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Class 12 NCERT Solutions- Mathematics Part I – Chapter 4 Determinants – Exercise 4.5

  • Last Updated : 18 Mar, 2021

Find adjoint of each of the matrices in Exercises 1 and 2.

Question 1. \begin{bmatrix} 1 &2 \\ 3 & 4 \end{bmatrix}

Solution: 

A = \begin{bmatrix} 1 &2 \\ 3 & 4 \end{bmatrix}

A11 = 4; A12 = -3; A21 = -2; A22 = 1

adj A = \begin{bmatrix} A_{11} &A_{21} \\ A_{12} & A_{22} \end{bmatrix}

adj A = \begin{bmatrix} 4 &-2 \\ -3 & 1 \end{bmatrix}

Question 2. \begin{bmatrix} 1 & -1 &2 \\ 2&3  &5 \\ -2& 0 &1 \end{bmatrix}

Solution: 

A = \begin{bmatrix} 1 & -1 &2 \\ 2&3  &5 \\ -2& 0 &1 \end{bmatrix}

A11\begin{vmatrix} 3 &5 \\ 0& 1 \end{vmatrix}

A11 = 3 – 0 = 3

A12-\begin{vmatrix} 2&5 \\ -2& 1 \end{vmatrix}

A12 = -(2 + 10) = -12

A13\begin{vmatrix} 2&3\\ -2& 0 \end{vmatrix}

A13 = 0 + 6 = 6

A21-\begin{vmatrix} -1&2\\ 0& 1 \end{vmatrix}

A21 = -(-1 – 0) = 1

A22\begin{vmatrix} 1&2\\ -2& 1 \end{vmatrix}

A22 = 1 + 4 = 5

A23-\begin{vmatrix} 1&-1\\ -2&0 \end{vmatrix}

A23 = -(0 – 2) = 2

A31\begin{vmatrix} -1&2\\ 3&5 \end{vmatrix}

A31 = -5 – 6 = -11

A32-\begin{vmatrix} 1&2\\ 2&5 \end{vmatrix}p>A32 = -(5 – 4) = -1

A33\begin{vmatrix} 1&-1\\ 2&3 \end{vmatrix}

A33 = 3 + 2 = 5

adj A = \begin{bmatrix} A_{11} & A_{21} &A_{31} \\ A_{12}&A_{22} &A_{32} \\ A_{13}& A_{23} &A_{33} \end{bmatrix}

adj A = \begin{bmatrix} 3& 1 &-11 \\ -12&5  &-1 \\ 6& 2 &5 \end{bmatrix}

Verify A(adj A) = (adj A)A = |A| I in exercises 3 and 4.

Question 3. \begin{bmatrix} 2 &3 \\ -4 & -6 \end{bmatrix}

Solution: 

|A| = -12 -(-12) 

|A| = -12 + 12 = 0

so, |A|*I = 0 * \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix}

|A|*I = \begin{bmatrix} 0&0 \\ 0&0 \end{bmatrix}     …̷

8230; (1)

Now, for adjoint of A

A11 = -6

A12 = 4

A21 = -3

A22 = 2

adj A = \begin{bmatrix} A_{11} &A_{21} \\ A_{12} & A_{22} \end{bmatrix}

adj A = \begin{bmatrix} -6&-3 \\ 4&2 \end{bmatrix}

Now, A(adj A) = \begin{bmatrix} 2&3 \\ -4&-6 \end{bmatrix} \begin{bmatrix} -6&-3 \\ 4&2 \end{bmatrix}

A(adj A) = \begin{bmatrix} -12+12&-6+6 \\ 24-24&12-12 \end{bmatrix}   [Tex]\begin{bmatrix} 11& 0  &0 \\ 0 & 11 &0 \\ 0 & 0 & 11 \end{bmatrix}[/Tex]

A(adj A) = \begin{bmatrix} 0&0 \\ 0&0 \end{bmatrix}     …………(2)

Now, (adj A)A = \begin{bmatrix} -6&-3 \\ 4&2 \end{bmatrix} \begin{bmatrix} 2&3 \\ -4&-6 \end{bmatrix}"231" style="vertical-align: -25px;"/>

(adj A)A = \begin{bmatrix} -12+12&-18+18 \\ 8-8&12-12 \end{bmatrix}

(adj A)A = \begin{bmatrix} 0&0 \\ 0&0 \end{bmatrix}     …………….(3)

From eq(1), (2), and (3), you can see that A(adj A) = (adj A)A = |A|I

Question 4. \begin{bmatrix} 1 &-1  &2 \\ 3 & 0 &-2 \\ 1 & 0 & 3 \end{bmatrix}

Solution: 

A = \begin{bmatrix} 1 &-1  &2 \\ 3 & 0 &-2 \\ 1 & 0 & 3 \end{bmatrix}

|A| = 1(0 – 0) + 1(9 + 2) + 2(0 – 0)

|A| = 11

|A| * I = 11\begin{bmatrix} 1& 0  &0 \\ 0 & 1 &0 \\ 0 & 0 & 1 \end{bmatrix}

|A| * I = \begin{bmatrix} 11& 0  &0 \\ 0 & 11 &0 \\ 0 & 0 & 11 \end{bmatrix}

Now, for adjoint of A

A11 = 0

A12 = -(9 + 2) = -11

A13 = 0

A21 = -(-3 – 0) = 3

A22 = 3 – 2 = 1

A23 = -(0 + 1) = -1

A31 = 2 – 0 = 2

A32 = -(-2 – 6) = 8

A33 = 0 + 3 = 3

adj A = \begin{bmatrix} 0& 3  &2 \\ -11 & 1 &8 \\ 0 & -1 & 3 \end{bmatrix}"160" style="vertical-align: -40px;"/>

Now,

A(adjA) = \begin{bmatrix} 1 &-1  &2 \\ 3 & 0 &-2 \\ 1 & 0 & 3 \end{bmatrix} \begin{bmatrix} 0& 3  &2 \\ -11 & 1 &8 \\ 0 & -1 & 3 \end{bmatrix}

A(adj A) = \begin{bmatrix} 0+11+0& 3-1-2  &2-8+6 \\ 0+0+0 & 9+0+2 &6+0-6 \\ 0+0+0 & 3+0-3 & 2+0+9 \end{bmatrix}

A(adj A) = \begin{bmatrix} 11& 0  &0 \\ 0 & 11 &0 \\ 0 & 0 & 11 \end{bmatrix}

Also,

(adj A).A = \begin{bmatrix} 0& 3  &2 \\ -11 & 1 &8 \\ 0 & -1 & 3 \end{bmatrix} \begin{bmatrix} 1 &-1  &2 \\ 3 & 0 &-2 \\ 1 & 0 & 3 \end{bmatrix}

(adj A).A = \begin{bmatrix} 0+9+2& 0+0+0  &0-6+6 \\ -11+3+8 & 11+0+0 &-22-2+24 \\ 0-3+3 & 0+0+0 & 0+2+9 \end{bmatrix}

(adj A).A = \begin{bmatrix} 11& 0  &0 \\ 0 & 11 &0 \\ 0 & 0 & 11 \end{bmatrix}

From above, y

an see,

A(adj A) = (adj A)A = I

Find the inverse of each of the matrices (if it exists) given in Exercises 5 to 11.

Question 5. \begin{bmatrix} 2 & -2\\ 4 & 3 \end{bmatrix}

Solution: 

|A| = 6 – (-8) = 14

|A| ≠ 0, So inverse exists.

A11 = 3

A12 = 2

A21 = -4

A22 = 2

adj A = \begin{bmatrix} 3 & 2\\ -4 & 2 \end{bmatrix}

A-1 = (adj A)/|A|

A-1\frac{1}{14}\begin{bmatrix} 3 & 2\\ -4 & 2 \end{bmatrix}"120" style="vertical-align: -25px;"/>

Question 6. \begin{bmatrix} -1 & 5\\ -3 & 2 \end{bmatrix}

Solution: 

A = \begin{bmatrix} -1 &5 \\ -3& 2 \end{bmatrix}

|A| = -2 + 15 = 13 ≠ 0

Hence, inverse exists.

A11 = 2

A12 = 3

A21 = -5

A22 = -1

adj A = \begin{bmatrix} 2 &-5 \\ 3& -1 \end{bmatrix}

Question 7. \begin{bmatrix} 1& 2  &3 \\ 0 & 2 &4 \\ 0 & 0 & 5 \end{bmatrix}

Solution: 

A = \begin{bmatrix} 1& 2  &3 \\ 0 & 2 &4 \\ 0 & 0 & 5 \end{bmatrix}

|A| = 1(10 – 0) – 2(0 – 0) + 3(0 – 0) = 10

For adj A

A11 = 10 – 0 = 0

A12 = -(0 – 0) = 0

A13 = 0 – 0 = 0

A21 = -(10 – 0) = -

p>

A22 = 5 – 0 = 5

A23 = -(0 – 0) = 0

A31 = 8 – 6 = 2

A32 = -(4 – 0) = -4

A33 = 2 – 0 = 2

adj A = \begin{bmatrix} 10& -10  &2 \\ 0 & 5 &-4 \\ 0 & 0 & 2 \end{bmatrix}

A-1 = (adj A)/|A|

A-1\frac{1}{10}\begin{bmatrix} 10& -10  &2 \\ 0 & 5 &-4 \\ 0 & 0 & 2 \end{bmatrix}

Question 8. \begin{bmatrix} 1& 0  &0 \\ 3 & 3 &0 \\ 5 & 2 & -1 \end{bmatrix}

Solution: 

A = \begin{bmatrix} 1& 0  &0 \\ 3 & 3 &0 \\ 5 & 2 & -1 \end{bmatrix}

|A|

-3 – 0) – 0 + 0 = -3 ≠ 0

Hence, inverse exists.

For adj A

A11 = -3 – 0 = -3

A12 = -(-3 – 0) = 3

A13 = 6 – 15 = -9

A21 = -(0 – 0) = 0

A22 = -1 – 0 = -1

A23 = -(2 – 0) = -2

A31 = 0 – 0 = 0 

A32 = -(0 – 0) = 0

A33 = 3 – 0 = 3

adj A = \begin{bmatrix} -3& 0  &0 \\ 3 & -1 &0 \\ -9 & -2 & 3 \end{bmatrix}

A-1 = (adj A)/|A|

A-1\frac{-1}{3}\begin{bmatrix} -3& 0  &0 \\ 3 & -1 &0 \\ -9 & -2 & 3 \end{bmatrix}

Question 9. \begin{bmatrix} 2& 1  &3 \\ 4 & -1 &0 \\ -7 & 2 & 1 \end{bmatrix}

Solution: 

A = \begin{bmatrix} 2& 1  &3 \\ 4 & -1 &0 \\ -7 & 2 & 1 \end{bmatrix}

|A| = 2(-1 – 0) – 1(4 – 0) + 3(8 – 7) = -3 ≠ 0

Hence, inverse exists.

For adj A

<

sub>11 = -1 – 0 = -1 

A12 = -(4 – 0) = -4

A13 = 8 – 7 = 1

A21 = -(1 – 6) = 5

A22 = 2 + 21 = 23

A23 = -(4 + 7) = -11

A31 = 0 + 3 = 3

A32 = -(0 – 12) = 12

A33 = -2 – 4 = -6

adj A = \begin{bmatrix} -1& 5  &3 \\ -4 & 23 &12 \\ 1 & -11 & -6 \end{bmatrix}

Question 10. \begin{bmatrix} 1& -1  &2 \\ 0 & 2 &-3 \\ 3 & -2 & 4 \end{bmatrix}

Solution: 

A = \begin{bmatrix} 1& -1  &2 \\ 0 & 2 &-3 \\ 3 & -2 & 4 \end{bmatrix}

|A| = 1(8 – 6) – 0 + 3(3 – 4) = -1

Now for adj A

A11 = 8 – 6 =2 

A12 = -(0 + 9) = -9

A13 = 0 – 6 = -6

A21 = -(-4 + 4) =

>

A22 = 4 – 6 = -2

A23 = -(-2 + 3) = -1

A31 = 3 – 4 = -1

A32 = -(-3 – 0) = 3

A33 = 2 – 0 = 2

adj A = \begin{bmatrix} 2& 0  &-1 \\ -9 & -2 &3 \\ -6 & -1 & 2 \end{bmatrix}

A-1 = (adj A)/|A|

A-1-\begin{bmatrix} 2& 0  &-1 \\ -9 & -2 &3 \\ -6 & -1 & 2 \end{bmatrix}

A-1\begin{bmatrix} -2& 0  &1 \\ 9 & 2 &-3 \\ 6 & 1 & -2 \end{bmatrix}

Question 11. \begin{bmatrix} 1& 0  &0 \\ 0 & cos\alpha  &sin\alpha  \\ 0 & sin\alpha & -cos\alpha \end{bmatrix}

Solution: 

A = \begin{bmatrix} 1& 0  &0 \\ 0 & cos\alpha  &sin\alpha  \\ 0 & sin\alpha & -cos\alpha \end{bmatrix}"208" style="vertical-align: -40px;"/>

|A| = 1(-cos2α – sin2α) = -1

Now,

A11 = -cos2α – sin2α = -1

A12 = 0

A13 = 0

A21 = 0

A22 = -cosα

A23 = -sinα

A31 = 0

A32 = -sinα

A33 = cosα

adj A = \begin{bmatrix} -1& 0  &0 \\ 0 & -cos\alpha  &-sin\alpha  \\ 0 & -sin\alpha & cos\alpha \end{bmatrix}

A-1 = (adj A)/|A|

A-1-\begin{bmatrix} -1& 0  &0 \\ 0 & -cos\alpha  &-sin\alpha  \\ 0 & -sin\alpha & cos\alpha \end{bmatrix}

A-1\begin{bmatrix} 1& 0  &0 \\ 0 & cos\alpha  &sin\alpha  \\ 0 & sin\alpha & -cos\alpha \end{bmatrix}

Question 12. Let A = \begin{bmatrix} 3 &7 \\ 2& 5 \end{bmatrix} and B = \begin{bmatrix} 6 &8 \\ 7& 9 \end{bmatrix}, verify that (AB) – 1 = B – 1A – 1 

Solution: 

A = \begin{bmatrix} 3 &7 \\ 2& 5 \end{bmatrix}

|A| = 15 – 14 = 1

A1 = 5

A12 = -2

A21 = -7

A22 = 3

A-1 = (adj A)/|A|

A-1\begin{bmatrix} 5 &-7 \\ -2& 3 \end{bmatrix}

B = \begin{bmatrix} 6 &8 \\ 7& 9 \end{bmatrix}

|B| = 54 – 56 = -2

adj B = \begin{bmatrix} 9 &-8 \\ -7& 6 \end{bmatrix}

B-1 = (adj B)/|B|

B-1\frac{-1}{2}\begin{bmatrix} 9 &-8 \\ -7& 6 \end{bmatrix}

B-1\begin{bmatrix} \frac{-9}{2} &4 \\ \frac{7}{2}& -3 \end{bmatrix}"106" style="vertical-align: -25px;"/>

Now,

B-1A-1\begin{bmatrix} \frac{-9}{2} &4 \\ \frac{7}{2}& -3 \end{bmatrix} \begin{bmatrix} 5 &-7 \\ -2& 3 \end{bmatrix}

B-1A-1\begin{bmatrix} \frac{-45}{2}-8 &\frac{63}{2}+12 \\ \frac{35}{2}+6& \frac{-49}{2}-9 \end{bmatrix}

B-1A-1\begin{bmatrix} \frac{-61}{2} &\frac{87}{2} \\ \frac{47}{2}& \frac{-67}{2} \end{bmatrix}

Now, AB = \begin{bmatrix} 3 &7 \\ 2& 5 \end{bmatrix} \begin{bmatrix} 6 &8 \\ 7& 9 \end{bmatrix}

AB = \begin{bmatrix} 18+49 &24+63 \\ 12+35& 16+45 \end{bmatrix}

AB = \begin{bmatrix} 67 &87 \\ 47& 61 \end{bmatrix}

|AB| = 67 * 61 – 87 * 47 = -2

adj (AB) = \begin{bmatrix} 61 &-87 \\ -47& 67 \end{bmatrix}>

(AB)-1 = (adj AB)/|AB|

(AB)-1\begin{bmatrix} 61 &-87 \\ -47& 67 \end{bmatrix}

(AB)-1\begin{bmatrix} \frac{-61}{2} &\frac{87}{2} \\ \frac{47}{2}& \frac{-67}{2} \end{bmatrix}

From above, you can see that (AB)-1 = B-1A-1.

Hence, it is proved.

Question 13. A = \begin{bmatrix} 3 &1 \\ -1& 2 \end{bmatrix}, show that A2 – 5A + 7I = O. Hence find A-1.

Solution: 

A = \begin{bmatrix} 3 &1 \\ -1& 2 \end{bmatrix}

A2\begin{bmatrix} 3 &1 \\ -1& 2 \end{bmatrix} \begin{bmatrix} 3 &1 \\ -1& 2 \end{bmatrix}

A2\begin{bmatrix} 9-1 &3+2\\ -3-2& -1+4 \end{bmatrix}

A2\begin{bmatrix} 8 &5\\ -5&3 \end{bmatrix}>

So, A2 – 5A + 7I

\begin{bmatrix} 8 &5\\ -5&3 \end{bmatrix} – 5\begin{bmatrix} 8 &5\\ -5&3 \end{bmatrix} + 7\begin{bmatrix} 1 &0\\ 0&1 \end{bmatrix}

\begin{bmatrix} 8 &5\\ -5&3 \end{bmatrix} – \begin{bmatrix} 15 &5 \\ -5& 10 \end{bmatrix} \begin{bmatrix} 7 &0 \\ 0& 7 \end{bmatrix}

\begin{bmatrix} 0 &0\\ 0&0 \end{bmatrix}

= O

Hence, A2 – 5A + 7I = O

It can be written as 

A.A – 5A = -7I

Multiplying by A-1 in both sides

A.A(A-1) – 5AA-1 = 7IA-1

A(AA1) – 5I = -7A-1

AI – 5I = -7A-1

A-1 = -(A – 5I)/7

A-1 =1/7(\begin{bmatrix} 5 &0\\ 0&5 \end{bmatrix} – \begin{bmatrix} 3 &1\\ -1&2 \end{bmatrix})

A-1\frac{1}{7}\begin{bmatrix} 2 &-1 \\ 1& 3 \end{bmatrix}

Question 14. For the matrix A =\begin{bmatrix} 3 &2 \\ 1& 1 \end{bmatrix}, find the numbers a and b such that A2 + aA + bI = O.

Solution: 

A = \begin{bmatrix} 3 &2 \\ 1& 1 \end{bmatrix}

A2\begin{bmatrix} 3 &2 \\ 1& 1 \end{bmatrix} \begin{bmatrix} 3 &2 \\ 1& 1 \end{bmatrix}

A2\begin{bmatrix} 9+2 &6+2 \\ 3+1& 2+1 \end{bmatrix}

A2 = \begin{bmatrix} 11 &8 \\ 4& 3 \end{bmatrix}"vertical-align: -25px;"/>

Now,

A2 – aA + bI = O

Multiplying by A-1 in both sides

(AA)A-1 + aAA-1 + bIA-1 = O

A(AA-1) + aI + b(IA-1) = O

AI + aI + bA-1 = O

A + aI = -bA-1

A-1 = -(A + aI)/b

Now,

A-1 = (adj A)/|A|

A-1\begin{bmatrix} 1 &-2 \\ -1& 3 \end{bmatrix}"107" style="vertical-align: -25px;"/>

Now, 

\begin{bmatrix} 1 &-2 \\ -1& 3 \end{bmatrix} = -1/b[\begin{bmatrix} 3 &2 \\ 1& 1 \end{bmatrix} + \begin{bmatrix} a &0 \\ 0& a \end{bmatrix}]

= -1/b\begin{bmatrix} 3+a &2 \\ 1& 1+a \end{bmatrix}

\begin{bmatrix} \frac{-3-a}{b} &\frac{-2}{b} \\ \frac{-1}{b}& \frac{-1-a}{b} \end{bmatrix}

On comparing elements you will get

-1/b = -1

b = 1

(-3 – a)/b

-3 – a = 1

a = -4

Hence, a = -4 and b = 1

Question 15. A = \begin{bmatrix} 1& 1  &1 \\ 1 & 2  &-3  \\ 2 & -1 & 3 \end{bmatrix}, show that A3 – 6A2 + 5A + 11I = O. Hence find A-1

Solution: 

A = \begin{bmatrix} 1& 1  &1 \\ 1 & 2  &-3  \\ 2 & -1 & 3 \end{bmatrix}

A2\begin{bmatrix} 1& 1  &1 \\ 1 & 2  &-3  \\ 2 & -1 & 3 \end{bmatrix} \begin{bmatrix} 1& 1  &1 \\ 1 & 2  &-3  \\ 2 & -1 & 3 \end{bmatrix}

A2\begin{bmatrix} 1+1+2& 1+2-1  &1-3+3 \\ 1+2-6 & 1+4+3  &1-6-9  \\ 2-1+6 & 2-2-3 & 2+3+9 \end{bmatrix}

A2\begin{bmatrix} 4& 2  &1 \\ -3 & 8  &-14  \\ 7 & -3 & 14 \end{bmatrix}

A3 = A2.A

A3\begin{bmatrix} 4& 2  &1 \\ -3 & 8  &-14  \\ 7 & -3 & 14 \end{bmatrix} \begin{bmatrix} 1& 1  &1 \\ 1 &2  &-3  \\ 2 &-1 & 3 \end{bmatrix}

A3\begin{bmatrix} 4+2+2& 4+4-1  &4-6+3 \\ -3+8-28 &-3+16+14 &-3-24-42  \\ 7-3+28 &7-6-14 & 7+9+42 \end{bmatrix}

A3\begin{bmatrix} 8& 7  &1 \\ -23 &27 &-69  \\ 32&-13 & 58 \end{bmatrix}"207" style="vertical-align: -40px;"/>

A3 – 6A2 + 5A + 11I

\begin{bmatrix} 8& 7  &1 \\ -23 &27 &-69  \\ 32&-13 & 58 \end{bmatrix} – 6\begin{bmatrix} 4& 2  &1 \\ -3 & 8  &-14  \\ 7 & -3 & 14 \end{bmatrix} + 5\begin{bmatrix} 1& 1  &1 \\ 1 & 2  &-3  \\ 2 & -1 & 3 \end{bmatrix} + 11\begin{bmatrix} 1& 0  &0 \\ 0 & 1  &0  \\ 0 & 0 & 1 \end{bmatrix}

\begin{bmatrix} 8& 7  &1 \\ -23 &27 &-69  \\ 32&-13 & 58 \end{bmatrix} – \begin{bmatrix} 24& 12  &6 \\ -18&48 &-84  \\ 42&-18 & 84 \end{bmatrix} + \begin{bmatrix} 5& 5  &5 \\ 5&10 &-15  \\ 10&-5 & 15 \end{bmatrix} + \begin{bmatrix} 11& 0  &0 \\ 0&11 &0  \\ 0&0 & 11 \end{bmatrix}

\begin{bmatrix} 0& 0  &0 \\ 0&0 &0  \\ 0&0 &0 \end{bmatrix}

= O

Hence, A3 – 6A2 + 5A + 11I = O

Now,

A3 – 6A2 + 5A + 11I = O

(AAA)A-1 – 6(AA)A-1 + 5(AA-1) + 11IA-1 = O

AA(AA-1) – 6A(AA-1) +

A-1) = -11(IA-1)

A2 – 6A + 5I = -11 A-1

A-1 = -1/11(A2 – 6A + 5I) ………….(1)

Now, A2 – 6A + 5I

=\begin{bmatrix} 4& 2  &1 \\ -3 & 8  &-14  \\ 7 & -3 & 14 \end{bmatrix}  – 6\begin{bmatrix} 1& 1  &1 \\ 1 & 2  &-3  \\ 2 & -1 & 3 \end{bmatrix} + 5\begin{bmatrix} 1& 0  &0 \\ 0 & 1  &0  \\ 0 & 0 & 1 \end{bmatrix}

=\begin{bmatrix} 4& 2  &1 \\ -3 & 8  &-14  \\ 7 & -3 & 14 \end{bmatrix} – \begin{bmatrix} 6& 6  &6 \\ 6&12 &-18  \\ 12&-6 & 18 \end{bmatrix} + \begin{bmatrix} 5& 0 &0 \\ 0&5 &0  \\ 0&0 &5 \end{bmatrix}

\begin{bmatrix} 3& -4 &-5 \\ -9&1 &4  \\ -5&3 &1 \end{bmatrix}

From eq(1) you have 

A-1 = -1/11 \begin{bmatrix} 3& -4 &-5 \\ -9&1 &4  \\ -5&3 &1 \end{bmatrix}

Question 16. A = \begin{bmatrix} 2& -1 &1 \\ -1&2 &-1  \\ 1&-1 &2 \end{bmatrix}, verify that A3 – 6A2 + 9A – 4I = O and hence fin A-1.

Solution: 

A = \begin{bmatrix} 2& -1 &1 \\ -1&2 &-1  \\ 1&-1 &2 \end{bmatrix}"168" style="vertical-align: -40px;"/>

A2\begin{bmatrix} 2& -1 &1 \\ -1&2 &-1  \\ 1&-1 &2 \end{bmatrix} \begin{bmatrix} 2& -1 &1 \\ -1&2 &-1  \\ 1&-1 &2 \end{bmatrix}

A2\begin{bmatrix} 4+1+1& -2-2-1 &2+1+2 \\ -2-2-1&1+4+1 &-1-2-2  \\ 2+1+2&-1-2-2 &1+1+4 \end{bmatrix}

A2\begin{bmatrix} 6& -5 &5 \\ -5&6 &-5  \\ 5&-5 &6 \end{bmatrix}

A3 = A2.A

A3\begin{bmatrix} 6& -5 &5 \\ -5&6 &-5  \\ 5&-5 &6 \end{bmatrix} \begin{bmatrix} 2& -1 &1 \\ -1&2 &-1  \\ 1&-1 &2 \end{bmatrix}

A3\begin{bmatrix} 12+5+5& -6-10-5 &6+5+10 \\ -10-6-5&5+12+5 &-5-6-10  \\ 10+5+6&-5-10-6 &5+5+12 \end{bmatrix}

A3\begin{bmatrix} 22& -21 &21 \\ -21&22 &-21  \\ 21&-21 &22 \end{bmatrix}

Now,

A3 – 6A2 + 9A – 4I

\begin{bmatrix} 22& -21 &21 \\ -21&22 &-21  \\ 21&-21 &22 \end{bmatrix} – 6\begin{bmatrix} 6& -5 &5 \\ -5&6 &-5  \\ 5&-5 &6 \end{bmatrix} + 9\begin{bmatrix} 2& -1 &1 \\ -1&2 &-1  \\ 1&-1 &2 \end{bmatrix} – 4\begin{bmatrix} 1& 0 &0 \\ 0&1 &0  \\ 0&0 &1 \end{bmatrix}"106" style="vertical-align: -40px;"/>

=\begin{bmatrix} 22& -21 &21 \\ -21&22 &-21  \\ 21&-21 &22 \end{bmatrix} – \begin{bmatrix} 36& -30 &30 \\ -30&36 &-30  \\ 30&-30 &36 \end{bmatrix} + \begin{bmatrix} 18& -9 &9 \\ -9&18 &-9  \\ 9&-9 &18 \end{bmatrix} – \begin{bmatrix} 4& 0 &0 \\ 0&4 &0  \\ 0&0 &4 \end{bmatrix}

=\begin{bmatrix} 40& -30 &30 \\ -30&40 &-30  \\ 30&-30 &40 \end{bmatrix} – \begin{bmatrix} 40& -30 &30 \\ -30&40 &-30  \\ 30&-30 &40 \end{bmatrix}

\begin{bmatrix} 0& 0 &0 \\ 0&0 &0  \\ 0&0 &0 \end{bmatrix}

= O

So, A3 – 6A2 + 9A – 4I = O

Now,

A3 – 6A2 + 9A – 4I = O

Multiplying by A-1 in both sides

(AAA)A-1 – 6(AA)A-1 + 9AA-1 – 4IA-1 = O

AA(AA-1) – 6A(AA-1)

(AA-1) = 4(IA-1)

AAI – 6AI +9I = 4A-1

A2 – 6A + 9I = 4A-1

A-1 = 1/4(A2 – 6A + 9I) ……….(1)

A2 – 6A + 9I

\begin{bmatrix} 6& -5 &5 \\ -5&6 &-5  \\ 5&-5 &6 \end{bmatrix} – 6\begin{bmatrix} 2& -1 &1 \\ -1&2 &-1  \\ 1&-1 &2 \end{bmatrix} + 9\begin{bmatrix} 1& 0 &0 \\ 0&1 &0  \\ 0&0 &1 \end{bmatrix}

\begin{bmatrix} 3& 1 &-1 \\ 1&3 &1  \\ -1&1 &3 \end{bmatrix}

From eq(1), you have 

A-1\frac{1}{4}\begin{bmatrix} 3& 1 &-1 \\ 1&3 &1  \\ -1&1 &3 \end{bmatrix}

Question 17. Let A be a non-singular matrix of order 3 * 3. Then |adj A| is equal to 

(A) |A|      (B) |A|2      (C) |A|3      (D) 3|A|

Solution:&nb

/strong>

You know,

(adj A)A = |A|I

(adj A)A = \begin{bmatrix} |A| &0  &0 \\ 0 & |A| &0 \\ 0 &0  &|A| \end{bmatrix}

|(adj A)A| = \begin{vmatrix} |A| &0  &0 \\ 0 & |A| &0\\ 0 &0  &|A| \end{vmatrix}

|(adj A)A| = |A|3\begin{vmatrix} 1 &0  &0 \\ 0 & 1 &0\\ 0 &0  &1 \end{vmatrix}

|(adj A)A| = |A|3 I

|adj A| = |A|3

Hence, option B is correct.

Question 18. If A is an invertible matrix of order 2, then det(A-1) is equal to 

(A) det(A)      (B) 1/(det A)      (C) 1      (D) 0

Solution: 

Since A is an invertible matrix then A-1 exist

p>

And A-1 = (adj A)/|A|

Suppose a 2 order matrix is A = \begin{bmatrix} a &b   \\ c &d \end{bmatrix}

Then |A| = ad – bc

and adj A = \begin{bmatrix} d &-b   \\ -c &a  \end{bmatrix}

A-1 = (adj A)/|A|

A-1\frac{1}{|A|}\begin{bmatrix} d &-b   \\ -c &a \end{bmatrix}

A-1\begin{bmatrix} \frac{d}{|A|} &\frac{-b}{|A|}  \\ \frac{c}{|A|}&\frac{a}{|A|}       \end{bmatrix}

|A-1| = \begin{vmatrix} \frac{d}{|A|} &\frac{-b}{|A|} \\ \frac{-c}{|A|} & \frac{a}{|A|} \end{vmatrix}

|A-1| = \frac{1}{|A|^{2}}\begin{vmatrix} d &-b \\ -c &a \end{vmatrix}

|A-1| = (ad – bc)/|A|2

|A-1| = |A|/|A|2

|A-1| = 1/|A|

det(A-1) = 1/(det A)

Hence, option B is correct.


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