)]:}andB={:[(5,-7),(4,5)]:}`, thenA. AB exists.B. BA exists.C. (A+B) exists.D. (A-B) exists.
Answer» Correct Answer - B Verify the addition and multiplication rules.

Discussion

675.	If the matrix `{:[(a+b,0,0),(0,b+c,0),(0,0,c+a)]:}`, (where a,b,c are positive integers) is a scalar matrix, then the value of (a+b+c) can beA. 6B. 8C. 5D. 7
Answer» Correct Answer - A In a scalar matrix, principal diagonal elements are same.

Discussion

676.	Find the inverse of the matrix A = `[{:(2, -4), (3, -5):}]`.
Answer» `\|A\| = \|{:(2, -4),(3, -5):}\| = -10 + 12 = 2 ne 0.` `therefore` A is non-singular and `A^(-1)` exists. `therefore A^(-1) = (1)/(ad-ac) [{:(d, -b), (-c, a):}] = (1)/(2) [{:(-5, 4),(-3, 2):}]` `A^(-1) = [{:(-(5)/(2), (4)/(2)), (-(3)/(2), (2)/(2)):}] = [{:(-(5)/(2), 2),(-(3)/(2), 1):}]`

Discussion

677.	`"If A" =[{:(p, q),(r, s):}]`, ps = 15 and det A = 21, then find the value of qr. (a) 6 (b) -6 (c) 5 (d) -8
Answer» `A = [{:(p, q), (r, s):}]` det A = ps - qr = 21 `rArr qr = ps - 21` qr = 15-21 `rArr qr = -6`.

Discussion

678.	`"If A" = ({:(3, -6),(-1, 2):}) " and B" = ({:(2, 6),(1, 3):})` are two matrices, then find AB + BA. (a) I (b) O (c) A (d) B
Answer» `AB = ({:(3, -6),(-1, 2):}) ({:(2, 6),(1, 3):})` `= ({:(6-6, 18-18),(-2+2, -6+6):}) = ({:(0, 0),(0, 0):})` `BA= ({:(2, 6),(1, 3):})({:(3, -6),(-1, 2):})` `=({:(6-6, -12+12),(3-3, -6+6):}) = ({:(0, 0),(0, 0):})` AB+BA = 0.

Discussion

679.	`"If A"= ({:(2, 3),(5, 1):}), B = ({:(a, b),(c, d):})` and AB = -13I, then find the value of a + b - c + d. (a) 5 (b) 3 (c) 2 (d) 1
Answer» AB = -13I `({:(2, 3),(5, 1):})({:(a, b),(c, d):}) = -13 ({:(1, 0),(0, 1):})` `({:(2a+ 3c, 2b+3d),(5a+c, 5b+d):}) = ({:(-13, 0),(0, -13):})` `rArr 2a + 3c =- 13 " "(1)` `5a +c = 0 " " (2)` `2a + 3d = 0 " " (3)` `5b +d = -13 " " (4)` Solving Eqs. (1) and (2), we get a = 1 and c=-5. Solving Eqs. (3) and (4), we get b = -3 and d=2. a + b - c + d = 1 - 3 + 5 + 2 = 5.

Discussion

680.	`"If" A = ({:(0, 2009),(-2009, 0):}), " then find " [A^(2009) + (A^(T))^(2009)].` (a) `({:(1, 0),(0, 1):})` (b) `({:(0, 0),(0, 0):})` (c) `({:(0, 2009),(-2009, 0):})` (d) None of these
Answer» Find `A^(2)` then calculate `A^(2009)`

Discussion

681.	` "If A" = [{:(2, 3),(-1, 4):}] " and B" = [{:(-3, 1),(4, -2):}] " then find" A-B.`
Answer» ` A -B= [{:(2, 3),(-1, 4):}] - [{:(-3, 1),(4, -2):}]` ` A -B= [{:(2-(-3), 3-1),(-1-4, 4-(-2)):}] = [{:(5, 2),(-5, 6):}]`

Discussion

682.	If A and B are symmetric matrices of same order, prove that (i) AB + BA is a symmetric matrix. (ii) AB – BA is a skew-symmetric matrix.
Answer» Given A and B are symmetric matrices ⇒ – A^T = A and B^T = B (i) To prove AB + BA is a symmetric matrix. Proof: Now (AB + BA)^T = (AB)^T + (BA)^T = B^TA^T + A^T B^T = BA + AB = AB + BA i.e. (AB + BA)^T = AB + BA ⇒ (AB + BA) is a symmetric matrix. (ii) To prove AB – BA is a skew symmetric matrix. Proof: (AB – BA)^T = (AB)^T – (BA)^T = B^T A^T – A^T B^T = BA – AB i.e. (AB – BA)^T = – (AB – BA) ⇒ AB – BA is a skew symmetric matrix.

682.

If A and B are symmetric matrices of same order, prove that (i) AB + BA is a symmetric matrix. (ii) AB – BA is a skew-symmetric matrix.

Answer»

Given A and B are symmetric matrices

⇒ – A^T = A and B^T = B

(i) To prove AB + BA is a symmetric matrix.

Proof: Now (AB + BA)^T = (AB)^T + (BA)^T = B^TA^T + A^T B^T

= BA + AB = AB + BA

i.e. (AB + BA)^T = AB + BA

⇒ (AB + BA) is a symmetric matrix.

(ii) To prove AB – BA is a skew symmetric matrix.

Proof: (AB – BA)^T = (AB)^T – (BA)^T = B^T A^T – A^T B^T = BA – AB

i.e. (AB – BA)^T = – (AB – BA)

⇒ AB – BA is a skew symmetric matrix.

Discussion

683.	Let A and B be two symmetric matrices. Prove that AB = BA if and only if AB is a symmetric matrix.
Answer» Let A and B be two symmetric matrices ⇒ A^T = A and B^T = B ... (1) Given that AB = BA (2) To prove AB is symmetric: Now (AB)^T = B^TA^T = BA (from(1)) But (AB)^T = AB by ... (2) ⇒ AB is symmetric. Conversely let AB be a symmetric matrix. ⇒ (AB)^T = AB i.e. B^T A^T = AB i.e. BA = AB (from (1)) ⇒ AB is symmetric

683.

Let A and B be two symmetric matrices. Prove that AB = BA if and only if AB is a symmetric matrix.

Answer»

Let A and B be two symmetric matrices

⇒ A^T = A and B^T = B ... (1)

Given that AB = BA (2)

To prove AB is symmetric:

Now (AB)^T = B^TA^T = BA

(from(1)) But (AB)^T = AB by ... (2)

⇒ AB is symmetric.

Conversely let AB be a symmetric matrix.

⇒ (AB)^T = AB

i.e. B^T A^T = AB

i.e. BA = AB (from (1))

⇒ AB is symmetric

Discussion

684.	If `{:A=[(1,0,0),(0,1,0),(a,b,-1)]:}`, then `A^2` is equal toA. a null matrixB. a unit matrixC. `-A`D. A
Answer» Correct Answer - B We have, `{:A^2=[(1,0,0),(0,1,0),(a,b,-1)][(1,0,0),(0,1,0),(a,b,-1)]=[(1,0,0),(0,1,0),(0,0,1)]:}`=a unit matrix.

Discussion

685.	Find the values of p, q, r and s if \(\begin{bmatrix} p^2 -1 & \ 0 &-31 - q^3 \\[0.3em] 7 &r + 1& 9\\[0.3em] -2 &8 & s - 1 \end{bmatrix}\)= \(\begin{bmatrix} 1& \ 0 &-4 \\[0.3em] 7 &3/2& 9\\[0.3em] -2 &8 & π \end{bmatrix}\)
Answer» When two matrices (of same order) are equal then their corresponding entries are equal. Here \(\begin{bmatrix} p^2 -1 & \ 0 &-31 - q^3 \\[0.3em] 7 &r + 1& 9\\[0.3em] -2 &8 & s - 1 \end{bmatrix}\)= \(\begin{bmatrix} 1& \ 0 &-4 \\[0.3em] 7 &3/2& 9\\[0.3em] -2 &8 & π \end{bmatrix}\) ⇒ p² – 1 = 1 ⇒ p² = 1 + 1 = 2 p = ± √2 -31 – q³ = -4 -q³ = -4 + 31 = 27 q³ = -27 = (-3)³ ⇒ q = -3 r + 1 = 3/2 ⇒ r = 3/2 – 1 = 3 - 2/2 = 1/2 s – 1 = π ⇒ s = – π + 1 (i.e.,) s = 1 – π So, p = ± √2, q = -3, r = 1/2 and s = 1 – π

685.

Find the values of p, q, r and s if \(\begin{bmatrix} p^2 -1 & \ 0 &-31 - q^3 \\[0.3em] 7 &r + 1& 9\\[0.3em] -2 &8 & s - 1 \end{bmatrix}\)= \(\begin{bmatrix} 1& \ 0 &-4 \\[0.3em] 7 &3/2& 9\\[0.3em] -2 &8 & π \end{bmatrix}\)

Answer»

When two matrices (of same order) are equal then their corresponding entries are equal.

Here \(\begin{bmatrix} p^2 -1 & \ 0 &-31 - q^3 \\[0.3em] 7 &r + 1& 9\\[0.3em] -2 &8 & s - 1 \end{bmatrix}\)= \(\begin{bmatrix} 1& \ 0 &-4 \\[0.3em] 7 &3/2& 9\\[0.3em] -2 &8 & π \end{bmatrix}\)

⇒ p² – 1 = 1

⇒ p² = 1 + 1 = 2

p = ± √2

-31 – q³ = -4

-q³ = -4 + 31 = 27

q³ = -27 = (-3)³

⇒ q = -3

r + 1 = 3/2

⇒ r = 3/2 – 1 = 3 - 2/2 = 1/2

s – 1 = π

⇒ s = – π + 1 (i.e.,) s = 1 – π

So, p = ± √2, q = -3, r = 1/2 and s = 1 – π

Discussion

686.	Let M be `a3xx3` matrix satisfying `{:M[(0),(1),(0)]=[(1),(-1),(6)]=[(1),(1),(-1)]and","M=[(1),(1),(1)]=[(0),(0),(12)]:}` Then the sum of the diagonal entries of M, isA. 7B. 8C. 9D. 6
Answer» Correct Answer - C Let `M={:[(a,b,c),(x,y,z),(l,m,n)]:}`. Then, `{:M[(0),(1),(0)]=[(1),(-1),(6)]rArr [(b),(y),(m)]=[(-1),(2),(3)]rArr b=-1,y=2,m=3:}` `{:M[(1),(-1),(0)]=[(1),(1),(1)]rArr a-b=1,x-y=1,l-m=-1:}` `rArra=0,x=3,l=2` and, `{:M[(1),(1),(1)]=[(0),(0),(12)]rArra+b+c=0,x+y+z=0,l+m+n=12:}` `rArr c=1,z=-1,n=7` `:. " Sum of diagonal elements of "M=a+y+n=0+2+7=9`

Discussion

687.	Determine the value of x + y if \(\begin{bmatrix} 2x + y &4x \\[0.3em] 5x - 7 &4x \end{bmatrix}\)= \(\begin{bmatrix} 7 &7x - 13 \\[0.3em] y &x + 6 \end{bmatrix}\)
Answer» \(\begin{bmatrix} 2x + y &4x \\[0.3em] 5x - 7 &4x \end{bmatrix}\)= \(\begin{bmatrix} 7 &7x - 13 \\[0.3em] y &x + 6 \end{bmatrix}\) ⇒ 2x + y = 7 ... (1) 4x = 7y – 13 ... (2) 5x – 7 = y … (3) 4x = x + 6 ... (4) From (4) 4x – x = 6 3x = 6 ⇒ x = 6/3 = 2 Substituting x = 2 in (1), we get 2(2) + y = 7 ⇒ 4 + y = 7 ⇒ y = 7 – 4 = 3 So x = 2 and y = 3 ∴ x + y = 2 + 3 = 5

687.

Determine the value of x + y if \(\begin{bmatrix} 2x + y &4x \\[0.3em] 5x - 7 &4x \end{bmatrix}\)= \(\begin{bmatrix} 7 &7x - 13 \\[0.3em] y &x + 6 \end{bmatrix}\)

Answer»

\(\begin{bmatrix} 2x + y &4x \\[0.3em] 5x - 7 &4x \end{bmatrix}\)= \(\begin{bmatrix} 7 &7x - 13 \\[0.3em] y &x + 6 \end{bmatrix}\)

⇒ 2x + y = 7 ... (1)

4x = 7y – 13 ... (2)

5x – 7 = y … (3)

4x = x + 6 ... (4)

From (4) 4x – x = 6

3x = 6 ⇒ x = 6/3 = 2

Substituting x = 2 in (1), we get

2(2) + y = 7 ⇒ 4 + y = 7 ⇒ y = 7 – 4 = 3

So x = 2 and y = 3

∴ x + y = 2 + 3 = 5

Discussion

688.	The value of x for which the metrix product `{:[(2,0,7),(0,1,0),(1,-2,1)][(-x,14x,7x),(0,1,0),(x,-4x,-2x)]:}`equal an identity matrix, isA. `1//2`B. `1//3`C. `1//4`D. `1//5`
Answer» Correct Answer - D We have, `{:[(2,0,7),(0,1,0),(1,-2,1)][(-x,14x,7x),(0,1,0),(x,-4x,-2x)]=[(1,0,0),(0,1,0),(0,0,1)]:}` `rArr{:[(5x,0,0),(0,1,0),(0,10x-2,5x)]=[(1,0,0),(0,1,0),(0,0,1)]:}` `rArr 5x=1,10x-2=0rArrx=1/5`

Discussion

689.	Find the matrix A such that\( \begin{bmatrix}1 & 1 \\[0.3em]0 & 1 \\[0.3em]\end{bmatrix}\)A = \( \begin{bmatrix}3 &3 & 5 \\[0.3em]1& 0 &1 \\[0.3em]\end{bmatrix}\)
Answer» \( \begin{bmatrix}1 & 1 \\[0.3em]0 & 1 \\[0.3em]\end{bmatrix}\)A = \( \begin{bmatrix}3 &3 & 5 \\[0.3em]1& 0 &1 \\[0.3em]\end{bmatrix}\) We know that the two matrices are eligible for their product only when the number of columns of first matrix is equal to the number of rows of the second matrix. So, \( \begin{bmatrix}1 & 1 \\[0.3em]0 & 1 \\[0.3em]\end{bmatrix}\) is 2 × 2 matrix, and \( \begin{bmatrix}3 &3 & 5 \\[0.3em]1& 0 &1 \\[0.3em]\end{bmatrix}\) is 3 × 2 matrix Now, In order to get a 3 × 2 matrix as solution 2 × 2 matrix should be multiplied by 2 × 3 matrix. Hence matrix A is 2 × 3 matrix. Let, A = \( \begin{bmatrix}a &b &c \\[0.3em]d& e &f \\[0.3em]\end{bmatrix}\) So the given question becomes, \( \begin{bmatrix}1 & 1 \\[0.3em]0 & 1 \\[0.3em]\end{bmatrix}\) \( \begin{bmatrix}a &b &c \\[0.3em]d& e &f \\[0.3em]\end{bmatrix}\) = \( \begin{bmatrix}3 &3 & 5 \\[0.3em]1& 0 &1 \\[0.3em]\end{bmatrix}\) [as c_ij = a_i1b_1j + a_i2b_2j + … + a_inb_nj] ⇒ \( \begin{bmatrix}a+d &b+e &c+f \\[0.3em]d& e &f \\[0.3em]\end{bmatrix}\) = \( \begin{bmatrix}3 &3 & 5 \\[0.3em]1& 0 &1 \\[0.3em]\end{bmatrix}\) To satisfy the above equality condition, corresponding entries of the matrices should be equal, i.e., d = 1, e = 0, f = 1 a + d = 3 ⇒ a + 1 = 3 ⇒ a = 2 b + e = 3 ⇒ b + 0 = 3 ⇒ b = 3 c + f = 5 ⇒ c + 1 = 5 ⇒ c = 4 Now, Substituting these values in matrix A, we get A = \( \begin{bmatrix}a &b &c \\[0.3em]d& e &f \\[0.3em]\end{bmatrix}\) = \( \begin{bmatrix}2 &3 &4 \\[0.3em]1& 0 &1\\[0.3em]\end{bmatrix}\) is the matrix A.

689.

Find the matrix A such that\( \begin{bmatrix}1 & 1 \\[0.3em]0 & 1 \\[0.3em]\end{bmatrix}\)A = \( \begin{bmatrix}3 &3 & 5 \\[0.3em]1& 0 &1 \\[0.3em]\end{bmatrix}\)

Answer»

\( \begin{bmatrix}1 & 1 \\[0.3em]0 & 1 \\[0.3em]\end{bmatrix}\)A = \( \begin{bmatrix}3 &3 & 5 \\[0.3em]1& 0 &1 \\[0.3em]\end{bmatrix}\)

We know that the two matrices are eligible for their product only when the number of columns of first matrix is equal to the number of rows of the second matrix.

So,

\( \begin{bmatrix}1 & 1 \\[0.3em]0 & 1 \\[0.3em]\end{bmatrix}\) is 2 × 2 matrix, and
\( \begin{bmatrix}3 &3 & 5 \\[0.3em]1& 0 &1 \\[0.3em]\end{bmatrix}\) is 3 × 2 matrix

Now,

In order to get a 3 × 2 matrix as solution 2 × 2 matrix should be multiplied by 2 × 3 matrix. Hence matrix A is 2 × 3 matrix.

Let,

A = \( \begin{bmatrix}a &b &c \\[0.3em]d& e &f \\[0.3em]\end{bmatrix}\)

So the given question becomes,

\( \begin{bmatrix}1 & 1 \\[0.3em]0 & 1 \\[0.3em]\end{bmatrix}\) \( \begin{bmatrix}a &b &c \\[0.3em]d& e &f \\[0.3em]\end{bmatrix}\) = \( \begin{bmatrix}3 &3 & 5 \\[0.3em]1& 0 &1 \\[0.3em]\end{bmatrix}\)

[as c_ij = a_i1b_1j + a_i2b_2j + … + a_inb_nj]

⇒ \( \begin{bmatrix}a+d &b+e &c+f \\[0.3em]d& e &f \\[0.3em]\end{bmatrix}\) = \( \begin{bmatrix}3 &3 & 5 \\[0.3em]1& 0 &1 \\[0.3em]\end{bmatrix}\)

To satisfy the above equality condition, corresponding entries of the matrices should be equal, i.e.,

d = 1, e = 0, f = 1

a + d = 3

⇒ a + 1 = 3

⇒ a = 2

b + e = 3

⇒ b + 0 = 3

⇒ b = 3

c + f = 5

⇒ c + 1 = 5

⇒ c = 4

Now,

Substituting these values in matrix A, we get

A = \( \begin{bmatrix}a &b &c \\[0.3em]d& e &f \\[0.3em]\end{bmatrix}\)

= \( \begin{bmatrix}2 &3 &4 \\[0.3em]1& 0 &1\\[0.3em]\end{bmatrix}\) is the matrix A.

Discussion

690.	If `A` is a square matrix of order `3` such that `\|A\|=5`, then `\|Adj(4A)\|=`A. `5^(3)xx4^(2)`B. `5^(2)xx4^(3)`C. `5^(2)xx16^(3)`D. `5^(3)xx16^(2)`
Answer» Correct Answer - C `(c )` `Adj(4A)=4^(2)Adj(A)=16Adj(A)` `implies\|Adj4A\|=16^(3)=\|AdjA\|=16^(3)*5^(2)`

Discussion

691.	If A and B are squar matrices of order 3 such that \|A\|=-1, \|B\|=3 then \|3AB\| is equal to
Answer» `\|3AB\|=>3^n\|AB\|` `3^3\|A\|\|B\|=>27(-1)*3` `\|3AB\|=-81` order=3.

Discussion

692.	Find the matrix A such that\( \begin{bmatrix}1 & 1 \\[0.3em]0 & 1 \\[0.3em]\end{bmatrix}\)A = \( \begin{bmatrix}3 &3 & 5 \\[0.3em]1& 0 &1 \\[0.3em]\end{bmatrix}\)
Answer» \( \begin{bmatrix}1 & 1 \\[0.3em]0 & 1 \\[0.3em]\end{bmatrix}\)A = \( \begin{bmatrix}3 &3 & 5 \\[0.3em]1& 0 &1 \\[0.3em]\end{bmatrix}\) We know that the two matrices are eligible for their product only when the number of columns of first matrix is equal to the number of rows of the second matrix. So, \( \begin{bmatrix}1 & 1 \\[0.3em]0 & 1 \\[0.3em]\end{bmatrix}\) is 2 × 2 matrix, and \( \begin{bmatrix}3 &3 & 5 \\[0.3em]1& 0 &1 \\[0.3em]\end{bmatrix}\) is 3 × 2 matrix Now, In order to get a 3 × 2 matrix as solution 2 × 2 matrix should be multiplied by 2 × 3 matrix. Hence matrix A is 2 × 3 matrix. Let, A = \( \begin{bmatrix}a &b &c \\[0.3em]d& e &f \\[0.3em]\end{bmatrix}\) So the given question becomes, \( \begin{bmatrix}1 & 1 \\[0.3em]0 & 1 \\[0.3em]\end{bmatrix}\) \( \begin{bmatrix}a &b &c \\[0.3em]d& e &f \\[0.3em]\end{bmatrix}\) = \( \begin{bmatrix}3 &3 & 5 \\[0.3em]1& 0 &1 \\[0.3em]\end{bmatrix}\) [as c_ij = a_i1b_1j + a_i2b_2j + … + a_inb_nj] ⇒ \( \begin{bmatrix}a+d &b+e &c+f \\[0.3em]d& e &f \\[0.3em]\end{bmatrix}\) = \( \begin{bmatrix}3 &3 & 5 \\[0.3em]1& 0 &1 \\[0.3em]\end{bmatrix}\) To satisfy the above equality condition, corresponding entries of the matrices should be equal, i.e., d = 1, e = 0, f = 1 a + d = 3 ⇒ a + 1 = 3 ⇒ a = 2 b + e = 3 ⇒ b + 0 = 3 ⇒ b = 3 c + f = 5 ⇒ c + 1 = 5 ⇒ c = 4 Now, Substituting these values in matrix A, we get A = \( \begin{bmatrix}a &b &c \\[0.3em]d& e &f \\[0.3em]\end{bmatrix}\) = \( \begin{bmatrix}2 &3 &4 \\[0.3em]1& 0 &1\\[0.3em]\end{bmatrix}\) is the matrix A.

692.

Find the matrix A such that\( \begin{bmatrix}1 & 1 \\[0.3em]0 & 1 \\[0.3em]\end{bmatrix}\)A = \( \begin{bmatrix}3 &3 & 5 \\[0.3em]1& 0 &1 \\[0.3em]\end{bmatrix}\)

Answer»

\( \begin{bmatrix}1 & 1 \\[0.3em]0 & 1 \\[0.3em]\end{bmatrix}\)A = \( \begin{bmatrix}3 &3 & 5 \\[0.3em]1& 0 &1 \\[0.3em]\end{bmatrix}\)

We know that the two matrices are eligible for their product only when the number of columns of first matrix is equal to the number of rows of the second matrix.

So,

\( \begin{bmatrix}1 & 1 \\[0.3em]0 & 1 \\[0.3em]\end{bmatrix}\) is 2 × 2 matrix, and
\( \begin{bmatrix}3 &3 & 5 \\[0.3em]1& 0 &1 \\[0.3em]\end{bmatrix}\) is 3 × 2 matrix

Now,

In order to get a 3 × 2 matrix as solution 2 × 2 matrix should be multiplied by 2 × 3 matrix. Hence matrix A is 2 × 3 matrix.

Let,

A = \( \begin{bmatrix}a &b &c \\[0.3em]d& e &f \\[0.3em]\end{bmatrix}\)

So the given question becomes,

\( \begin{bmatrix}1 & 1 \\[0.3em]0 & 1 \\[0.3em]\end{bmatrix}\) \( \begin{bmatrix}a &b &c \\[0.3em]d& e &f \\[0.3em]\end{bmatrix}\) = \( \begin{bmatrix}3 &3 & 5 \\[0.3em]1& 0 &1 \\[0.3em]\end{bmatrix}\)

[as c_ij = a_i1b_1j + a_i2b_2j + … + a_inb_nj]

⇒ \( \begin{bmatrix}a+d &b+e &c+f \\[0.3em]d& e &f \\[0.3em]\end{bmatrix}\) = \( \begin{bmatrix}3 &3 & 5 \\[0.3em]1& 0 &1 \\[0.3em]\end{bmatrix}\)

To satisfy the above equality condition, corresponding entries of the matrices should be equal, i.e.,

d = 1, e = 0, f = 1

a + d = 3

⇒ a + 1 = 3

⇒ a = 2

b + e = 3

⇒ b + 0 = 3

⇒ b = 3

c + f = 5

⇒ c + 1 = 5

⇒ c = 4

Now,

Substituting these values in matrix A, we get

A = \( \begin{bmatrix}a &b &c \\[0.3em]d& e &f \\[0.3em]\end{bmatrix}\)

= \( \begin{bmatrix}2 &3 &4 \\[0.3em]1& 0 &1\\[0.3em]\end{bmatrix}\) is the matrix A.

Discussion

693.	Without using the concept of the inverse of a matrix, find the matrix \( \begin{bmatrix} x & y \\[0.3em] z & u\\[0.3em] \end{bmatrix}\) such that : \( \begin{bmatrix} 5 & -7 \\[0.3em] -2 & 3\\[0.3em] \end{bmatrix}\)\( \begin{bmatrix} x & y \\[0.3em] z & u\\[0.3em] \end{bmatrix}\)=\( \begin{bmatrix} -16 & -6 \\[0.3em] 7 & 2\\[0.3em] \end{bmatrix}\)
Answer» Given, \( \begin{bmatrix}5 & -7 \\[0.3em]-2 & 3\\[0.3em]\end{bmatrix}\)\( \begin{bmatrix}x & y \\[0.3em]z & u\\[0.3em]\end{bmatrix}\)=\( \begin{bmatrix}-16 & -6 \\[0.3em]7 & 2\\[0.3em]\end{bmatrix}\) Multiplying we get, \( \begin{bmatrix}5x-7z & 5y-7u \\[0.3em]-2x+3z & -2y+3u\\[0.3em]\end{bmatrix}\)= \( \begin{bmatrix}-16 & -6 \\[0.3em]7 & 2\\[0.3em]\end{bmatrix}\) From above we can see that, 5x – 7z = – 16 …(1) –2x + 3z = 7 ….(2) 5y – 7u = – 6 …..(3) –2y + 3u = 2 ……(4) Now, We have to solve these equations to find values of x, y, z and u Multiplying eq (1) by 2 and eq (2) by 5 and adding the equations we get, 10x – 14z + 10x + 15z = – 32 + 35 Z = 3 Putting this value in eq (1) we get, 5x – 21 = – 16 5x = 5 x = 1 Now, Multiplying eq (3) by 2 and eq (4) by 5 and adding we get, 10y – 14u + 10y + 15u = – 12 + 10 u = – 2 Putting value of u in equation (3) we get, 5 y + 14 = – 6 5 y = – 20 y = – 4 Therefore now we have, \( \begin{bmatrix}x & y \\[0.3em]z & u\\[0.3em]\end{bmatrix}\)= \( \begin{bmatrix}1 &-4 \\[0.3em]3 &-2\\[0.3em]\end{bmatrix}\)

693.

Without using the concept of the inverse of a matrix, find the matrix \( \begin{bmatrix} x & y \\[0.3em] z & u\\[0.3em] \end{bmatrix}\) such that : \( \begin{bmatrix} 5 & -7 \\[0.3em] -2 & 3\\[0.3em] \end{bmatrix}\)\( \begin{bmatrix} x & y \\[0.3em] z & u\\[0.3em] \end{bmatrix}\)=\( \begin{bmatrix} -16 & -6 \\[0.3em] 7 & 2\\[0.3em] \end{bmatrix}\)

Answer»

Given,
\( \begin{bmatrix}5 & -7 \\[0.3em]-2 & 3\\[0.3em]\end{bmatrix}\)\( \begin{bmatrix}x & y \\[0.3em]z & u\\[0.3em]\end{bmatrix}\)=\( \begin{bmatrix}-16 & -6 \\[0.3em]7 & 2\\[0.3em]\end{bmatrix}\)

Multiplying we get,

\( \begin{bmatrix}5x-7z & 5y-7u \\[0.3em]-2x+3z & -2y+3u\\[0.3em]\end{bmatrix}\)= \( \begin{bmatrix}-16 & -6 \\[0.3em]7 & 2\\[0.3em]\end{bmatrix}\)

From above we can see that,

5x – 7z = – 16 …(1)

–2x + 3z = 7 ….(2)

5y – 7u = – 6 …..(3)

–2y + 3u = 2 ……(4)

Now,

We have to solve these equations to find values of x, y, z and u Multiplying eq (1) by 2 and eq (2) by 5 and adding the equations we get,

10x – 14z + 10x + 15z = – 32 + 35

Z = 3

Putting this value in eq (1) we get,

5x – 21 = – 16

5x = 5

x = 1

Now,

Multiplying eq (3) by 2 and eq (4) by 5 and adding we get,

10y – 14u + 10y + 15u = – 12 + 10

u = – 2

Putting value of u in equation (3) we get,

5 y + 14 = – 6

5 y = – 20

y = – 4

Therefore now we have,

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

Discussion

694.	Which of the following is an orthogonal matrix ?A. `[(6//7,2//7,-3//7),(2//7,3//7,6//7),(3//7,-6//7,2//7)]`B. `[(6//7,2//7,3//7),(2//7,-3//7,6//7),(3//7,6//7,-2//7)]`C. `[(-6//7,-2//7,-3//7),(2//7,3//7,6//7),(-3//7,6//7,2//7)]`D. `[(6//7,-2//7,3//7),(2//7,2//7,-3//7),(-6//7,2//7,3//7)]`
Answer» Correct Answer - A Matrix `[(a_(1),a_(2),a_(3)),(b_(1),b_(2),b_(3)),(c_(1),c_(2),c_(3))]` is orthogonal if `sum a_(i)^(2)=sum b_(i)^(2)=sun c_(i)^(2)=1, sum a_(i)b_(i)=sum b_(i)b_(i)=sum c_(i)a_(i)=0`

Discussion

695.	If `A=[(0,x),(y,0)]` and `A^(3)+A=O` then sum of possible values of xy is
Answer» Correct Answer - B `A^(3)+A=O` `implies A(A^(2)+I)=O` `implies \|A(A^(2)+I)\|=0` `implies \|A\|=0` or `\|A^(2)+I\|=0` If `\|A\|=0`, then `xy=0` `A^(2)+I=[(0,x),(y,0)][(0,x),(y,0)]+[(1,0),(0,1)]` `=[(xy,0),(0,xy)]+[(1,0),(0,1)]` `=[(xy+1,0),(0,xy+1)]` `\|A^(2)+I\|=0` `implies (xy+1)^(2)=0` `implies xy=-1` So, sum of possible values of `xy` is `-1`.

Discussion

696.	If `A ,B ,A+I ,A+B`are idempotent matrices, then `A B`is equal toA. `BA`B. `-BA`C. `I`D. `O`
Answer» Correct Answer - B Given `A, B, A + I, A+B` are idempotent. Hence, `A^(2)=A, B^(2)=B, (A+I)^(2)=A+I` and `(A+B)^(2)=A+B` `implies A^(2)+B^(2)+AB+BA=A+B` `implies A+B+AB+BA=A+B` `implies AB+BA=O`

Discussion

697.	The value of x for which the matrix `\|(-x,x,2),(2,x,-x),(x,-2,-x)\|` will be non-singular, areA. `-2le x le2`B. for all x other than 2 and -2C. `xge2`D. `xle-2`0
Answer» Correct Answer - B

Discussion

698.	Let A and B be matrices of order `3 xx 3.` If `AB = 0,` then which of the following can be concluded?A. `A=Oand B=O`B. `absA=OandabsB=O`C. either absA=Oor absB=O`D. `A=Oor B=O`
Answer» Correct Answer - C

Discussion

699.	How many different diagonal matrices of order n can be formed which are idempotent ?
Answer» Correct Answer - `2^(n)` Matrix A is diagonal matrix. `:. A`=dia. `(a_(1), a_(2), ..., a_(n))` `implies A^(2)=` dia. `((a_(1))^(2), (a_(2))^(2),..,(a_(n))^(2))` since A involuntary, we have `:. A^(2)=A` `implies (a_(1))^(2)=a_(1), (a_(2))^(2)=a_(2), ..., (a_(n))^(2)=a_(2),...,(a_(n))^(2)=a_(n)`. `implies a_(1), a_(2), ..., a_(n)=0, 1`. Thus, number of required matrix A is `2xx2xx2xx...n` times `=2^(n)`

Discussion

700.	For what value of x, matrix [(3 - 2x, x + 1), (2, 4)] is singular?
Answer» For a singular matrix \|A\| = 0 12 - 8x - 2x - 2 = 0 - 10x = - 10 x = 1

700.

For what value of x, matrix [(3 - 2x, x + 1), (2, 4)] is singular?

Answer»

For a singular matrix |A| = 0

12 - 8x - 2x - 2 = 0

- 10x = - 10

x = 1

Discussion

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Let A and B be matrices of order `3 xx 3.` If `AB = 0,` then which of the following can be concluded?A. `A=Oand B=O`B. `absA=OandabsB=O`C. either absA=Oor absB=O`D. `A=Oor B=O`

How many different diagonal matrices of order n can be formed which are idempotent ?

For what value of x, matrix [(3 - 2x, x + 1), (2, 4)] is singular?

If `[{:(x, 0), (0, y):}] + [{:(-y, 4),(7, x):}] = [{:(4, 4), (7, 6):}]` then x = __ and y=_.

Find the values of p, q, r and s if \(\begin{bmatrix} p^2 -1 & \ 0 &-31 - q^3 \\[0.3em] 7 &r + 1& 9\\[0.3em] -2 &8 & s - 1 \end{bmatrix}\)= \(\begin{bmatrix} 1& \ 0 &-4 \\[0.3em] 7 &3/2& 9\\[0.3em] -2 &8 & π \end{bmatrix}\)

Determine the value of x + y if \(\begin{bmatrix} 2x + y &4x \\[0.3em] 5x - 7 &4x \end{bmatrix}\)= \(\begin{bmatrix} 7 &7x - 13 \\[0.3em] y &x + 6 \end{bmatrix}\)

Find the matrix A such that\( \begin{bmatrix}1 & 1 \\[0.3em]0 & 1 \\[0.3em]\end{bmatrix}\)A = \( \begin{bmatrix}3 &3 & 5 \\[0.3em]1& 0 &1 \\[0.3em]\end{bmatrix}\)

Find the matrix A such that\( \begin{bmatrix}1 & 1 \\[0.3em]0 & 1 \\[0.3em]\end{bmatrix}\)A = \( \begin{bmatrix}3 &3 & 5 \\[0.3em]1& 0 &1 \\[0.3em]\end{bmatrix}\)