class 12 maths exercise 4.5 ncert solution

EXERCISE 4.5 (Determinant)

class 12 maths exercise 4.5 ncert solution

Question 1: Find the adjoint of the matrix \begin{bmatrix}1&2\\3&4\end{bmatrix}

Solution: LetA=\begin{bmatrix}1&2\\3&4\end{bmatrix}

Then,A_{11}=4,A_{12}=-3,A_{21}=-2,A_{22}=1
Thus

adjA=\begin{bmatrix}A_{11}&A_{12}\\A_{21}&A_{22}\end{bmatrix}'

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

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

Question 2: Find the adjoint of the matrix \begin{bmatrix}1&-1&2\\2&3&5\\-2&0&1\end{bmatrix}

Solution: Let A=\begin{bmatrix}1&-1&2\\2&3&5\\-2&0&1\end{bmatrix}
Then,

A_{11}=\begin{vmatrix}3&5\\0&1\end{vmatrix}=3-0=3

A_{12}=-\begin{vmatrix}2&5\\-2&1\end{vmatrix}=-(2+10)=-12

A_{13}=\begin{vmatrix}2&3\\-2&0\end{vmatrix}=0+6=6

A_{21}=-\begin{vmatrix}-1&2\\0&1\end{vmatrix}=-(-1-0)=1

A_{22}=\begin{vmatrix}1&2\\-2&1\end{vmatrix}=1+4=5

A_{11}=\begin{vmatrix}3&5\\0&1\end{vmatrix}=3-0=3

A_{23}=-\begin{vmatrix}1&-1\\-2&0\end{vmatrix}=-(0-2)=2

A_{31}=\begin{vmatrix}-1&2\\3&5\end{vmatrix}=-5-6=-11

A_{32}=-\begin{vmatrix}1&2\\2&5\end{vmatrix}=-(5-4)=-1

A_{33}=\begin{vmatrix}1&-1\\2&3\end{vmatrix}=(3+2)=5

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

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

Question 3:  Verify A(\operatorname{adj} A)=(\operatorname{adj} A) A=|A| I for

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

Solution: Let A=\begin{bmatrix}2 & 3 \\ -4 & -6\end{bmatrix}

Then,

|A| =-12-(-12)

=0

Also,

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

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

Now,
A_{11}=-6, A_{12}=4

A_{21}=-3 , A_{22}=2

Hence,

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

Now,

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

=\begin{bmatrix} -12+12 & -6+6 \\ 24-24 & 12-12 \end{bmatrix}

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

Also,

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

=\begin{bmatrix} -12+12 & -18+18 \\ 8-8 & 12-12 \end{bmatrix}

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

Hence,

A(\operatorname{adj} A)=(\operatorname{adj} A) A=|A| I.

Question 4: verify A(adjA)=(adjA)A=|A|I for

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

Solution : Let A=\begin{bmatrix}1&-1&2\\3&0&-2\\1&0&3\end{bmatrix}
Then

|A|=1(0-0)+1(9+2)+2(0-0) =11
Also,

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

=\begin{bmatrix}11&0&0\\0&11&0\\0&0&11\end{bmatrix}
Now,

A_{11}=(0-0)=0

A_{12}=-(9+2)=-11`

A_{13}=(0-0)=0

A_{21}=-(-3-0)=3

A_{22}=(3-2)=1

A_{23}=-(0+1)=-1

A_{31}=(2-0)=2

A_{32}=-(-2-6)=8

A_{33}=(0+3)=3

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

=\begin{bmatrix}0&3&2\\-11&1&8\\0&-1&3\end{bmatrix}

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}

=\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}

=\begin{bmatrix}11&0&0\\0&11&0\\0&0&11\end{bmatrix}

Also,

(adjA)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}

=\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&2+0+9\end{bmatrix}

\begin{bmatrix}11&0&0\\0&11&0\\0&0&11\end{bmatrix}

Hence, A(adjA)=(adjA)A=|A|I

Question 5: Find the inverse of each of the matrix \begin{bmatrix}1&-2\\4&3\end{bmatrix}(if it exist).

Solution: Let A=\begin{bmatrix}1&-2\\4&3\end{bmatrix}
Then,

|A|=6+8=14
Now,

A_{11}=3,A_{12}=-4,A_{21}=2,A_{22}=2

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

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

A^{-1}=\frac{adjA}{|A|}

=\frac{1}{14}\begin{bmatrix}3&2\\-4&2\end{bmatrix}

Question 6: Find the inverse of the matrix \begin{bmatrix}-1&5\\-3&2\end{bmatrix}(if it exists)

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

Then |A|=-2+15=13
Now,

A_{11}=2,A_{12}=3,A_{21}=-5,A_{22}=-1

Therefore,

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

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

Hence,

A^{-1}=\frac{1}{13}\begin{bmatrix}2&-5\\3&-1\end{bmatrix}

Question 7: Find the inverse of the matrix \begin{bmatrix}1&2&3\\0&2&4\\0&0&5\end{bmatrix}(if it exists)

Solution: Let A=\begin{bmatrix}1&2&3\\0&2&4\\0&0&5\end{bmatrix}
Then,

|A|=1(10-0)-2(0-0)+3(0-0)
Now,

A_{11}=(10-0)=10,A_{12}=-(0-0)=0,A_{13}=(0-0)=0

A_{21}=-(10-0)=-10,A_{22}=(5-0)=5,A_{23}=-(0-0)=0

A_{31}=(8-6)=2,A_{32}=-(4-0)=-4,A_{33}=(2-0)=2

Therefore,

adjA=\begin{bmatrix}10&0&0\\-10&5&0\\2&-4&2\end{bmatrix}'

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

Hence,

A^{-1}=\frac{adjA}{|A|}

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

Question 8: Find the inverse of each of the matrix\begin{bmatrix}1&0&0\\3&3&0\\5&2&-1\end{bmatrix}(if it exists)

Solution: Let A=\begin{bmatrix}1&0&0\\3&3&0\\5&2&-1\end{bmatrix}
Then

|A|=1(-3-0)-0-0
Now,

A_{11}=(-3-0)=-3,A_{12}=-(-3-0)=3'A_{13}=(6-15)=-9

A_{21}=-(0-0)=0,A_{22}=(-1-0)=-1,A_{23}=-(2-0)=-2

A_{31}=(0-0)=0,A_{32}=-(0-0)=0,A_{33}=(3-0)=3

adjA=\begin{bmatrix}-3&3&-9\\0&-1&-2\\0&0&3\end{bmatrix}'

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

A^{-1}=\frac{adjA}{|A|}

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

Question 9: Find the inverse of each of the matrix\begin{bmatrix}2&1&3\\4&-1&0\\-7&2&1\end{bmatrix}(if it exists)

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

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

=-3
Now,

A_{11}=(-1-0)=-1,A_{12}=-(4-0)=-4,A_{13}=(8-7)=1

A_{21}=-(1-6)=5,A_{22}=(2+21)=23,A_{23}=-(4+7)=-11

A_{31}=(0+3)=3,A_{32}=-(0-12)=12,A_{33}=(-2-4)=-6

adjA=\begin{bmatrix}-1&-4&1\\5&12&-11\\3&12&-6\end{bmatrix}'

adjA=\begin{bmatrix}-1&5&3\\-4&23&12\\1&-11&-6\end{bmatrix}
Hence,

A^{-1}=\frac{adjA}{|A|}

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

Question 10:  Find the inverse of the matrix \begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix} (if it exists).

Solution: Let A=\begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix}
Then, expanding

|A|=1(8-6)+1(0+9)+2(0-6)

=2+9-12

=-1
Now,

A_{11}=(8-6)=2,A_{12}=-(0+9)=-9,A_{13}=(0-6)=-6

A_{21}=-(-4+4)=0,A_{22}=(4-6)=-2,A_{23}=-(-2+3)=-1

A_{31}=(3-4)=5,A_{32}=-(-3-0)=3,A_{33}=(2-0)=2

adjA=\begin{bmatrix}2&-9&-6\\0&-2&-1\\-1&3&2\end{bmatrix}'
Therefore,

=\begin{bmatrix}2&0&-1\\-9&-2&3\\-6&-1&2\end{bmatrix}
Hence,

A^{-1}=\frac{adjA}{|A|}

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

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

Question 11: Find the inverse of the matrix \begin{bmatrix}1&0&0\\0&\cos\alpha&\sin\alpha\\0&\sin\alpha&-\cos\alpha\end{bmatrix} (if it exists)

Solution: Let =\begin{bmatrix}1&0&0\\0&\cos\alpha&\sin\alpha\\0&\sin\alpha&-\cos\alpha\end{bmatrix}
Then,

|A|=1(-\cos^2\alpha-\sin^2\alpha)

=-(\cos^2\alpha+\sin^2\alpha)

=-1
Now,

A_{11}=(-\cos^2\alpha-\sin^2\alpha)=-1,A_{12}=0,_{13}=0

A_{21}=0,A_{22}=-\cos\alpha,A_{23}=-\sin\alpha

A_{31}=0,A_{32}=-\sin\alpha,A_{33}=\cos\alpha
Therefore,

adjA=\begin{bmatrix}-1&0&0\\0&-\cos\alpha&-\sin\alpha\\0&-\sin\alpha&\cos\alpha\end{bmatrix}'

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

A^{-1}=\frac{adjA}{|A|}

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

=\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^{-1}A^{-1}

Solution:  Let A=\begin{bmatrix}3&7\\2&5\end{bmatrix}
Then,

|A|=15-14=1
Now,

A_{11}=5,A_{12}=-2

A_{21}=-7,A_{22}=3

adjA=\begin{bmatrix}5&-2\\-7&3\end{bmatrix}'

=\begin{bmatrix}5&-7\\-2&3\end{bmatrix}
Therefore,

A^{-1}=\frac{adjA}{|A|}

=\begin{bmatrix}5&-7\\-2&3\end{bmatrix}
Now for B

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

|B|=54-56=-2

Now, B_{11}=9,B_{12}-7

B_{21}=-8,B_{22}=6

adjB=\begin{bmatrix}9&-7\\-8&6\end{bmatrix}'

=\begin{bmatrix}9&-8\\-7&6\end{bmatrix}
Then,

B^{-1}=\frac{adjB}{|B|}

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

=\begin{bmatrix}-\frac{9}{2}&4\\\frac{7}{2}&-3\end{bmatrix}

Now,

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

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

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

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

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

=\begin{bmatrix}67&87\\47&61\end{bmatrix}
Then we have,

|AB|=67\times61-87\times47

=4087-4089=-2

and

A_{11}=61,A_{12}=-47

A_{21}=-87,A_{22}=67

Therefore,

\operatorname{adj}(A B)=\begin{bmatrix} 61 & -47 \\ -87 & 67 \end{bmatrix}'

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

Thus,

(A B)^{-1} =\frac{1}{|A B|} \operatorname{adj}(A B)

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

=\begin{bmatrix} -\frac{61}{2} & \frac{87}{2} \\ \frac{47}{2} & -\frac{67}{2} \end{bmatrix}

From (1) and (2),

(A B)^{-1}=B^{-1} A^{-1}

Hence, proved.

Question 13: If A=\begin{bmatrix}3 & 1 \\ -1 & 2\end{bmatrix}, show that A^{2}-5 A+7 I=0. Hence find A^{-1}.

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

Therefore,

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

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

=\begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}

Now,

A^{2}-5 A+7 I = \begin{bmatrix} 8 & 5\\ -5 & 3 \end{bmatrix}-5\begin{bmatrix} 3 & 1 \\ -1 & 2 \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} -7 & 0 \\ 0 & -7 \end{bmatrix}+\begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}

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

Hence, A^{2}-5 A+7 I=0.

Now,

\Rightarrow A \cdot A-5 A=-7 I

\left.\Rightarrow A \cdot A\left(A^{-1}\right)-5 A \cdot A^{-1}=-7 I A^{-1} \quad \quad \quad \text { post-multiplying by } A^{-1} \text { as }|A| \neq 0\right]

\Rightarrow A\left(A A^{-1}\right)-5 I=-7 A^{-1}

\Rightarrow A I-5 I=-7 A^{-1}

\Rightarrow A^{-1}=-\frac{1}{7}(A-5 I)

\Rightarrow A^{-1}=\frac{1}{7}(5 I-A)

\Rightarrow A^{-1}=\frac{1}{7}\left[\begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}-\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\right]

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

Thus,

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 A^{2}+a A+b I=0.

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

Therefore,

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

=\begin{bmatrix} 9+2 & 6+2 \\ 3+1 & 2+1 \end{bmatrix}=\begin{bmatrix} 11 & 8 \\ 4 & 3 \end{bmatrix}

Now, A^{2}+a A+b I=0.

Hence,

\Rightarrow(A \cdot A) A^{-1}+a A \cdot A^{-1}+b I A^{-1}=0 \quad\left[\text { post-multiplying by } A^{-1} \text { as }|A| \neq 0\right]

\Rightarrow A\left(A A^{-1}\right)+a I+b\left(I A^{-1}\right)=0

\Rightarrow A I+a I+b A^{-1}=0

\Rightarrow A+a I=-b A^{-1}

\Rightarrow A^{-1}=-\frac{1}{b}(A+a I) \quad \text { _..(1) }

Now,

A^{-1} =\frac{1}{|A|} \text { adjA }

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

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

From (1) and (2), we have,

\Rightarrow \begin{bmatrix} 1 & -2 \\ -1 & 3 \end{bmatrix}=\frac{1}{b}\left[\begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}+\begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix}\right]

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

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

Comparing the corresponding elements of the two matrices, we have:

\Rightarrow-\frac{1}{b}=-1

\Rightarrow h=1

Also,

\Rightarrow \frac{-3-a}{b}=1

\Rightarrow-3-a=1

\Rightarrow a=-4

Thus, a=-4 and b=1.

Question 15:  A=\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix}

show that

A^{3}-6 A^{2}+5 A+11 I=0

Hence, find A^{-1} .

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

Therefore,

A^{2} &=A \cdot A=\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}

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

And,

A^{3} =A^{2} \cdot A=\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}

=\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}=\begin{bmatrix} 8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58 \end{bmatrix}

Hence,

A^{3}-6 A^{2}+5 A+11 I

=\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}{ccc}11 & 0 & 0 \\0 & 11 & 0 \\0 & 0 & 11\end{bmatrix}

=\begin{bmatrix}24 & 12 & 6 \\-18 & 48 & -84 \\42 & -18 & 84\end{bmatrix}-\begin{bmatrix}24 & 12 & 6 \\-18 & 48 & -84 \\42 & -18 & 84\end{bmatrix}

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

=0

Thus, A^{3}-6 A^{2}+5 A+11 I=0

Now,

\Rightarrow A^{3}-6 A^{2}+5 A+11 I=0

\left.\Rightarrow(A A A) A^{-1}-6(A A) A^{-1}+5 A A^{-1}+11 I A^{-1}=0 \quad \text { [post-multiplying by } A^{-1} \text { as }|A| \neq 0\right]

\Rightarrow A A\left(A A^{-1}\right)-6 A\left(A A^{-1}\right)+5\left(A A^{-1}\right)=-11\left(I A^{-1}\right)

\Rightarrow A^{2}-6 A+5 I=-11 A^{-1}

\Rightarrow A^{-1}=-\frac{1}{11}\left(A^{2}-6 A+5 I\right)

Now,

A^{2}-6 A+5 I

=\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}9 & 2 & 1 \\-3 & 13 & -14 \\7 & -3 & 19\end{bmatrix}-\begin{bmatrix}6 & 6 & 6 \\6 & 12 & -18 \\12 & -6 & 18\end{bmatrix}

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

From equation (1) and (2)

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

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

Question 16: If A=\begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix}

verify that

A^{3}-6 A^{2}+9 A-4 I=0 . Hence, find A^{-1}.

Solution: Let A=\begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix}

Therefore,

A^{2} =A . A

=\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}

=\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}

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

And

A^{3} =A^{2} \cdot A

=\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}

=\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}

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

Now,

A^{3}-6 A^{2}+9 A-4 I

=\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}

=\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}

=0

Thus,

A^{3}-6 A^{2}+9 A-4 I=0

Now,

\Rightarrow A^{3}-6 A^{2}+9 A-4 I=0

\left.\Rightarrow(A A A) A^{-1}-6(A A) A^{-1}+9 A A^{-1}-4 I A^{-1}=0 \quad \text { [post-multiplying by } A^{-1} \text { as }|A| \neq 0\right]

\Rightarrow A A\left(A A^{-1}\right)-6 A\left(A A^{-1}\right)+9\left(A A^{-1}\right)=4\left(I A^{-1}\right)

\Rightarrow A A I-6 A I+9 I=4 A^{-1}

\Rightarrow A^{2}-6 A+9 I=4 A^{-1}

\Rightarrow A^{-1}=\frac{1}{4}\left(A^{2}-6 A+9 I\right) \quad \ldots(1)

Now,

A^{2}-6 A+9 I

=\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} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{bmatrix}-\begin{bmatrix} 12 & -6 & 6 \\ -6 & 12 & -6 \\ 6 & -6 & 12 \end{bmatrix}+\begin{bmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{bmatrix}

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

From equations (1) and (2)

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 square matrix of order 3 \times 3. Then |\operatorname{adj} A| is equal to:

(A) |A|

(B) |A|^{2}

(C) |A|^{3}

(D) 3|A|

Solution: the correct option is B.

Since A be a non-singular square matrix of order 3 \times 3

(\operatorname{adj} A) A =|A| I

=\begin{bmatrix} |A| & 0 & 0 \\ 0 & |A| & 0 \\ 0 & 0 & |A| \end{bmatrix}

Therefore,

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

|\operatorname{adj} A||A| =|A|^{3}\begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}

=|A|^{3} I

|a d j A| =|A|^{2}

Thus, the correct option is B.

Question 18: If A is an invertible matrix of order 2, the \operatorname{det}\left(A^{-1}\right) is equal to:

(A) \operatorname{det}(A)

(B) \frac{1}{\operatorname{det}(A)}

(C) 1

(D) 0

Solution: Thus, the correct option is B.

Since A is an invertible matrix, A^{-1} exists and A^{-1}=\frac{1}{|A|} adj A.

As matrix A is of order 2, let A=\begin{bmatrix}a & b \\ c & d\end{bmatrix} Then,

|A|=a d-b c

And

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

Now,

A^{-1} =\frac{1}{|A|} \text { adjA }

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

Hence,

\left|A^{-1}\right| =\mid \begin{bmatrix} \frac{d}{|A|} & \frac{-b}{|A|} \mid \\ \frac{-c}{|A|} & \frac{a}{|A|} \mid \end{bmatrix}

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

=\frac{1}{|A|^{2}}(a d-b c)

=\frac{1}{|A|^{2}} \cdot|A|

=\frac{1}{|A|}

Hence,

\operatorname{det}\left(A^{-1}\right)=\frac{1}{\operatorname{det}(A)}

Thus, the correct option is B.


Class 12 determinants multiple choice question

Case study problem determinant 4
On the occasion of children’s day. Class teacher of class XII shri singh,
Case study problem determinant 3
The monthly incomes of two sister Reshma and Ritam are in the ratio 3:4 and their

For solution

Team Gmath

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