NCERT Solutions for Class 12 Maths Chapter 3 - Matrices

Question 1(i): In the matrix A=[2519−735−2521231−517] , write: The order of the matrix

Answer:

A=[2519−735−2521231−517]

(i) The order of the matrix = number of row × number of columns =3×4 .

Question 1(ii): In the matrix A=[2519−735−2521231−517] , write:

The number of elements

Answer:

A=[2519−735−2521231−517]

(ii) The number of elements 3×4=12.

Question 1(iii): In the matrix A=[2519−735−2521231−517] , write:

Write the elements a ₁₃ , a ₂₁ , a ₃₃ , a ₂₄ , a ₂₃

Answer:

A=[2519−735−2521231−517]

(iii) An element aij implies the element in row number i and column number j.

a13=19 a21=35

a33=−5 a24=12

a23=52

Question 2: If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements?

Answer:

A matrix has 24 elements.

The possible orders are :

1×24,24×1,2×12,12×2,3×8,8×3,4×6and6×4 .

If it has 13 elements, then the possible orders are :

1×13and13×1 .

Question 3: If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?

Answer:

A matrix has 18 elements.

The possible orders are as given below

1×18,18×1,2×9,9×2,3×6and6×3

If it has 5 elements, then possible orders are :

1×5and5×1 .

Question 4(i): Construct a 2 × 2 matrix, A=[aij] whose elements are given by:

aij=(i+j)22

Answer:

A=[aij]

(i) aij=(i+j)22

Each element of this matrix is calculated as follows

a11=(1+1)22=222=42=2

a22=(2+2)22=422=162=8

a12=(1+2)22=322=92=4.5

a21=(2+1)22=322=92=4.5

Matrix A is given by

A=[24.54.58]

Question 4(ii): Construct a 2 × 2 matrix, A=[aij] , whose elements are given by:

aij=ij

Answer:

A 2 × 2 matrix, A=[aij]

(ii) aij=ij

a11=11=1

a22=22=1

a12=12

a21=21=2

Hence, the matrix is

A=[11221]

Question 4(iii): Construct a 2 × 2 matrix, A=[aij] , whose elements are given by:

aij=(i+2j)22

Answer:

(iii)

aij=(i+2j)22

a11=(1+(2×1))22=(1+2)22=322=92

a22=(2+(2×2))22=(2+4)22=622=362=18

a21=(2+(2×1))22=(2+2)22=422=162=8

a12=(1+(2×2))22=(1+4)22=522=252

Hence, the matrix is given by

A=[92252818]

Question 5(i): Construct a 3 × 4 matrix, whose elements are given by:

aij=12|−3i+j|

Answer:

(i)

aij=12|−3i+j|

a11=|−3+1|2=22=1

a12=|(−3×1)+2|2=12

a13=|(−3×1)+3|2=0

a21=|(−3×2)+1|2=52

a22=|(−3×2)+2|2=42=2

a23=|(−3×2)+3|2=|−6+3|2=|−3|2=32

a31=|(−3×3)+1|2=82=4

a32=|(−3×3)+2|2=72

a33=|(−3×3)+3|2=|−9+3|2=|−6|2=62=3

a14=|(−3×1)+4|2=|−3+4|2=|1|2=12

a24=|(−3×2)+4|2=|−6+4|2=|−2|2=22=1

a34=|(−3×3)+4|2=|−9+4|2=|−5|2=52

Hence, the required matrix of the given order is

A=[112012522321472352]

Question 5(ii): Construct a 3 × 4 matrix, whose elements are given by:

aij=2i−j

Answer:

A 3 × 4 matrix,

(ii) aij=2i−j

a11=2×1−1=2−1=1

a12=2×1−2=2−2=0

a13=2×1−3=2−3=−1

a21=2×2−1=4−1=3

a22=2×2−2=4−2=2

a23=2×2−3=4−3=1

a31=2×3−1=6−1=5

a32=2×3−2=6−2=4

a33=2×3−3=6−3=3

a14=2×1−4=2−4=−2

a24=2×2−4=4−4=0

a34=2×3−4=6−4=2

Hence, the matrix is

A=[10−1−2 32105432]

Question 6(i): Find the values of x, y, and z from the following equations:

[43x5]=[yz15]

Answer:

(i) [43x5]=[yz15]

If two matrices are equal, then their corresponding elements are also equal.

∴ x=1,y=4andz=3

Question 6(ii): Find the values of x, y and z from the following equations:

[x+y25+zxy]=[6258]

Answer:

(ii)

[x+y25+zxy]=[6258]

If two matrices are equal, then their corresponding elements are also equal.

∴ x+y=6 ⋅⋅⋅⋅⋅⋅⋅⋅⋅(i)

x=6−y

xy=8 ⋅⋅⋅⋅⋅⋅⋅⋅⋅(ii)

Solving equation (i) and (ii),

(6−y)y=8

6y−y2=8

y2−6y+8=0

solving this equation we get,

y=4andy=2

Putting the values of y, we get

x=2andx=4

And also equating the first element of the second raw

5+z=5 , z=0

Hence,

x=2,y=4,z=0andx=4,y=2,z=0

Question 6(iii): Find the values of x, y, and z from the following equations

[x+y+zx+zy+z]=[957]

Answer:

(iii)

[x+y+zx+zy+z]=[957]

If two matrices are equal, then their corresponding elements are also equal

x+y+z=9........(1)

x+z=5..............(2)

y+z=7..............(3)

subtracting (2) from (1) we will get y=4

substituting the value of y in equation (3) we will get z=3

now substituting the value of z in equation (2) we will get x=2

therefore,

x=2 , y=4 and z=3

Question 7: Find the value of a, b, c, and d from the equation:

[a−b2a+c2a−b3c+d]=[−15013]

Answer:

[a−b2a+c2a−b3c+d]=[−15013]

If two matrices are equal, then their corresponding elements are also equal

a−b=−1 .............................1

2a+c=5 .............................2

2a−b=0 .............................3

3c+d=13 .............................4

Solving equation 1 and 3 , we get

a=1andb=2

Putting the value of a in equation 2, we get

c=3

Putting the value of c in equation 4 , we get

d=4

Question 8: A=[aij]m×n is a square matrix, if

(A) m<n

(B) m>n

(C) m=n

(D) None of these

Answer:

A square matrix has the number of rows and columns equal.

Thus, for A=[aij]m×n to be a square matrix m and n should be equal.

Option (c) is correct.

Question 9: Which of the given values of x and y make the following pair of matrices equal

[3x+75y+12−3x] , [0y−284]

(A) x=−13,y=7

(B) Not possible to find

(C) y=7,x=−23

(D) x=−13,y=−23

Answer:

Given, [3x+75y+12−3x] =[0y−284]

If two matrices are equal, then their corresponding elements are also equal

3x+7=0⇒x=−73

y−2=5⇒y=5+2=7

y+1=8⇒y=8−1=7

2−3x=4⇒3x=2−4⇒3x=−2⇒x=−23

Here, the value of x is not unique, so option B is correct.

Question 10: The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is:

(A) 27
(B) 18
(C) 81
(D) 512

Answer:

Total number of elements in a 3 × 3 matrix

=3×3=9

If each entry is 0 or 1 then for every entry there are 2 permutations.

The total permutations for 9 elements

=29=512

Thus, option (D) is correct.

Question 1(i): Let A=[2432] , B=[13−25] , C=[−2534]

Find each of the following:

A + B

Answer:

A=[2432] B=[13−25]

(i) A + B

The addition of matrix can be done as follows

A+B=[2432] +[13−25]

A+B=[2+14+33+(−2)2+5]

A+B=[3717]

Question 1(ii): Let A=[2432] , B=[13−25] , C=[−2534]

Find each of the following:

A - B

Answer:

A=[2432] B=[13−25]

(ii) A - B

A−B=[2432] −[13−25]

A−B=[2−14−33−(−2)2−5]

A−B=[115−3]

Question 1(iii): Let A=[2432] , B=[13−25] , C=[−2534]

Find each of the following:

3A - C

Answer:

A=[2432] C=[−2534]

(iii) 3A - C

First, multiply each element of A with 3 and then subtract C

3A−C=3[2432] −[−2534]

3A−C=[61296] −[−2534]

3A−C=[6−(−2)12−59−36−4]

3A−C=[8762]

Question 1(iv): Let A=[2432] , B=[13−25] , C=[−2534]

Find each of the following:

AB

Answer:

A=[2432] B=[13−25]

(iv) AB

AB=[2432] ×[13−25]

AB=[2×1+4×−22×3+4×53×1+2×−23×3+2×5]

AB=[−626−119]

Question 1(v): Let A=[2432] , B=[13−25] , C=[−2534]

Find each of the following:

BA

Answer:

The multiplication is performed as follows

A=[2432] , B=[13−25]

BA=[13−25] ×[2432]

BA=[1×2+3×31×4+3×2−2×2+5×3−2×4+2×5]

BA=[1110112]

Question 2(i): Compute the following:

[ab−ba]+[abba]

Answer:

(i) [ab−ba]+[abba]

=[a+ab+b−b+ba+a]

=[2a2b02a]

Question 2(ii): Compute the following:

[a2+b2b2+c2a2+c2a2+b2]+[2ab2bc−2ac−2ab]

Answer:

(ii) The addition operation can be performed as follows

[a2+b2b2+c2a2+c2a2+b2]+[2ab2bc−2ac−2ab]

=[a2+b2+2abb2+c2+2bca2+c2−2aca2+b2−2ab]

=[(a+b)2(b+c)2(a−c)2(a−b)2]

Question 2(iii): Compute the following:

[−14−68516285]+[1276805324]

Answer:

(iii) The addition of the given three-by-three matrix is performed as follows

[−14−68516285]+[1276805324]

=[−1+124+7−6+68+85+016+52+38+25+4]

=[11110165215109]

Question 2(iv): Compute the following:

[cos2⁡xsin2⁡xsin2⁡xcos2⁡x]+[sin2⁡xcos2⁡xcos2⁡xsin2⁡x]

Answer:

(iv) The addition is done as follows

[cos2⁡xsin2⁡xsin2⁡xcos2⁡x]+[sin2⁡xcos2⁡xcos2⁡xsin2⁡x]

=[cos2+sin2⁡xsin2⁡x+cos2⁡xsin2⁡x+cos2⁡xcos2⁡x+sin2⁡x] since sin2⁡x+cos2⁡x=1

=[1111]

Question 3(i): Compute the indicated products.

[ab−ba][a−bba]

Answer:

(i) The multiplication is performed as follows

[ab−ba][a−bba]

=[ab−ba]×[a−bba]

=[a×a+b×ba×−b+b×a−b×a+a×b−b×−b+a×a]

=[a2+b200b2+a2]

Question 3(ii): Compute the indicated products.

[123][234]

Answer:

(ii) the multiplication can be performed as follows

[123][234]

=[1×21×31×42×22×32×43×23×33×4]

=[2344686912]

Question 3(iii): Compute the indicated products.

[1−223][123231]

Answer:

(iii) The multiplication can be performed as follows

[1−223][123231]

=[1×1+(−2)×21×2+(−2)×31×3+(−2)×12×1+3×22×2+3×32×3+3×1]

Question 3(iv): Compute the indicated products.

[234345456][1−35024305]

Answer:

(iv) The multiplication is performed as follows

[234345456][1−35024305]

=[234345456]×[1−35024305]

=[2×1+3×0+4×32×(−3)+3×2+4×02×5+3×4+4×53×1+4×0+5×33×(−3)+4×2+5×03×5+4×4+5×54×1+5×0+6×34×(−3)+5×2+6×04×5+5×4+6×5]

=[1404218−15622−270]

Question 3(v): Compute the indicated products.

[2132−11][101−121]

Answer:

(v) The product can be computed as follows

[2132−11][101−121]

=[2132−11]×[101−121]

=[2×1+1×(−1)2×0+1×(2)2×1+1×(1)3×1+2×(−1)3×0+2×(2)3×1+2×(1)(−1)×1+1×(−1)(−1)×0+1×(2)(−1)×1+1×(1)]

=[123145−220]

Question 3(vi): Compute the indicated products.

[3−13−102][2−31031]

Answer:

(vi) The given product can be computed as follows

[3−13−102][2−31031]

=[3−13−102]×[2−31031]

=[3×2+(−1)×1+3×33×(−3)+(−1)×0+3×1(−1)×2+0×1+2×3(−1)×−3+0×0+2×1]

=[14−645]

Question 4: If A=[12−35021−11] , B=[3−12425203] and C=[4120321−23] , then compute (A+B) and (B-C). Also verify that A + (B - C) = (A + B) - C

Answer:

A=[12−35021−11] , B=[3−12425203] and C=[4120321−23]

A+B=[12−35021−11] +[3−12425203]

A+B=[1+32+(−1)−3+25+40+22+51+2−1+01+3]

A+B=[41−19273−14]

B−C=[3−12425203] −[4120321−23]

B−C=[3−4−1−12−24−02−35−22−10−(−2)3−3]

B−C=[−1−204−13120]

Now, to prove A + (B - C) = (A + B) - C

L.H.S:A+(B−C)

A+(B−C)=[12−35021−11] +[−1−204−13120] (Puting value of B−C from above)

A+(B−C)=[1−12−2−3+05+40+(−1)2+31+1−1+21+0]

A+(B−C)=[00−39−15211]

R.H.S:(A+B)−C

(A+B)−C=[41−19273−14] −[4120321−23]

(A+B)−C=[4−41−1−1−29−02−37−23−1−1−(−2)4−3]

(A+B)−C=[00−39−15211]

Hence, we can see L.H.S = R.H.S = [00−39−15211]

Question 5: If A=[2315313234373223] and B=[25351152545756525] , then compute 3A - 5B

Answer:

A=[2315313234373223] and B=[25351152545756525]

3A−5B=3×[2315313234373223] −5×[25351152545756525]

3A−5B=[235124762] −[235124762]

3A−5B=[000000000]

3A−5B=0

Question 6: Simplify cos⁡θ[cos⁡θsin⁡θ−sin⁡θcos⁡θ]+sin⁡θ[sin⁡θ−cos⁡θcos⁡θsin⁡θ] .

Answer:

The simplification is explained in the following step

cos⁡θ[cos⁡θsin⁡θ−sin⁡θcos⁡θ]+sin⁡θ[sin⁡θ−cos⁡θcos⁡θsin⁡θ]

=[cos2⁡θsin⁡θcos⁡θ−sin⁡θcos⁡θcos2⁡θ]+[sin2⁡θ−sin⁡θcos⁡θsin⁡θcos⁡θsin2⁡θ]

=[cos2⁡θ+sin2⁡θsin⁡θcos⁡θ−sin⁡θcos⁡θ−sin⁡θcos⁡θ+sin⁡θcos⁡θcos2⁡θ+sin2⁡θ]

=[1001]=I

the final answer is an identity matrix of order 2

Question 7(i): Find X and Y, if

X+Y=[7025] and X−Y=[3003]

Answer:

(i) The given matrices are

X+Y=[7025] and X−Y=[3003]

X+Y=[7025].............................1

X−Y=[3003].............................2

Adding equation 1 and 2, we get

2X=[7025] +[3003]

2X=[7+30+02+05+3]

2X=[10028]

X=[5014]

Putting the value of X in equation 1, we get

[5014] +Y=[7025]

Y=[7025]− [5014]

Y=[7−50−02−15−4]

Y=[2011]

Question 7(ii): Find X and Y, if

2X+3Y=[2340] and 3X+2Y=[2−2−15]

Answer:

(ii) 2X+3Y=[2340] and 3X+2Y=[2−2−15]

2X+3Y=[2340]..........................1

3X+2Y=[2−2−15]......................2

Multiply equation 1 by 3 and equation 2 by 2 and subtract them,

3(2X+3Y)−2(3X+2Y)=3×[2340] −2×[2−2−15]

6X+9Y−6X−4Y=[69120] −[4−4−210]

9Y−4Y=[6−49−(−4)12−(−2)0−10]

5Y=[21314−10]

Y=[25135145−2]

Putting value of Y in equation 1 , we get

2X+3Y=[2340]

2X+3[25135145−2]=[2340]

2X+[65395425−6]=[2340]

2X=[2340]−[65395425−6]

2X=[2−653−3954−4250−(−6)]

2X=[45−245−2256]

X=[25−125−1153]

Question 8: Find X, if Y=[3214] and 2X+Y=[10−32]

Answer:

Y=[3214]

2X+Y=[10−32]

Substituting the value of Y in the above equation

2X+[3214]=[10−32]

2X=[10−32]−[3214]

2X=[1−30−2−3−12−4]

2X=[−2−2−4−2]

X=[−1−1−2−1]

Question 9: Find x and y, if 2[130x]+[y012]=[5618]

Answer:

2[130x]+[y012]=[5618]

[2602x]+[y012]=[5618]

[2+y6+00+12x+2]=[5618]

[2+y612x+2]=[5618]

Now equating LHS and RHS we can write the following equations

2+y=5 2x+2=8

y=5−2 2x=8−2

y=3 2x=6

x=3

Question 10: Solve the equation for x, y, z and t, if 2[xzyt]+3[1−102]=3[3546]

Answer:

2[xzyt]+3[1−102]=3[3546]

Multiplying with constant terms and rearranging we can rewrite the matrix as

[2x2z2y2t]=[9151218]−3[1−102]

[2x2z2y2t]=[9151218]−[3−306]

[2x2z2y2t]=[9−315−(−3)12−018−6]

[2x2z2y2t]=[6181212]

Dividing by 2 on both sides

[xzyt]=[3966]

x=3,y=6,z=9andt=6

Question 11: If x[23]+y[−11]=[105] , find the values of x and y.

Answer:

x[23]+y[−11]=[105]

[2x3x]+[−yy]=[105]

Adding both the matrix in LHS and rewriting

[2x−y3x+y]=[105]

2x−y=10........................1

3x+y=5........................2

Adding equation 1 and 2, we get

5x=15

x=3

Put the value of x in equation 2, we have

3x+y=5

3×3+y=5

9+y=5

y=5−9

y=−4

Question 12: Given 3[xyzw]=[x6−12w]+[4x+yz+w3] , find the values of x, y, z and w.

Answer:

3[xyzw]=[x6−12w]+[4x+yz+w3]

[3x3y3z3w]=[x+46+x+y−1+z+w2w+3]

If two matrices are equal then corresponding elements are also equal.

Thus, we have

3x=x+4

3x−x=4

2x=4

x=2

3y=6+x+y

Put the value of x

3y−y=6+2

2y=8

y=4

3w=2w+3

3w−2w=3

w=3

3z=−1+z+w

3z−z=−1+3

2z=2

z=1

Hence, we have x=2,y=4,z=1andw=3.

Question 13: If F(x)=[cos⁡x−sin⁡x0sin⁡xcos⁡x0001] , show that F(x)F(y)=F(x+y) .

Answer:

F(x)=[cos⁡x−sin⁡x0sin⁡xcos⁡x0001]

To prove : F(x)F(y)=F(x+y)

R.H.S:F(x+y)

F(x+y)=[cos⁡(x+y)−sin⁡(x+y)0sin⁡(x+y)cos⁡(x+y)0001]

L.H.S:F(x)F(y)

F(x)F(y)=[cos⁡x−sin⁡x0sin⁡xcos⁡x0001]×[cos⁡y−sin⁡y0sin⁡ycos⁡y0001]

F(x)F(y)=[cos⁡xcos⁡y−sin⁡xsin⁡y+0−cos⁡xsin⁡y−sin⁡xcos⁡y+00+0+0 sinxcos⁡y+cos⁡xsin⁡y+0−sin⁡xsin⁡y+cos⁡xcos⁡y+00+0+00+0+00+0+00+0+1]

F(x)F(y)=[cos⁡(x+y)−sin⁡(x+y)0sin⁡(x+y)cos⁡(x+y)0001]

Hence, we have L.H.S. = R.H.S i.e. F(x)F(y)=F(x+y) .

Question 14(i): Show that

[5−167][2134]≠[2134][5−167]

Answer:

To prove:

[5−167][2134]≠[2134][5−167]

L.H.S:[5−167][2134]

=[5×2+(−1)×35×1+(−1)×46×2+7×36×1+7×4]

=[713334]

R.H.S:[2134][5−167]

=[2×5+1×62×(−1)+1×73×5+4×63×(−1)+4×7]

=[1653925]

Hence, the right-hand side is not equal to the left-hand side, that is

Question 14(ii): Show that

[123010110][−1100−11234]≠[−1100−11234][123010110]

Answer:

To prove the following multiplication of three by three matrices is not equal

[123010110][−1100−11234]≠[−1100−11234][123010110]

L.H.S:[123010110][−1100−11234]

=[1×(−1)+2×0+3×21×(1)+2×(−1)+3×31×(0)+2×1+3×40×(−1)+1×0+0×20×(1)+1×(−1)+0×30×(0)+1×1+0×41×(−1)+1×0+0×21×(1)+1×(−1)+0×31×(0)+1×1+0×4]

=[58140−11−101]

R.H.S:[−1100−11234][123010110]

=[−1×(1)+1×0+0×1−1×(2)+1×(1)+0×1−1×(3)+1×0+0×00×(1)+−(1)×0+1×10×(2)+(−1)×(1)+1×10×(3)+(−1)×0+1×02×(1)+3×0+4×12×(2)+3×(1)+4×12×(3)+3×0+4×0]

=[−1−1−31006116]

Hence, L.H.S≠R.H.S i.e. [123010110][−1100−11234]≠[−1100−11234][123010110] .

Question 15: Find A2−5A+6I , if

A=[2012131−10]

Answer:

A=[2012131−10]

First, we will find out the value of the square of matrix A

A×A=[2012131−10]×[2012131−10]

A2=[2×2+0×2+1×12×0+0×1+1×−12×1+0×3+1×02×2+1×2+3×12×0+1×1+3×−12×1+1×3+3×01×2+(−1)×2+0×11×0+(−1)×1+0×−11×1+(−1)×3+0×0]

A2=[5−129−250−1−2]

I=[100010001]

∴ A2−5A+6I

=[5−129−250−1−2] −5[2012131−10] +6[100010001]

=[5−129−250−1−2] −[1005105155−50] +[600060006]

=[5−10+6−1−0+02−5+09−10+0−2−5+65−15+00−5+0−1−(−5)+0−2−0+6]

=[1−1−3−1−1−10−544]

Question16: If A=[102021203] prove that A3−6A2+7A+2I=0 .

Answer:

A=[102021203]

First, find the square of matrix A and then multiply it with A to get the cube of matrix A

A×A=[102021203] ×[102021203]

A2=[1+0+40+0+02+0+60+0+20+4+00+2+32+0+60+0+04+0+9]

A2=[5082458013]

A3=A2×A

A2×A=[5082458013] ×[102021203]

A3=[5+0+160+0+010+0+242+0+100+8+04+4+158+0+260+0+016+0+39]

A3=[210341282334055]

I=[100010001]

∴ A3−6A2+7A+2I=0

L.H.S :

[210341282334055] −6[5082458013] +7[102021203] +2[100010001]

=[210341282334055] −[3004812243048078] +[7014014714021] +[200020002]

=[21−30+7+20−0+0+034−48+14+012−12+0+08−24+14+223−30+7+034−48+14+00−0+0+055−78+21+2]

=[30−30048−4812−1224−2430−3048−48078−78]

=[000000000]=0

Hence, L.H.S = R.H.S

i.e. A3−6A2+7A+2I=0 .

Question 17: If A=[3−24−2] and I=[1001] , find k so that A2=kA−2I .

Answer:

A=[3−24−2]

I=[1001]

A×A=[3−24−2] ×[3−24−2]

A2=[9−8−6+412−8−8+4]

A2=[1−24−4]

A2=kA−2I

[1−24−4]= k[3−24−2]− 2[1001]

[1−24−4]= k[3−24−2]− [2002]

[1−24−4]+ [2002] =k[3−24−2]

[1+2−2+04+0−4+2] =[3k−2k4k−2k]

[3−24−2] =[3k−2k4k−2k]

We have, 3=3k

k=33=1

Hence, the value of k is 1.

Question 18: If A=[0−tan⁡α2tan⁡α20] and I is the identity matrix of order 2, show that I+A=(I−A)[cos⁡α−sin⁡αsin⁡αcos⁡α]

Answer:

A=[0−tan⁡α2tan⁡α20]

I=[1001]

To prove : I+A=(I−A)[cos⁡α−sin⁡αsin⁡αcos⁡α]

L.H.S : I+A

I+A=[1001] +[0−tan⁡α2tan⁡α20]

I+A=[1+00−tan⁡α20+tan⁡α21+0]

I+A=[1−tan⁡α2tan⁡α21]

R.H.S : (I−A)[cos⁡α−sin⁡αsin⁡αcos⁡α]

(I−A)[cos⁡α−sin⁡αsin⁡αcos⁡α] =([1001]− [0−tan⁡α2tan⁡α20]) ×[cos⁡α−sin⁡αsin⁡αcos⁡α]

(I−A)[cos⁡α−sin⁡αsin⁡αcos⁡α] =[1−00−(−tan⁡α2)0−tan⁡α21−0] ×[cos⁡α−sin⁡αsin⁡αcos⁡α]

(I−A)[cos⁡α−sin⁡αsin⁡αcos⁡α] =[1tan⁡α2−tan⁡α21] ×[cos⁡α−sin⁡αsin⁡αcos⁡α]

=[cos⁡α+sin⁡αtan⁡α2−sin⁡α+cos⁡αtan⁡α2−tan⁡α2cos⁡α+sin⁡αtan⁡α2sin⁡α+cos⁡α]

=[1−2sin2⁡α2+2sin⁡α2 cosα2tan⁡α2−2sin⁡α2 cosα2+(2cos2⁡α2−1)tan⁡α2−tan⁡α2(2cos2⁡α2−1)+2sin⁡α2 cosα2tan⁡α22sin⁡α2 cosα2+1−2sin2⁡α2]

=[1−2sin2⁡α2+2sin2⁡α2−2sin⁡α2 cosα2+2sin⁡α2 cosα2−tan⁡α2−2sin⁡α2 cosα2+tan⁡α2+2sin⁡α2 cosα22sin2⁡α2+1−2sin2⁡α2]

=[1−tan⁡α2tan⁡α21]

Hence, we can see L.H.S = R.H.S

i.e. I+A=(I−A)[cos⁡α−sin⁡αsin⁡αcos⁡α] .

Question 19(i): A trust fund has Rs. 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide Rs 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of:

Rs. 1800

Answer:

Let Rs. x be invested in the first bond.

Money invested in second bond = Rs (3000-x)

The first bond pays 5% interest per year and the second bond pays 7% interest per year.

To obtain an annual total interest of Rs. 1800, we have

[x(30000−x)] [51007100] =1800 (simple interest for 1 year =pricipal×rate100 )

5100x+7100(30000−x)=1800

5x+210000−7x=180000

210000−180000=7x−5x

30000=2x

x=15000

Thus, to obtain an annual total interest of Rs. 1800, the trust fund should invest Rs 15000 in the first bond and Rs 15000 in the second bond.

Question 19(ii): A trust fund has Rs. 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide Rs 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of:

Rs. 2000

Answer:

Let Rs. x be invested in the first bond.

Money invested in second bond = Rs (3000-x)

The first bond pays 5% interest per year and the second bond pays 7% interest per year.

To obtain an annual total interest of Rs. 1800, we have

[x(30000−x)] [51007100] =2000 (simple interest for 1 year =pricipal×rate100 )

5100x+7100(30000−x)=2000

5x+210000−7x=200000

210000−200000=7x−5x

10000=2x

x=5000

Thus, to obtain an annual total interest of Rs. 2000, the trust fund should invest Rs 5000 in the first bond and Rs 25000 in the second bond.

Question 20: The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are Rs 80, Rs 60 and Rs 40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.

Answer:

The bookshop has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books.

Their selling prices are Rs 80, Rs 60 and Rs 40 each respectively.

The total amount the bookshop will receive from selling all the books:

12 [10810] [806040]

=12(10×80+8×60+10×40)

=12(800+480+400)

=12(1680)

=20160

The total amount the bookshop will receive from selling all the books is 20160.

Question 21: Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k , respectively. Choose the correct answer in Exercises 21 and 22.

The restriction on n, k and p so that PY + WY will be defined are:
(A) k=3,p=n

(B) k is arbitrary, p=2

(C) p is arbitrary, k=3

(D) k=2,p=3

Answer:

P and Y are of order p∗k and 3∗k respectively.

∴ PY will be defined only if k=3, i.e. order of PY is p∗k .

W and Y are of order n∗3 and 3∗k respectively.

∴ WY is defined because the number of columns of W is equal to the number of rows of Y which is 3, i.e. the order of WY is n∗k

Matrices PY and WY can only be added if they both have same order i.e = n∗k implies p=n.

Thus, k=3,p=n are restrictions on n, k, and p so that PY + WY will be defined.

Option (A) is correct.

Question 22: Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k,
respectively. Choose the correct answer in Exercises 21 and 22. If n = p , then the order of the matrix 7X−5Z is:
(A) p × 2
(B) 2 × n
(C) n × 3
(D) p × n

Answer:

X has of order 2∗n .

∴ 7X also has of order 2∗n .

Z has of order 2∗p .

∴ 5Z also has of order 2∗p .

Mtarices 7X and 5Z can only be subtracted if they both have same order i.e 2∗n = 2∗p and it is given that p=n.

We can say that both matrices have order of 2∗n .

Thus, order of 7X−5Z is 2∗n .

Option (B) is correct.

Question 1(i): Find the transpose of each of the following matrices:

[512−1]

Answer:

A=[512−1]

The transpose of the given matrix is

AT=[512−1]

Question 1(ii): Find the transpose of each of the following matrices:

[1−123]

Answer:

A=[1−123]

interchanging the rows and columns of the matrix A we get

AT=[12−13]

Question 1(iii): Find the transpose of each of the following matrices:

[−15635623−1]

Answer:

A=[−15635623−1]

Transpose is obtained by interchanging the rows and columns of matrix

AT=[−13255366−1]

Question 2(i): If A=[−123579−211] and B=[−41−5120131] , then verify

(A+B)′=A′+B′

Answer:

A=[−123579−211] and B=[−41−5120131]

(A+B)′=A′+B′

L.H.S : (A+B)′

A+B=[−123579−211] +[−41−5120131]

A+B=[−1+(−4)2+13+(−5)5+17+29+0−2+11+31+1]

A+B=[−53−2699−142]

(A+B)′=[−56−1394−292]

R.H.S : A′+B′

A′+B′=[−15−2271391] +[−411123−501]

A′+B′=[−1+(−4)5+1−2+12+17+21+33+(−5)9+01+1]

A′+B′=[−56−1394−292]

Thus we find that the LHS is equal to RHS and hence verified.

Question 2(ii): If A=[−123579−211] and B=[−41−5120131] , then verify

(A−B)′=A′−B′

Answer:

A=[−123579−211] and B=[−41−5120131]

(A−B)′=A′−B′

L.H.S : (A−B)′

A−B=[−123579−211] −[−41−5120131]

A−B=[−1−(−4)2−13−(−5)5−17−29−0−2−11−31−1]

A−B=[318459−3−20]

(A−B)′=[34−315−2890]

R.H.S : A′−B′

A′−B′=[−15−2271391] −[−411123−501]

A′−B′=[−1−(−4)5−1−2−12−17−21−33−(−5)9−01−1]

A′−B′=[34−315−2890]

Hence, L.H.S = R.H.S. so verified that

(A−B)′=A′−B′ .

Question 3(i): If A′=[34−1201] and B=[−121123] , then verify

(A+B)′=A′+B′

Answer:

A′=[34−1201] B=[−121123]

A=(A′)′=[3−10421]

To prove: (A+B)′=A′+B′

L.H.S:(A+B)′=

A+B=[3−10421] +[−121123]

A+B=[3+(−1)−1+(−1)0+14+12+21+3]

A+B=[2−21544]

∴(A+B)′=[251414]

R.H.S: A′+B′

A′+B′=[34−1201] +[−112213]

A′+B′=[251414]

Hence, L.H.S = R.H.S i.e. (A+B)′=A′+B′ .

Question 3(ii): If A=[34−1201] and B=[−121123] , then verify

(A−B)′=A′−B′

Answer:

A′=[34−1201] B=[−121123]

A=(A′)′=[3−10421]

To prove: (A−B)′=A′−B′

L.H.S:(A−B)′=

A−B=[3−10421] −[−121123]

A−B=[3−(−1)−1−(2)0−14−12−21−3]

A−B=[4−3−130−2]

∴(A−B)′=[43−30−1−2]

R.H.S: A′−B′

A′−B′=[34−1201] −[−112213]

A′−B′=[43−30−1−2]

Hence, L.H.S = R.H.S i.e. (A−B)′=A′−B′ .

Question 4: If A′=[−2312] and B=[−1012] , then find (A+2B)′

Answer:

B=[−1012]

A′=[−2312]

A=(A′)′=[−2132]

(A+2B)′ :

A+2B=[−2132] +2[−1012]

A+2B=[−2132] +[−2024]

A+2B=[−2+(−2)1+03+22+4]

A+2B=[−4156]

Transpose is obtained by interchanging rows and columns and the transpose of A+2B is

(A+2B)′=[−4516]

Question 5(i): For the matrices A and B, verify that (AB)′=B′A′ , where

A=[1−43] , B=[−121]

Answer:

A=[1−43] , B=[−121]

To prove : (AB)′=B′A′

L.H.S:(AB)′

AB=[1−43] [−121]

AB=[−1214−8−4−363]

(AB)′=[−14−32−861−43]

R.H.S:B′A′

B′=[−121]

A′=[1−43]

B′A′=[−121] [1−43]

B′A′=[−14−32−861−43]

Hence, L.H.S =R.H.S

so it is verified that (AB)′=B′A′ .

Question 5(ii): For the matrices A and B, verify that (AB)′=B′A′ , where

A=[012] , B=[157]

Answer:

A=[012] , B=[157]

To prove : (AB)′=B′A′

L.H.S:(AB)′

AB=[012] [157]

AB=[00015721014]

(AB)′=[01205100714]

R.H.S:B′A′

B′=[157]

A′=[012]

B′A′=[157] [012]

B′A′=[01205100714]

Heence, L.H.S =R.H.S i.e. (AB)′=B′A′ .

Question 6(i): If A=[cos⁡αsin⁡α−sin⁡αcos⁡α] , then verify that A′A=I

Answer:

A=[cos⁡αsin⁡α−sin⁡αcos⁡α]

By interchanging rows and columns we get transpose of A

A′=[cos⁡α−sin⁡αsin⁡αcos⁡α]

To prove: A′A=I

L.H.S : A′A

A′A=[cos⁡α−sin⁡αsin⁡αcos⁡α] [cos⁡αsin⁡α−sin⁡αcos⁡α]

A′A=[cos2⁡α+sin2⁡αsin⁡αcos⁡α−sin⁡α cosαsin⁡αcos⁡α−sin⁡αcos⁡α sin2α+cos2⁡α]

A′A=[1001]=I=R.H.S

Question 6(ii): If A=[sin⁡αcos⁡α−cos⁡αsin⁡α] , then verify that A′A=I

Answer:

A=[sin⁡αcos⁡α−cos⁡αsin⁡α]

By interchanging columns and rows of the matrix A we get the transpose of A

A′=[sin⁡α−cos⁡αcos⁡αsin⁡α]

To prove: A′A=I

L.H.S : A′A

A′A=[sin⁡α−cos⁡αcos⁡αsin⁡α] [sin⁡αcos⁡α−cos⁡αsin⁡α]

A′A=[cos2⁡α+sin2⁡αsin⁡αcos⁡α−sin⁡α cosαsin⁡αcos⁡α−sin⁡αcos⁡α sin2α+cos2⁡α]

A′A=[1001]=I=R.H.S

Question 7(i): Show that the matrix A=[1−15−121513] is a symmetric matrix.

Answer:

A=[1−15−121513]

the transpose of A is

A′=[1−15−121513]

Since, A′=A so given matrix is a symmetric matrix.

Question 7(ii): Show that the matrix A=[01−1−1011−10] is a skew-symmetric matrix.

Answer:

A=[01−1−1011−10]

The transpose of A is

A′=[0−1110−1−110]

A′=−[01−1−1011−10]

A′=−A

Since, A′=−A so given matrix is a skew-symmetric matrix.

Question 8(i): For the matrix A=[1567] , verify that

(A+A′) is a symmetric matrix.

Answer:

A=[1567]

A′=[1657]

A+A′=[1567] +[1657]

A+A′=[1+15+66+57+7]

A+A′=[2111114]

(A+A′)′=[2111114]

We have A+A′=(A+A′)′

Hence, (A+A′) is a symmetric matrix.

Question 8(ii): For the matrix A=[1567] , verify that

(A−A′) is a skew symmetric matrix.

Answer:

A=[1567]

A′=[1657]

A−A′=[1567] −[1657]

A−A′=[1−15−66−57−7]

A−A′=[0−110]

(A−A′)′=[01−10]=−(A−A′)

We have A−A′=−(A−A′)′

Hence, (A−A′) is a skew-symmetric matrix.

Question 9: Find 12(A+A′) and 12(A−A′) , when A=[0ab−a0c−b−c0]

Answer:

A=[0ab−a0c−b−c0]

the transpose of the matrix is obtained by interchanging rows and columns

A′=[0−a−ba0−cbc0]

12(A+A′)=12([0ab−a0c−b−c0] +[0−a−ba0−cbc0])

12(A+A′)=12([0+0a+(−a)b+(−b)−a+a0+0c+(−c)−b+b−c+c0+0])

12(A+A′)=12[000000000]

12(A+A′)=[000000000]

12(A+A′)=0

12(A−A′)=12([0ab−a0c−b−c0] −[0−a−ba0−cbc0])

12(A−A′)=12([0−0a−(−a)b−(−b)−a−a0−0c−(−c)−b−b−c−c0−0])

12(A−A′)=12[02a2b−2a02c−2b−2c0]

12(A−A′)=[0ab−a0c−b−c0]

Question 10(i): Express the following matrices as the sum of a symmetric and a skew-symmetric matrix:

[351−1]

Answer:

A=[351−1]

A′=[315−1]

A+A′=[351−1] +[315−1]

A+A′=[666−2]

Let

B=12(A+A′)=12[666−2] =[333−1]

B′=[333−1]=B

Thus, 12(A+A′) is a symmetric matrix.

A−A′=[351−1] −[315−1]

A−A′=[04−40]

Let

C=12(A−A′)=12[04−40] =[02−20]

C′=[0−220]

C=−C′

Thus, 12(A−A′) is a skew symmetric matrix.

Represent A as sum of B and C.

B+C=[333−1] +[02−20] =[351−1]=A

Question:10(ii): Express the following matrices as the sum of a symmetric and a skew-symmetric matrix:

[6−22−23−12−13]

Answer:

A=[6−22−23−12−13]

A′=[6−22−23−12−13]

A+A′=[6−22−23−12−13] +[6−22−23−12−13]

A+A′=[12−44−46−24−26]

Let

B=12(A+A′)=12[12−44−46−24−26] =[6−22−23−12−13]

B′=[6−22−23−12−13]=B

Thus, 12(A+A′) is a symmetric matrix.

A−A′=[6−22−23−12−13] −[6−22−23−12−13]

A−A′=[000000000]

Let

C=12(A−A′)=12[000000000] =[000000000]

C′=[000000000]

C=−C′

Thus, 12(A−A′) is a skew-symmetric matrix.

Represent A as the sum of B and C.

B+C=[6−22−23−12−13] +[000000000] =[6−22−23−12−13]=A

Question 10(iii): Express the following matrices as the sum of a symmetric and a skew-symmetric matrix:

[33−1−2−21−4−52]

Answer:

A=[33−1−2−21−4−52]

A′=[3−2−43−2−5−112]

A+A′=[33−1−2−21−4−52] +[3−2−43−2−5−112]

A+A′=[61−51−4−4−5−44]

Let

B=12(A+A′)=12[61−51−4−4−5−44] =[312−5212−2−2−52−22]

B′=[312−5212−2−2−52−22]=B

Thus, 12(A+A′) is a symmetric matrix.

A−A′=[33−1−2−21−4−52] −[3−2−43−2−5−112]

A−A′=[053−506−3−60]

Let

C=12(A−A′)=12[053−506−3−60] =[05232−5203−32−30]

C′=[0−52−32520−33230]

C=−C′

Thus, 12(A−A′) is a skew-symmetric matrix.

Represent A as the sum of B and C.

B+C=[312−5212−2−2−52−22] +[05232−5203−32−30] =[33−1−2−21−4−52]=A

Question 10(iv): Express the following matrices as the sum of a symmetric and a skew-symmetric matrix:

[15−12]

Answer:

A=[15−12]

A′=[1−152]

A+A′=[15−12] +[1−152]

A+A′=[2444]

Let

B=12(A+A′)=12[2444] =[1222]

B′=[1222]=B

Thus, 12(A+A′) is a symmetric matrix.

A−A′=[15−12] −[1−152]

A−A′=[06−60]

Let

C=12(A−A′)=12[06−60] =[03−30]

C′=[0−330]

C=−C′

Thus, 12(A−A′) is a skew-symmetric matrix.

Represent A as the sum of B and C.

B+C=[1222] −[0−330] =[15−12]=A

Question 11: Choose the correct answer in the Exercises 11 and 12.

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:

If A, B are symmetric matrices then

A′=A and B′=B

we have, (AB−BA)′=(AB)′−(BA)′=B′A′−A′B′

=BA−AB

=−(AB−BA)

Hence, we have (AB−BA)=−(AB−BA)′

Thus,( AB-BA)' is skew symmetric.

Option A is correct.

Question 12: Choose the correct answer in the Exercises 11 and 12.

If A=[cos⁡α−sin⁡αsin⁡αcos⁡α] and A+A′=I , then the value of α is

(A) π6

(B) π3

(C) π

(D) 3π2

Answer:

A=[cos⁡α−sin⁡αsin⁡αcos⁡α]

A′=[cos⁡αsin⁡α−sin⁡αcos⁡α]

A+A′=[cos⁡α−sin⁡αsin⁡αcos⁡α] +[cos⁡αsin⁡α−sin⁡αcos⁡α] =[1001]

A+A′=[2cos⁡α002cos⁡α] =[1001]

2cosα=1

cosα=12

α=π3

Option B is correct.

Question 1: Matrices A and B will be inverse of each other only if

(A) AB=BA

(B) AB=BA=0

(C) AB=0,BA=I

(D) AB=BA=I

Answer:

We know that if A is a square matrix of order n and there is another matrix B of same order n, such that AB=BA=I , then B is inverse of matrix A.

In this case, it is clear that A is inverse of B.

Hence, matrices A and B will be inverse of each other only if AB=BA=I .

Option D is correct.

Question 1: If A and B are symmetric matrices, prove that AB−BA is a skew symmetric matrix.

Answer:

If A, B are symmetric matrices then

A′=A and B′=B

we have, (AB−BA)′=(AB)′−(BA)′=B′A′−A′B′

=BA−AB

=−(AB−BA)

Hence, we have (AB−BA)=−(AB−BA)′

Thus,( AB-BA)' is skew symmetric.

Question 2: Show that the matrix B′AB is symmetric or skew symmetric according as A is symmetric or skew symmetric.

Answer:

Let be a A is symmetric matrix, then A′=A

Consider, (B′AB)′=B′(AB)′

=(AB)′(B′)′

=B′A′(B)

=B′(A′B)

Replace A′ by A

=B′(AB)

i.e. (B′AB)′ =B′(AB)

Thus, if A is a symmetric matrix than B′(AB) is a symmetric matrix.

Now, let A be a skew-symmetric matrix, then A′=−A.

(B′AB)′=B′(AB)′

=(AB)′(B′)′

=B′A′(B)

=B′(A′B)

Replace A′ by - A ,

=B′(−AB)

=−B′AB

i.e. (B′AB)′ =−B′AB .

Thus, if A is a skew-symmetric matrix then −B′AB is a skew-symmetric matrix.

Hence, the matrix B′AB is symmetric or skew-symmetric according to as A is symmetric or skew-symmetric.

Question 3: Find the values of x , y , z if the matrix A=[02yzxy−zx−yz] satisfy the equation A′A=I

Answer:

A=[02yzxy−zx−yz]

A′=[0xx2yy−yz−zz]

A′A=I

[0xx2yy−yz−zz] [02yzxy−zx−yz] =[100010001]

[x2+x2xy−xy−xz+xzxy−xy4y2+y2+y22yz−yz−yz−zx+zx2yz−yz−yzz2+z2+z2] =[100010001]

[2x20006y20003z2] =[100010001]

Thus equating the terms elementwise

2x2=1 6y2=1 3z2=1

x2=12 y2=16 z2=13

x=±12 y=±16 z=±13

Question 4: For what values of x: [121][120201102][02x]=O ?

Answer:

[121][120201102][02x]=O

[1+4+12+0+00+2+2][02x]=O

[624][02x]=O

[0+4+4x]=O

4+4x=0

4x=−4

x=−1

Thus, value of x is -1.

Question 5: If A=[31−12] , show that A2−5A+7I=0 .

Answer:

A=[31−12]

A2=[31−12] [31−12]

A2=[9−13+2−3−2−1+4]

A2=[85−53]

I=[1001]

To prove: A2−5A+7I=0

L.H.S : A2−5A+7I

=[85−53] −5[31−12] +7[1001]

=[8−15+75−5+0−5+5+03−10+7]

=[0000]=0=R.H.S

Hence, we proved that

A2−5A+7I=0 .

Question 6: Find x, if [x−5−1][102021203][x41]=0 .

Answer:

[x−5−1][102021203][x41]=0

[x+0−20−10+02x−5−3][x41]=0

[x−2−102x−8][x41]=0

[x(x−2)−40+(2x−8)]=0

[x2−2x−40+2x−8]=0

∴x2−48=0

x2=48

thus the value of x is

x=±43

Question 7(a): A manufacturer produces three products x, y, z which he sells in two markets.
Annual sales are indicated below:

Market Products
I 10,000 2,000 18,000
II 6,000 20,000 8,000

If unit sale prices of x, y and z are ` 2.50, ` 1.50 and ` 1.00, respectively, find the total revenue in each market with the help of matrix algebra.

Answer:

The unit sale prices of x, y and z are ` 2.50, ` 1.50 and ` 1.00, respectively.

The total revenue in the market I with the help of matrix algebra can be represented as :

[10000200018000][2.501.501.00]

=10000×2.50+2000×1.50+18000×1.00

=25000+3000+18000

=46000

The total revenue in market II with the help of matrix algebra can be represented as :

[6000200008000][2.501.501.00]

=6000×2.50+20000×1.50+8000×1.00

=15000+30000+8000

=53000

Hence, total revenue in the market I is 46000 and total revenue in market II is 53000.

Question 7(b): A manufacturer produces three products x, y, z which he sells in two markets.
Annual sales are indicated below:

Market Products
I 10,000 2,000 18,000
II 6,000 20,000 8,000

If the unit costs of the above three commodities are ` 2.00, ` 1.00, and 50 paise respectively. Find the gross profit.

Answer:

The unit costs of the above three commodities are ` 2.00, ` 1.00 and 50 paise respectively.

The total cost price in market I with the help of matrix algebra can be represented as :

[10000200018000][2.001.000.50]

=10000×2.00+2000×1.00+18000×0.50

=20000+2000+9000

=31000

Total revenue in the market I is 46000 , gross profit in the market is =46000−31000 =Rs.15000

The total cost price in market II with the help of matrix algebra can be represented as :

[6000200008000][2.001.000.50]

=6000×2.0+20000×1.0+8000×0.50

=12000+20000+4000

=36000

Total revenue in market II is 53000, gross profit in the market is =53000−36000=Rs.17000

Question 8: Find the matrix X so that X[123456]=[−7−8−9246]

Answer:

X[123456]=[−7−8−9246]

The matrix given on R.H.S is 2×3 matrix and on LH.S is 2×3 matrix.Therefore, X has to be 2×2 matrix.

Let X be [acbd]

[acbd] [123456]=[−7−8−9246]

[a+4c2a+5c3a+6cb+4d2b+5d3b+6d]=[−7−8−9246]

a+4c=−7 2a+5c=−8 3a+6c=−9

b+4d=2 2b+5d=4 3b+6d=6

Taking, a+4c=−7

a=−4c−7

2a+5c=−8

−8c−14+5c=−8

−3c=6

c=−2

a=−4×−2−7

a=8−7=1

b+4d=2

b=−4d+2

2b+5d=4

⇒ −8d+4+5d=4

⇒−3d=0

⇒d=0

b=−4d+2

⇒b=−4×0+2=2

Hence, we have a=1,b=2,c=−2,d=0

Matrix X is [1−220] .

Question 9: Choose the correct answer in the following questions:

If A=[αβγ−α] is such that A2=I

(A) 1+α2+βγ=0

(B) 1−α2+βγ=0

(C) 1−α2−βγ=0

(D) 1+α2−βγ=0

Answer:

A=[αβγ−α]

A2=I

[αβγ−α] [αβγ−α] =[1001]

[α2+βγαβ−αβαγ−αγβγ+α2] =[1001]

[α2+βγ00βγ+α2] =[1001]

Thus we obtained that

α2+βγ=1

⇒1−α2−βγ=0

Option C is correct.

Question 10: If the matrix A is both symmetric and skew-symmetric, then

(A) A is a diagonal matrix
(B) A is a zero matrix
(C) A is a square matrix
(D) None of these

Answer:

If the matrix A is both symmetric and skew-symmetric, then

A′=A and A′=−A

A′=A′

⇒A=−A

⇒A+A=0

⇒2A=0

⇒A=0

Hence, A is a zero matrix.

Option B is correct.

Question 11: 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:

A is a square matrix such that A2=A

(I+A)3−7A

=I3+A3+3I2A+3IA2−7A

=I+A2.A+3A+3A2−7A

=I+A.A+3A+3A−7A (Replace A2 by A )

=I+A2+6A−7A

=I+A−A

=I

Hence, we have (I+A)3−7A=I

Option C is correct.