# NCERT Solutions For Class 12 Maths Chapter 4 Exercise 4.5

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## Chapter 4 Ex.4.5 Question 1

Find the adjoint of the matrix $$\left( {\begin{array}{*{20}{c}}1&2\\3&4\end{array}} \right)$$

### Solution

Let $$A = \left( {\begin{array}{*{20}{c}}1&2\\3&4\end{array}} \right)$$

Then,

\begin{align}&{A_{11}} = 4\;\;\;\;\;\;\;\;{A_{12}} = - 3\\&{A_{21}} = - 2\;\;\;\;\;\;{A_{22}} = 1\end{align}

Therefore,

\begin{align}adj\;A &= \left( {\begin{array}{*{20}{c}}{{A_{11}}}&{{A_{12}}}\\{{A_{21}}}&{{A_{22}}}\end{array}} \right)\\& = \left( {\begin{array}{*{20}{c}}4&{ - 2}\\{ - 3}&1\end{array}} \right)\end{align}

## Chapter 4 Ex.4.5 Question 2

Find the adjoint of the matrix $$\left( {\begin{array}{*{20}{c}}1&{ - 1}&2\\2&3&5\\{ - 2}&0&1\end{array}} \right)$$

### Solution

Let $$A = \left( {\begin{array}{*{20}{c}}1&{ - 1}&2\\2&3&5\\{ - 2}&0&1\end{array}} \right)$$

Then,

\begin{align}&{A_{11}} = \left| {\begin{array}{*{20}{c}}3&5\\0&1\end{array}} \right| = 3\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{A_{12}} = - \left| {\begin{array}{*{20}{c}}2&5\\{ - 2}&1\end{array}} \right| = - 12\;\;\;\;\;\;\;\;{A_{13}} = \left| {\begin{array}{*{20}{c}}2&3\\{ - 2}&0\end{array}} \right| = 6\\&{A_{21}} = - \left| {\begin{array}{*{20}{c}}{ - 1}&2\\0&1\end{array}} \right| = 1\;\;\;\;\;\;\;\;\;\;{A_{22}} = \left| {\begin{array}{*{20}{c}}1&2\\{ - 2}&1\end{array}} \right| = 5\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{A_{23}} = - \left| {\begin{array}{*{20}{c}}1&{ - 1}\\{ - 2}&0\end{array}} \right| = 2\\&{A_{31}} = \left| {\begin{array}{*{20}{c}}{ - 1}&2\\3&5\end{array}} \right| = - 11\;\;\;\;\;\;\;\;{A_{32}} = - \left| {\begin{array}{*{20}{c}}1&2\\2&5\end{array}} \right| = - 1\;\;\;\;\;\;\;\;\;\;\;\;\;{A_{33}} = \left| {\begin{array}{*{20}{c}}1&{ - 1}\\2&3\end{array}} \right| = 5\end{align}

Therefore,

$$adj\;A = \left( {\begin{array}{*{20}{c}}3&1&{ - 11}\\{ - 12}&5&{ - 1}\\6&2&5\end{array}} \right)$$

## Chapter 4 Ex.4.5 Question 3

Verify $$A\left( {adj\;A} \right) = \left( {adj\;A} \right)A = \left| A \right|I$$ for

$$\left( {\begin{array}{*{20}{c}}2&3\\{ - 4}&{ - 6}\end{array}} \right)$$

### Solution

Let $$A = \left( {\begin{array}{*{20}{c}}2&3\\{ - 4}&{ - 6}\end{array}} \right)$$

Then,

\begin{align}\left| A \right| &= - 12 - \left( { - 12} \right)\\ &= 0\end{align}

Also,

\begin{align}\left| A \right|I &= 0\left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)\\& = \left( {\begin{array}{*{20}{c}}0&0\\0&0\end{array}} \right)\end{align}

Now,

\begin{align}{A_{11}} &= - 6\;\;\;\;\;\;\;\;\;\;\;\;\;{A_{12}} = 4\\{A_{21}} &= - 3\;\;\;\;\;\;\;\;\;\;\;\;\;{A_{22}} = 2\end{align}

Hence,

$$adj\;A = \left( {\begin{array}{*{20}{c}}{ - 6}&{ - 3}\\4&2\end{array}} \right)$$

Now,

\begin{align}A\left( {adj\;A} \right) &= \left( {\begin{array}{*{20}{c}}2&3\\{ - 4}&{ - 6}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - 6}&{ - 3}\\4&2\end{array}} \right)\\& = \left( {\begin{array}{*{20}{c}}{ - 12 + 12}&{ - 6 + 6}\\{24 - 24}&{12 - 12}\end{array}} \right)\\& = \left( {\begin{array}{*{20}{c}}0&0\\0&0\end{array}} \right)\end{align}

Also,

\begin{align}\left( {adj\;A} \right)A &= \left( {\begin{array}{*{20}{c}}{ - 6}&{ - 3}\\4&2\end{array}} \right)\left( {\begin{array}{*{20}{c}}2&3\\{ - 4}&{ - 6}\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{ - 12 + 12}&{ - 18 + 18}\\{8 - 8}&{12 - 12}\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}0&0\\0&0\end{array}} \right)\end{align}

Hence, $$A\left( {adj\;A} \right) = \left( {adj\;A} \right)A = \left| A \right|I$$.

## Chapter 4 Ex.4.5 Question 4

Verify $$A\left( {adj\;A} \right) = \left( {adj\;A} \right)A = \left| A \right|I$$ for

$$\left( {\begin{array}{*{20}{c}}1&{ - 1}&2\\3&0&{ - 2}\\1&0&3\end{array}} \right)$$

### Solution

Let $$A = \left( {\begin{array}{*{20}{c}}1&{ - 1}&2\\3&0&{ - 2}\\1&0&3\end{array}} \right)$$

Then,

\begin{align}\left| A \right|& = 1\left( {0 - 0} \right) + 1\left( {9 + 2} \right) + 2\left( {0 - 0} \right)\\& = 11\end{align}

Also,

\begin{align}\left| A \right|I &= 11\left( {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right)\\& = \left( {\begin{array}{*{20}{c}}{11}&0&0\\0&{11}&0\\0&0&{11}\end{array}} \right)\end{align}

Now,

\begin{align}&{A_{11}} = 0\;\;\;\;\;\;\;\;\;\;\;{A_{12}} = - 11\;\;\;\;\;\;\;\;\;\;\;{A_{13}} = 0\\&{A_{21}} = 3\;\;\;\;\;\;\;\;\;\;\;{A_{22}} = 1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{A_{23}} = - 1\\&{A_{31}} = 2\;\;\;\;\;\;\;\;\;\;\;{A_{32}} = 8\;\;\;\;\;\;\;\;\;\;\;\;\;\;{A_{33}} = 3\end{align}

Hence,

$$adj\;A = \left( {\begin{array}{*{20}{c}}0&3&2\\{ - 11}&1&8\\0&{ - 1}&3\end{array}} \right)$$

Now,

\begin{align}A\left( {adj\;A} \right) &= \left( {\begin{array}{*{20}{c}}1&{ - 1}&2\\3&0&{ - 2}\\1&0&3\end{array}} \right)\left( {\begin{array}{*{20}{c}}0&3&2\\{ - 11}&1&8\\0&{ - 1}&3\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{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{array}} \right) = \left( {\begin{array}{*{20}{c}}{11}&0&0\\0&{11}&0\\0&0&{11}\end{array}} \right)\end{align}

Also,

\begin{align}\left( {adj\;A} \right)A &= \left( {\begin{array}{*{20}{c}}0&3&2\\{ - 11}&1&8\\0&{ - 1}&3\end{array}} \right)\left( {\begin{array}{*{20}{c}}1&{ - 1}&2\\3&0&{ - 2}\\1&0&3\end{array}} \right)\\& = \left( {\begin{array}{*{20}{c}}{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{array}} \right)\\& = \left( {\begin{array}{*{20}{c}}{11}&0&0\\0&{11}&0\\0&0&{11}\end{array}} \right)\end{align}

Hence, $$A\left( {adj\;A} \right) = \left( {adj\;A} \right)A = \left| A \right|I$$.

## Chapter 4 Ex.4.5 Question 5

Find the inverse of each of the matrix $$\left( {\begin{array}{*{20}{c}}2&{ - 2}\\4&3\end{array}} \right)$$ (if it exists).

### Solution

Let $$A = \left( {\begin{array}{*{20}{c}}2&{ - 2}\\4&3\end{array}} \right)$$

Then,

\begin{align}\left| A \right| &= 6 + 8\\ &= 14\end{align}

Now,

\begin{align}{A_{11}} &= 3\;\;\;\;\;\;\;\;\;\;\;{A_{12}} = - 4\\{A_{21}}& = 2\;\;\;\;\;\;\;\;\;\;\;{A_{22}} = 2\end{align}

Therefore,

$$adj\;A = \left( {\begin{array}{*{20}{c}}3&2\\{ - 4}&2\end{array}} \right)$$

Hence,

\begin{align}{A^{ - 1}} &= \frac{1}{{\left| A \right|}}adj\;A\\ &= \frac{1}{{14}}\left( {\begin{array}{*{20}{c}}3&2\\{ - 4}&2\end{array}} \right)\end{align}

## Chapter 4 Ex.4.5 Question 6

Find the inverse of each of the matrix $$\left( {\begin{array}{*{20}{c}}{ - 1}&5\\{ - 3}&2\end{array}} \right)$$ (if it exists)

### Solution

Let $$A = \left( {\begin{array}{*{20}{c}}{ - 1}&5\\{ - 3}&2\end{array}} \right)$$

Then,

$$\left| A \right| = - 2 + 15 = 13$$

Now,

\begin{align}{A_{11}} &= 2\;\;\;\;\;\;\;\;\;\;{A_{12}} = 3\\{A_{21}}& = - 5\;\;\;\;\;\;\;\;{A_{22}} = - 1\end{align}

Therefore,

$$adj\;A = \left( {\begin{array}{*{20}{c}}2&{ - 5}\\3&{ - 1}\end{array}} \right)$$

Hence,

\begin{align}{A^{ - 1}} &= \frac{1}{{\left| A \right|}}adj\;A\\ &= \frac{1}{{13}}\left( {\begin{array}{*{20}{c}}2&{ - 5}\\3&{ - 1}\end{array}} \right)\end{align}

## Chapter 4 Ex.4.5 Question 7

Find the inverse of each of the matrix $$\left( {\begin{array}{*{20}{c}}1&2&3\\0&2&4\\0&0&5\end{array}} \right)$$ (if it exists)

### Solution

Let $$A = \left( {\begin{array}{*{20}{c}}1&2&3\\0&2&4\\0&0&5\end{array}} \right)$$

Then,

\begin{align}\left| A \right| &= 1\left( {10 - 0} \right) - 2\left( {0 - 0} \right) + 3\left( {0 - 0} \right)\\& = 10\end{align}

Now,

$\begin{array}{l}{A_{11}} = 10\;\;\;\;\;\;\;\;\;\;\;{A_{12}} = 0\;\;\;\;\;\;\;\;\;\;\;{A_{13}} = 0\\{A_{21}} = -10\;\;\;\;\;\;\;\;{A_{22}} = 5\;\;\;\;\;\;\;\;\;\;\;{A_{23}} = 0\\{A_{31}} = 2\;\;\;\;\;\;\;\;\;\;\;\;{A_{32}} = -4\;\;\;\;\;\;\;\;\;{A_{33}} = 2 \end{array}$

Therefore,

$$adj\;A = \left( {\begin{array}{*{20}{c}}{10}&{ - 10}&2\\0&5&{ - 4}\\0&0&2\end{array}} \right)$$

Hence,

\begin{align}{A^{ - 1}} &= \frac{1}{{\left| A \right|}}adj\;A\\ &= \frac{1}{{10}}\left( {\begin{array}{*{20}{c}}{10}&{ - 10}&2\\0&5&{ - 4}\\0&0&2\end{array}} \right)\end{align}

## Chapter 4 Ex.4.5 Question 8

Find the inverse of each of the matrix $$\left( {\begin{array}{*{20}{c}}1&0&0\\3&3&0\\5&2&{ - 1}\end{array}} \right)$$ (if it exists)

### Solution

Let $$A = \left( {\begin{array}{*{20}{c}}1&0&0\\3&3&0\\5&2&{ - 1}\end{array}} \right)$$

Then,

$$\left| A \right| = 1\left( { - 3 - 0} \right) - 0 + 0 = - 3$$

Now,

\begin{align}{A_{11}} &= - 3\;\;\;\;\;\;\;\;\;\;\;{A_{12}} = 3\;\;\;\;\;\;\;\;\;\;\;{A_{13}} = -9\\{A_{21}} = 0\;\;\;\;\;\;\;\;\;\;\;\;\;{A_{22}} = -1\;\;\;\;\;\;\;\;\;\;\;{A_{23}} = -2\\{A_{31}} &= 0\;\;\;\;\;\;\;\;\;\;\;\;\;{A_{32}} = 0\;\;\;\;\;\;\;\;\;\;\;{A_{33}} = 3\end{align}

Therefore,

$adj\;A = \left( {\begin{array}{*{20}{c}}{ - 3}&0&0\\3&{ - 1}&0\\{ - 9}&{ - 2}&3\end{array}} \right)$

Hence,

\begin{align}{A^{ - 1}} &= \frac{1}{{\left| A \right|}}adj\;A\\& = \frac{{ - 1}}{3}\left( {\begin{array}{*{20}{c}}{ - 3}&0&0\\3&{ - 1}&0\\{ - 9}&{ - 2}&3\end{array}} \right)\end{align}

## Chapter 4 Ex.4.5 Question 9

Find the inverse of each of the matrix $$\left( {\begin{array}{*{20}{c}}2&1&3\\4&{ - 1}&0\\{ - 7}&2&1\end{array}} \right)$$ (if it exists)

### Solution

Let $$A = \left( {\begin{array}{*{20}{c}}2&1&3\\4&{ - 1}&0\\{ - 7}&2&1\end{array}} \right)$$

Then,

\begin{align}\left| A \right| &= 2\left( { - 1 - 0} \right) - 1\left( {4 - 0} \right) + 3\left( {8 - 7} \right)\\ &= 2\left( { - 1} \right) - 1\left( 4 \right) + 3\left( 1 \right)\\& = - 3\end{align}

Now,

\begin{align}{A_{11}} &= - 1\;\;\;\;\;\;\;\;\;\;\;{A_{12}} = - 4\;\;\;\;\;\;\;\;\;\;\;{A_{13}} = 1\\{A_{21}} = 5\;\;\;\;\;\;\;\;\;\;\;\;\;{A_{22}} = 23\;\;\;\;\;\;\;\;\;\;\;{A_{23}} = - 11\\{A_{31}} &= 3\;\;\;\;\;\;\;\;\;\;\;\;\;{A_{32}} = 12\;\;\;\;\;\;\;\;\;\;\;{A_{33}} = - 6\end{align}

Therefore,

$adjA = \left( {\begin{array}{*{20}{c}}{ - 1}&5&3\\{ - 4}&{23}&{12}\\&1&{ - 11}&{ - 6}\end{array}} \right)$

Hence,

\begin{align}{A^{ - 1}}& = \frac{1}{{\left| A \right|}}adjA\\ &= - \frac{1}{3}\left( {\begin{array}{*{20}{c}}{ - 1}&5&3\\{ - 4}&{23}&{12}\\1&{ - 11}&{ - 6}\end{array}} \right)\end{align}

## Chapter 4 Ex.4.5 Question 10

Find the inverse of each of the matrix $$\left( {\begin{array}{*{20}{c}}1&{ - 1}&2\\0&2&{ - 3}\\3&{ - 2}&4\end{array}} \right)$$ (if it exists)

### Solution

Let $$A = \left( {\begin{array}{*{20}{c}}1&{ - 1}&2\\0&2&{ - 3}\\3&{ - 2}&4\end{array}} \right)$$

Then, expanding along $${C_1}$$,

\begin{align}\left| A \right| &= 1\left( {8 - 6} \right) - 0 + 3\left( {3 - 4} \right) = 2 - 3\\ &= - 1\end{align}

Now,

\begin{align}{A_{11}} &= 2\;\;\;\;\;\;\;\;\;\;\;{A_{12}} = - 9\;\;\;\;\;\;\;\;\;\;\;{A_{13}} = - 6\\{A_{21}} &= 0\;\;\;\;\;\;\;\;\;\;\;{A_{22}} = - 2\;\;\;\;\;\;\;\;\;\;\;{A_{23}} = - 1\\{A_{31}}& = - 1\;\;\;\;\;\;\;\;\;{A_{32}} = 3\;\;\;\;\;\;\;\;\;\;\;{A_{33}} = 2\end{align}

Therefore,

$$adjA = \left( {\begin{array}{*{20}{c}}2&0&{ - 1}\\{ - 9}&{ - 2}&3\\&{ - 6}&{ - 1}&2\end{array}} \right)$$

Hence,

\begin{align}{A^{ - 1}}& = \frac{1}{{\left| A \right|}}adjA\\ &= - 1\left( {\begin{array}{*{20}{c}}2&0&{ - 1}\\{ - 9}&{ - 2}&3\\{ - 6}&{ - 1}&2\end{array}} \right)\\& = \left( {\begin{array}{*{20}{c}}{ - 2}&0&1\\9&2&{ - 3}\\6&1&{ - 2}\end{array}} \right)\end{align}

## Chapter 4 Ex.4.5 Question 11

Find the inverse of each of the matrix $$\left( {\begin{array}{*{20}{c}}1&0&0\\0&{\cos \alpha }&{\sin \alpha }\\0&{\sin \alpha }&{ - \cos \alpha }\end{array}} \right)$$ (if it exists)

### Solution

Let $$A = \left( {\begin{array}{*{20}{c}}1&0&0\\0&{\cos \alpha }&{\sin \alpha }\\0&{\sin \alpha }&{ - \cos \alpha }\end{array}} \right)$$

Then,

\begin{align}\left| A \right| &= 1\left( { - {{\cos }^2}\alpha - {{\sin }^2}\alpha } \right) = - \left( {{{\cos }^2}\alpha + {{\sin }^2}\alpha } \right)\\ &= - 1\end{align}

Now,

\begin{align}{A_{11}} &= - {\cos ^2}\alpha - {\sin ^2}\alpha = - 1\;\;\;\;\;{A_{12}} = 0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{A_{13}} = 0\\{A_{21}}& = 0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{A_{22}} = - \cos \alpha \;\;\;\;\;\;\;\;\;\;{A_{23}} = - \sin \alpha \\{A_{31}} &= 0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{A_{32}} = - \sin \alpha \;\;\;\;\;\;\;\;\;\;{A_{33}} = \cos \alpha \end{align}

Therefore,

$${adj} A=\left(\begin{array}{ccc}-1 & 0 & 0 \\ 0 & -\cos \alpha & -\sin \alpha \\ 0 & -\sin \alpha & \cos \alpha\end{array}\right)$$

Hence,

\begin{align}{A^{ - 1}} &= \frac{1}{{\left| A \right|}}adjA\\ &= - 1\left( {\begin{array}{*{20}{c}}{ - 1}&0&0\\0&{ - \cos \alpha }&{ - \sin \alpha }\\0&{ - \sin \alpha }&{\cos \alpha }\end{array}} \right)\\& = \left( {\begin{array}{*{20}{c}}1&0&0\\0&{\cos \alpha }&{\sin \alpha } \ \\0&{\sin \alpha }&{ - \cos \alpha }\end{array}} \right)\end{align}

## Chapter 4 Ex.4.5 Question 12

Let $$A = \left( {\begin{array}{*{20}{c}}3&7\\2&5\end{array}} \right)$$ and $$B = \left( {\begin{array}{*{20}{c}}6&8\\7&9\end{array}} \right)$$. Verify that $${\left( {AB} \right)^{ - 1}} = {B^{ - 1}}{A^{ - 1}}$$.

### Solution

Let $$A = \left( {\begin{array}{*{20}{c}}3&7\\2&5\end{array}} \right)$$

Then,

\begin{align}\left| A \right| &= 15 - 14\\& = 1\end{align}

Now,

\begin{align}{A_{11}}& = 5\;\;\;\;\;\;\;\;\;\;\;{A_{12}} = - 2\\{A_{21}} &= - 7\;\;\;\;\;\;\;\;\;\;\;{A_{22}} = 3\end{align}

Then,

$$adjA \left( {\begin{array}{*{20}{c}}5&{ - 7}\\&{ - 2}&3\end{array}} \right)$$

Therefore,

\begin{align}{A^{ - 1}}& = \frac{1}{{\left| A \right|}}adjA\\& = \left( {\begin{array}{*{20}{c}}5&{ - 7}\\{ - 2}&3\end{array}} \right)\end{align}

Now,

Let $$B = \left( {\begin{array}{*{20}{c}}6&8\\7&9\end{array}} \right)$$

Then,

\begin{align}\left| B \right|& = 54 - 56\\ &= - 2\end{align}

Now,

\begin{align}{A_{11}}& = 9\;\;\;\;\;\;\;\;\;\;\;{A_{12}} = - 7\\{A_{21}}& = - 8\;\;\;\;\;\;\;\;\;{A_{22}} = 6\end{align}

Then,

$$adjB = \left( {\begin{array}{*{20}{c}}9&{ - 8}\\{ - 7}&6\end{array}} \right)$$

Therefore,

\begin{align}{B^{ - 1}}& = \frac{1}{{\left| B \right|}}adjB\\ &= - \frac{1}{2}\left( {\begin{array}{*{20}{c}}9&{ - 8}\\{ - 7}&6\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{ - \frac{9}{2}}&4\\{\frac{7}{2}}&{ - 3}\end{array}} \right)\end{align}

Now,

\begin{align}{B^{ - 1}}{A^{ - 1}} &= \left( {\begin{array}{*{20}{c}}{ - \frac{9}{2}}&4\\{\frac{7}{2}}&{ - 3}\end{array}} \right)\left( {\begin{array}{*{20}{c}}5&{ - 7}\\{ - 2}&3\end{array}} \right)\\& = \left( {\begin{array}{*{20}{c}}{ - \frac{{45}}{2} - 8}&{\frac{{63}}{2} + 12}\\{\frac{{35}}{2} + 6}&{ - \frac{{49}}{2} - 9}\end{array}} \right)\\& = \left( {\begin{array}{*{20}{c}}{ - \frac{{61}}{2}}&{\frac{{87}}{2}}\\{\frac{{47}}{2}}&{\frac{{ - 67}}{2}}\end{array}} \right)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}

Also,

\begin{align}AB &= \left( {\begin{array}{*{20}{c}}3&7\\2&5\end{array}} \right)\left( {\begin{array}{*{20}{c}}6&8\\7&9\end{array}} \right)\\& = \left( {\begin{array}{*{20}{c}}{18 + 49}&{24 + 63}\\{12 + 35}&{16 + 45}\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{67}&{87}\\{47}&{61}\end{array}} \right)\end{align}

Then, we have

\begin{align}\left| {AB} \right|& = 67\left( {61} \right) - 87\left( {47} \right)\\& = 4087 - 4089\\ &= - 2\end{align}

Therefore,

$$adj\left( {AB} \right) = \left( {\begin{array}{*{20}{c}}{61}&{ - 87}\\&{ - 47}&{67}\end{array}} \right)$$

Thus,

\begin{align}{\left( {AB} \right)^{ - 1}} &= \frac{1}{{\left| {AB} \right|}}adj\left( {AB} \right)\\& = - \frac{1}{2}\left( {\begin{array}{*{20}{c}}{61}&{ - 87}\\{ - 47}&{67}\end{array}} \right)\\& = \left( {\begin{array}{*{20}{c}}{ - \frac{{61}}{2}}&{\frac{{87}}{2}}\\{\frac{{47}}{2}}&{ - \frac{{67}}{2}}\end{array}} \right)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

From $$\left( 1 \right)$$ and $$\left( 2 \right)$$,

$${\left( {AB} \right)^{ - 1}} = {B^{ - 1}}{A^{ - 1}}$$

Hence, proved.

## Chapter 4 Ex.4.5 Question 13

If $$A = \left( {\begin{array}{*{20}{c}}3&1\\{ - 1}&2\end{array}} \right)$$, show that $${A^2} - 5A + 7I = 0$$.

Hence find $${A^{ - 1}}$$.

### Solution

Let $$A = \left( {\begin{array}{*{20}{c}}3&1\\{ - 1}&2\end{array}} \right)$$

Therefore,

\begin{align}{A^2} = A.A &= \left( {\begin{array}{*{20}{c}}3&1\\{ - 1}&2\end{array}} \right)\left( {\begin{array}{*{20}{c}}3&1\\{ - 1}&2\end{array}} \right)\\& = \left( {\begin{array}{*{20}{c}}{9 - 1}&{3 + 2}\\{ - 3 - 2}&{ - 1 + 4}\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}8&5\\{ - 5}&3\end{array}} \right)\end{align}

Now,

\begin{align}{A^2} - 5A + 7I &= \left( {\begin{array}{*{20}{c}}8&5\\{ - 5}&3\end{array}} \right) - 5\left( {\begin{array}{*{20}{c}}3&1\\{ - 1}&2\end{array}} \right) + 7\left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)\\& = \left( {\begin{array}{*{20}{c}}8&5\\{ - 5}&3\end{array}} \right) - \left( {\begin{array}{*{20}{c}}{15}&5\\{ - 5}&{10}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}7&0\\0&7\end{array}} \right)\\& = \left( {\begin{array}{*{20}{c}}{ - 7}&0\\0&{ - 7}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}7&0\\0&7\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}0&0\\0&0\end{array}} \right)\end{align}

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

Now,

\begin{align}& \Rightarrow \; A.A - 5A = - 7I\\ &\Rightarrow \;A.A\left( {{A^{ - 1}}} \right) - 5A.{A^{ - 1}} = - 7I{A^{ - 1}}\;\;\;\;\;\;\;\;\;\;\left[ {{\text{post - multiplying by }}{A^{ - 1}}{\text{ as }}\left| A \right| \ne 0} \right]\\ &\Rightarrow\; A\left( {A{A^{ - 1}}} \right) - 5I = - 7{A^{ - 1}}\\ &\Rightarrow \;AI - 5I = - 7{A^{ - 1}}\\ &\Rightarrow {A^{ - 1}} = - \frac{1}{7}\left( {A - 5I} \right)\\ &\Rightarrow \;{A^{ - 1}} = \frac{1}{7}\left( {5I - A} \right)\\ &\Rightarrow \; {A^{ - 1}} = \frac{1}{7}\left[ {\left( {\begin{array}{*{20}{c}}5&0\\0&5\end{array}} \right) - \left( {\begin{array}{*{20}{c}}3&1\\{ - 1}&2\end{array}} \right)} \right]\\& \Rightarrow\; {A^{ - 1}} = \frac{1}{7}\left( {\begin{array}{*{20}{c}}2&{ - 1}\\1&3\end{array}} \right)\end{align}

Thus,

$${A^{ - 1}} = \frac{1}{7}\left( {\begin{array}{*{20}{c}}2&{ - 1}\\1&3\end{array}} \right)$$

## Chapter 4 Ex.4.5 Question 14

For the matrix $$A = \left( {\begin{array}{*{20}{c}}3&2\\1&1\end{array}} \right)$$, find the numbers $$a$$ and $$b$$ such that $${A^2} + aA + bI = 0$$.

### Solution

Let $$A = \left( {\begin{array}{*{20}{c}}3&2\\1&1\end{array}} \right)$$

Therefore,

\begin{align}{A^2} = A.A &= \left( {\begin{array}{*{20}{c}}3&2\\1&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}3&2\\1&1\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{9 + 2}&{6 + 2}\\{3 + 1}&{2 + 1}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{11}&8\\4&3\end{array}} \right)\end{align}

Now, $${A^2} + aA + bI = 0$$.

Hence,

\begin{align}& \Rightarrow \left( {A.A} \right){A^{ - 1}} + aA.{A^{ - 1}} + bI{A^{ - 1}} = 0\;\;\;\;\;\;\;\;\;\;\;\left[ {{\rm{post - multiplying by }}{A^{ - 1}}{\rm{ as }}\left| A \right| \ne 0} \right]\\ &\Rightarrow A\left( {A{A^{ - 1}}} \right) + aI + b\left( {I{A^{ - 1}}} \right) = 0\\& \Rightarrow AI + aI + b{A^{ - 1}} = 0\\ &\Rightarrow A + aI = - b{A^{ - 1}}\\ &\Rightarrow {A^{ - 1}} = - \frac{1}{b}\left( {A + aI} \right)\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}

Now,

\begin{align}{A^{ - 1}}& = \frac{1}{{\left| A \right|}}adjA\\& = \frac{1}{1}\left( {\begin{array}{*{20}{c}}1&{ - 2}\\{ - 1}&3\end{array}} \right)\\& = \left( {\begin{array}{*{20}{c}}1&{ - 2}\\{ - 1}&3\end{array}} \right)\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

From $$\left( 1 \right)$$and $$\left( 2 \right)$$, we have,

\begin{align} &\Rightarrow \left( {\begin{array}{*{20}{c}}1&{ - 2}\\{ - 1}&3\end{array}} \right) = \frac{1}{b}\left[ {\left( {\begin{array}{*{20}{c}}3&2\\1&1\end{array}} \right) + \left( {\begin{array}{*{20}{c}}a&0\\0&a\end{array}} \right)} \right]\\& \Rightarrow \left( {\begin{array}{*{20}{c}}1&{ - 2}\\{ - 1}&3\end{array}} \right) = - \frac{1}{b}\left( {\begin{array}{*{20}{c}}{3 + a}&2\\1&a\end{array}} \right)\\ &\Rightarrow \left( {\begin{array}{*{20}{c}}1&{ - 2}\\{ - 1}&3\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{\frac{{ - 3 - a}}{b}}&{ - \frac{2}{b}}\\{ - \frac{1}{b}}&{\frac{{ - 1 - a}}{b}}\end{array}} \right)\end{align}

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

\begin{align}& \Rightarrow - \frac{1}{b} = - 1\\& \Rightarrow b = 1\end{align}

Also,

\begin{align}& \Rightarrow \frac{{ - 3 - a}}{b} = 1\\& \Rightarrow - 3 - a = 1\\& \Rightarrow a = - 4\end{align}

Thus, $$a = - 4$$ and $$b = 1$$.

## Chapter 4 Ex.4.5 Question 15

For the matrix $$A = \left( {\begin{array}{*{20}{c}}1&1&1\\1&2&{ - 3}\\2&{ - 1}&3\end{array}} \right)$$, show that $${A^3} - 6{A^2} + 5A + 11I = 0$$. Hence, find $${A^{ - 1}}$$.

### Solution

Let $$A = \left( {\begin{array}{*{20}{c}}1&1&1\\1&2&{ - 3}\\2&{ - 1}&3\end{array}} \right)$$

Therefore,

\begin{align}{A^2} &= A.A = \left( {\begin{array}{*{20}{c}}1&1&1\\1&2&{ - 3}\\2&{ - 1}&3\end{array}} \right)\left( {\begin{array}{*{20}{c}}1&1&1\\1&2&{ - 3}\\2&{ - 1}&3\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{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{array}} \right) = \left( {\begin{array}{*{20}{c}}4&2&1\\{ - 3}&8&{ - 14}\\7&{ - 3}&{14}\end{array}} \right)\end{align}

And,

\begin{align}{A^3} = {A^2}.A& = \left( {\begin{array}{*{20}{c}}4&2&1\\{ - 3}&8&{ - 14}\\7&{ - 3}&{14}\end{array}} \right)\left( {\begin{array}{*{20}{c}}1&1&1\\1&2&{ - 3}\\2&{ - 1}&3\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{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{array}} \right) = \left( {\begin{array}{*{20}{c}}8&7&1\\{ - 23}&{27}&{ - 69}\\{32}&{ - 13}&{58}\end{array}} \right)\end{align}

Hence,

\begin{align}{A^3} - 6{A^2} + 5A + 11I &= \left( {\begin{array}{*{20}{c}}8&7&1\\{ - 23}&{27}&{ - 69}\\{32}&{ - 13}&{58}\end{array}} \right) - 6\left( {\begin{array}{*{20}{c}}4&2&1\\{ - 3}&8&{ - 14}\\7&{ - 3}&{14}\end{array}} \right)\\& \qquad + 5\left( {\begin{array}{*{20}{c}}1&1&1\\1&2&{ - 3}\\2&{ - 1}&3\end{array}} \right) + 11\left( {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right)\\\\& = \left( {\begin{array}{*{20}{c}}8&7&1\\{ - 23}&{27}&{ - 69}\\{32}&{ - 13}&{58}\end{array}} \right) - \left( {\begin{array}{*{20}{c}}{24}&{12}&6\\{ - 18}&{48}&{ - 84}\\{42}&{ - 18}&{84}\end{array}} \right)\\& \qquad + \left( {\begin{array}{*{20}{c}}5&5&5\\5&{10}&{ - 15}\\{10}&{ - 5}&{15}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{11}&0&0\\0&{11}&0\\0&0&{11}\end{array}} \right)\\\\& = \left( {\begin{array}{*{20}{c}}{24}&{12}&6\\{ - 18}&{48}&{ - 84}\\{42}&{ - 18}&{84}\end{array}} \right) - \left( {\begin{array}{*{20}{c}}{24}&{12}&6\\{ - 18}&{48}&{ - 84}\\{42}&{ - 18}&{84}\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}0&0&0\\0&0&0\\0&0&0\end{array}} \right)\\& = 0\end{align}

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

Now,

\begin{align}& \Rightarrow {A^3} - 6{A^2} + 5A + 11I = 0\\ &\Rightarrow \left( {AAA} \right){A^{ - 1}} - 6\left( {AA} \right){A^{ - 1}} + 5A{A^{ - 1}} + 11I{A^{ - 1}} = 0\\& \qquad \quad \left[ {{\text{post - multiplying by }}{A^{ - 1}}{\rm{ as }}\left| A \right| \ne 0} \right]\\\\ &\Rightarrow AA\left( {A{A^{ - 1}}} \right) - 6A\left( {A{A^{ - 1}}} \right) + 5\left( {A{A^{ - 1}}} \right) = - 11\left( {I{A^{ - 1}}} \right)\\ &\Rightarrow {A^2} - 6A + 5I = - 11{A^{ - 1}}\\& \Rightarrow {A^{ - 1}} = - \frac{1}{{11}}\left( {{A^2} - 6A + 5I} \right)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}

Now,

\begin{align}{A^2} - 6A + 5I& = \left( {\begin{array}{*{20}{c}}4&2&1\\{ - 3}&8&{ - 14}\\7&{ - 3}&{14}\end{array}} \right) - 6\left( {\begin{array}{*{20}{c}}1&1&1\\1&2&{ - 3}\\2&{ - 1}&3\end{array}} \right) + 5\left( {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}4&2&1\\{ - 3}&8&{ - 14}\\7&{ - 3}&{14}\end{array}} \right) - \left( {\begin{array}{*{20}{c}}6&6&6\\6&{12}&{ - 18}\\{12}&{ - 6}&{18}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}5&0&0\\0&5&0\\0&0&5\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}9&2&1\\{ - 3}&{13}&{ - 14}\\7&{ - 3}&{19}\end{array}} \right) - \left( {\begin{array}{*{20}{c}}6&6&6\\6&{12}&{ - 18}\\{12}&{ - 6}&{18}\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}3&{ - 4}&{ - 5}\\{ - 9}&1&4\\{ - 5}&3&1\end{array}} \right)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

From equation $$\left( 1 \right)$$ and

\begin{align}{A^{ - 1}} &= - \frac{1}{{11}}\left( {\begin{array}{*{20}{c}}3&{ - 4}&{ - 5}\\{ - 9}&1&4\\{ - 5}&3&1\end{array}} \right)\\& = \frac{1}{{11}}\left( {\begin{array}{*{20}{c}}{ - 3}&4&5\\9&{ - 1}&{ - 4}\\5&{ - 3}&{ - 1}\end{array}} \right)\end{align}

## Chapter 4 Ex.4.5 Question 16

If $$A = \left( {\begin{array}{*{20}{c}}2&{ - 1}&1\\{ - 1}&2&{ - 1}\\1&{ - 1}&2\end{array}} \right)$$, verify that $${A^3} - 6{A^2} + 9A - 4I = 0$$. Hence, find $${A^{ - 1}}$$.

### Solution

Let $$A = \left( {\begin{array}{*{20}{c}}2&{ - 1}&1\\{ - 1}&2&{ - 1}\\1&{ - 1}&2\end{array}} \right)$$

Therefore,

\begin{align}{A^2} &= A.A\\ &= \left( {\begin{array}{*{20}{c}}2&{ - 1}&1\\{ - 1}&2&{ - 1}\\1&{ - 1}&2\end{array}} \right)\left( {\begin{array}{*{20}{c}}2&{ - 1}&1\\{ - 1}&2&{ - 1}\\1&{ - 1}&2\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{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{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}6&{ - 5}&5\\{ - 5}&6&{ - 5}\\5&{ - 5}&6\end{array}} \right)\end{align}

And

\begin{align}{A^3} &= {A^2}.A\\ &= \left( {\begin{array}{*{20}{c}}6&{ - 5}&5\\{ - 5}&6&{ - 5}\\5&{ - 5}&6\end{array}} \right)\left( {\begin{array}{*{20}{c}}2&{ - 1}&1\\{ - 1}&2&{ - 1}\\1&{ - 1}&2\end{array}} \right)\\& = \left( {\begin{array}{*{20}{c}}{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{array}} \right)\\& = \left( {\begin{array}{*{20}{c}}{22}&{ - 21}&{21}\\{ - 21}&{22}&{ - 21}\\{21}&{ - 21}&{22}\end{array}} \right)\end{align}

Now,

\begin{align}{A^3} - 6{A^2} + 9A - 4I& = \left( {\begin{array}{*{20}{c}}{22}&{ - 21}&{21}\\{ - 21}&{22}&{ - 21}\\{21}&{ - 21}&{22}\end{array}} \right) - 6\left( {\begin{array}{*{20}{c}}6&{ - 5}&5\\{ - 5}&6&{ - 5}\\5&{ - 5}&6\end{array}} \right) \\& \qquad + 9\left( {\begin{array}{*{20}{c}}2&{ - 1}&1\\{ - 1}&2&{ - 1}\\1&{ - 1}&2\end{array}} \right) - 4\left( {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right)\\\\&= \left( {\begin{array}{*{20}{c}}{22}&{ - 21}&{21}\\{ - 21}&{22}&{ - 21}\\{21}&{ - 21}&{22}\end{array}} \right) - \left( {\begin{array}{*{20}{c}}{36}&{ - 30}&{30}\\{ - 30}&{36}&{ - 30}\\{30}&{ - 30}&{36}\end{array}} \right)\\& \qquad + \left( {\begin{array}{*{20}{c}}{18}&{ - 9}&9\\{ - 9}&{18}&{ - 9}\\9&{ - 9}&{18}\end{array}} \right) - \left( {\begin{array}{*{20}{c}}4&0&0\\0&4&0\\0&0&4\end{array}} \right)\\\\& = \left( {\begin{array}{*{20}{c}}{40}&{ - 30}&{30}\\{ - 30}&{40}&{ - 30}\\{30}&{ - 30}&{40}\end{array}} \right) - \left( {\begin{array}{*{20}{c}}{40}&{ - 30}&{30}\\{ - 30}&{40}&{ - 30}\\{30}&{ - 30}&{40}\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}0&0&0\\0&0&0\\0&0&0\end{array}} \right)\\ &= 0\end{align}

Thus,

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

Now,

\begin{align}& \Rightarrow {A^3} - 6{A^2} + 9A - 4I = 0\\ &\Rightarrow \left( {AAA} \right){A^{ - 1}} - 6\left( {AA} \right){A^{ - 1}} + 9A{A^{ - 1}} - 4I{A^{ - 1}} = 0\\&\qquad\left[ {{\text{post - multiplying by }}{A^{ - 1}}{\text{ as }}\left| A \right| \ne 0} \right]\\\\ &\Rightarrow AA\left( {A{A^{ - 1}}} \right) - 6A\left( {A{A^{ - 1}}} \right) + 9\left( {A{A^{ - 1}}} \right) = 4\left( {I{A^{ - 1}}} \right)\\& \Rightarrow AAI - 6AI + 9I = 4{A^{ - 1}}\\& \Rightarrow {A^2} - 6A + 9I = 4{A^{ - 1}}\\& \Rightarrow {A^{ - 1}} = \frac{1}{4}\left( {{A^2} - 6A + 9I} \right)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}

Now,

\begin{align}{A^2} - 6A + 9I& = \left( {\begin{array}{*{20}{c}}6&{ - 5}&5\\{ - 5}&6&{ - 5}\\5&{ - 5}&6\end{array}} \right) - 6\left( {\begin{array}{*{20}{c}}2&{ - 1}&1\\{ - 1}&2&{ - 1}\\1&{ - 1}&2\end{array}} \right) + 9\left( {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}6&{ - 5}&5\\{ - 5}&6&{ - 5}\\5&{ - 5}&6\end{array}} \right) - \left( {\begin{array}{*{20}{c}}{12}&{ - 6}&6\\{ - 6}&{12}&{ - 6}\\6&{ - 6}&{12}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}9&0&0\\0&9&0\\0&0&9\end{array}} \right)\\& = \left( {\begin{array}{*{20}{c}}3&1&{ - 1}\\1&3&1\\{ - 1}&1&3\end{array}} \right)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

From equations $$\left( 1 \right)$$ and $$\left( 2 \right)$$,

$${A^{ - 1}} = \frac{1}{4}\left( {\begin{array}{*{20}{c}}3&1&{ - 1}\\1&3&1\\{ - 1}&1&3\end{array}} \right)$$

## Chapter 4 Ex.4.5 Question 17

Let $$A$$be a non-singular square matrix of order $$3 \times 3$$. Then $$\left| {adjA} \right|$$ is equal to:

(A) $$\left| A \right|$$

(B) $${\left| A \right|^2}$$

(C) $${\left| A \right|^3}$$

(D) $$3\left| A \right|$$

### Solution

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

\begin{align}\left( {adjA} \right)A = \left| A \right|I\\ = \left( {\begin{array}{*{20}{c}}{\left| A \right|}&0&0\\0&{\left| A \right|}&0\\0&0&{\left| A \right|}\end{array}} \right)\\\end{align}

Therefore,

\begin{align}\left| {\left( {adjA} \right)A} \right|& = \left| {\begin{array}{*{20}{c}}{\left| A \right|}&0&0\\0&{\left| A \right|}&0\\0&0&{\left| A \right|}\end{array}} \right|\\\left| {adjA} \right|\left| A \right| &= {\left| A \right|^3}\left| {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right|\\ &= {\left| A \right|^3}I\\\left| {adjA} \right| &= {\left| A \right|^2}\end{align}

Thus, the correct option is B.

## Chapter 4 Ex.4.5 Question 18

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

(A) $$\det \left( A \right)$$

(B) $$\frac{1}{{\det \left( A \right)}}$$

(C) $$1$$

(D) $$0$$

### Solution

Since $$A$$ is an invertible matrix, $${A^{ - 1}}$$exists and $${A^{ - 1}} = \frac{1}{{\left| A \right|}}adjA$$.

As matrix $$A$$is of order $$2$$, let $$A = \left( {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right)$$

Then,

$$\left| A \right| = ad - bc$$

And

$$adjA = \left( {\begin{array}{*{20}{c}}d&{ - b}\\{ - c}&a\end{array}} \right)$$

Now,

\begin{align}{A^{ - 1}}& = \frac{1}{{\left| A \right|}}adjA\\&= \left( {\begin{array}{*{20}{c}}{\frac{d}{{\left| A \right|}}}&{\frac{{ - b}}{{\left| A \right|}}}\\{\frac{{ - c}}{{\left| A \right|}}}&{\frac{a}{{\left| A \right|}}}\end{array}} \right)\end{align}

Hence,

$$\left| {{A^{ - 1}}} \right| = \left| {\begin{array}{*{20}{c}}{\frac{d}{{\left| A \right|}}}&{\frac{{ - b}}{{\left| A \right|}}}\\{\frac{{ - c}}{{\left| A \right|}}}&{\frac{a}{{\left| A \right|}}}\end{array}} \right|$$

\begin{align}\left| {{A^{ - 1}}} \right| &= \frac{1}{{{{\left| A \right|}^2}}}\left| {\begin{array}{*{20}{c}}d&{ - b}\\{ - c}&a\end{array}} \right|\\ &= \frac{1}{{{{\left| A \right|}^2}}}\left( {ad - bc} \right)\\ &= \frac{1}{{{{\left| A \right|}^2}}}.\left| A \right|\\& = \frac{1}{{\left| A \right|}}\end{align}

Hence, $$\det \left( {{A^{ - 1}}} \right) = \frac{1}{{\det \left( A \right)}}$$

Thus, the correct option is B.

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