Exercise 1.5

Solution:

(i) A = $\left[ \begin{matrix} -1 & 1 \\ 2 & 0 \\ \end{matrix} \right]$

Determinant of matrix A is calculated as:

det A = $\left| \text{A} \right|$

= $\left| \begin{matrix} -1 & 1 \\ 2 & 0 \\ \end{matrix} \right|$

= $\left( -1 \right)\times 0-2\times 1$

= $0-2$

= $-2$

 

Therefore,

$\left| \text{A} \right|$ = $-2$

 

(ii) B = $\left[ \begin{matrix} 1 & 3 \\ 2 & -2 \\ \end{matrix} \right]$

Determinant of matrix B is calculated as:

det B = $\left| \text{B} \right|$

= $\left| \begin{matrix} 1 & 3 \\ 2 & -2 \\ \end{matrix} \right|$

= $1\times \left( -2 \right)-3\times 2$

= $-2-6$

= $-8$

 

Therefore,

$\left| \text{B} \right|$ = $-8$

 

(iii) C = $\left[ \begin{matrix} 3 & 2 \\ 3 & 2 \\ \end{matrix} \right]$

Determinant of matrix C is calculated as:

det C = $\left| \text{C} \right|$

= $\left| \begin{matrix} 3 & 2 \\ 3 & 2 \\ \end{matrix} \right|$

= $3 \times 2 - 2 \times 3$

= $6 - 6$

 

Therefore, $\left| \text{C} \right|$ = $0$

 

(iv) D = $\left[ \begin{matrix} 3 & 2 \\ 1 & 4 \\ \end{matrix} \right]$

Determinant of matrix D is calculated as:

det D = $\left| \text{D} \right|$

= $\left| \begin{matrix} 3 & 2 \\ 1 & 4 \\ \end{matrix} \right|$

= $3 \times 4 - 2 \times 1$

= $12 - 2$

 

Therefore,

$\left| \text{D} \right|$ = $10$

Solution:

A matrix is said to be singular if its determinant is equal to zero. i.e., $\left| \text{A} \right|=0$

 

(i) A = $\left[ \begin{matrix} 3 & 6 \\ 2 & 4 \\ \end{matrix} \right]$

Determinant of matrix A is calculated as:

det A = $\left| \text{A} \right|$

= $\left| \begin{matrix} 3 & 6 \\ 2 & 4 \\ \end{matrix} \right|$

= $3 \times 4 - 2 \times 6$

= $12 – 12$

= $0$

 

 

Therefore, $\left| \text{A} \right|$ = 0

As determinant of A is equal to zero so, A is a singular matrix.

 

(ii) B = $\left[ \begin{matrix} 4 & 1 \\ 3 & 2 \\ \end{matrix} \right]$

Determinant of matrix B is calculated as:

det B = $\left| \text{B} \right|$

= $\left| \begin{matrix} 4 & 1 \\ 3 & 2 \\ \end{matrix} \right|$

= $4 \times 2 - 1 \times 3$

= $8 - 3$

= $5$

 

Therefore, $\left| \text{B} \right|$ $\ne $ $0$

As determinant of B is not equal to zero, so B is a non-singular matrix.

 

(iii) C = $\left[ \begin{matrix} 7 & -9 \\ 3 & 5 \\ \end{matrix} \right]$

Determinant of matrix C is calculated as:

det C = $\left| \text{C} \right|$

= $\left| \begin{matrix} 7 & -9 \\ 3 & 5 \\ \end{matrix} \right|$

= $7 \times 5 - 3 \times (-9)$

= $35 + 27$

= $62$

 

Therefore, $\left| \text{C} \right|$ $~\ne $ $0$

As determinant of C is not equal to zero, so C is a non-singular matrix.

 

(iv) D = $\left[ \begin{matrix} 5 & -10 \\ -2 & 4 \\ \end{matrix} \right]$

Determinant of matrix D is calculated as:

det D = $\left| \text{D} \right|$

= $\left[ \begin{matrix} 5 & -10 \\ -2 & 4 \\ \end{matrix} \right]$

= $5 \times 4 - (-10) \times (-2)$

= $20 - 20$

= $0$

 

Therefore, $\left| \text{D} \right|$ = $0$

As determinant of D is equal to zero so, D is a singular matrix.

Solution:

(i) A = $\left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]$

The multiplicative inverse of matrix A is calculated as:

${{\text{A}}^{-1}}$ = \[\frac{Adj~\left( \text{A} \right)}{\left| \text{A} \right|}\]

 

$\text{A}$ = $\left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]$

$Adj\text{ }\!\!~\!\!\text{ }\left( \text{A} \right)$ = $\left[ \begin{matrix} 0 & -3 \\ -2 & -1 \\ \end{matrix} \right]$

 

$\left| \text{A} \right|$ = $\left| \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right|$

$\left| \text{A} \right|$ = $\left( -1 \right)\times 0-2\times 3$

$\left| \text{A} \right|$ = $0-6$

$\left| \text{A} \right|$ = $-6\ne 0$

Since A is a non-singular matrix therefore solution is possible

 

${{\text{A}}^{-1}}$ = $\frac{\left[ \begin{matrix} 0 & -3 \\ -2 & -1 \\ \end{matrix} \right]}{-6}$

= $\left[ \begin{matrix} \frac{0}{-6} & \frac{-3}{-6} \\ \frac{-2}{-6} & \frac{-1}{-6} \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 0 & \frac{1}{2} \\ \frac{1}{3} & \frac{1}{6} \\ \end{matrix} \right]$

${{\text{A}}^{-1}}$ = $\left[ \begin{matrix} 0 & \frac{1}{2} \\ \frac{1}{3} & \frac{1}{6} \\ \end{matrix} \right]$

 

(ii) B = $\left[ \begin{matrix} 1 & 2 \\ -3 & -5 \\ \end{matrix} \right]$

The multiplicative inverse of matrix B is calculated as:

${{\text{B}}^{-1}}$ = $\frac{Adj~\text{B}}{\left| \text{B} \right|}$

 

$\text{B}$ = $\left[ \begin{matrix} 1 & 2 \\ -3 & -5 \\ \end{matrix} \right]$

$Adj\text{ }\!\!~\!\!\text{ B}$ = $\left[ \begin{matrix} -5 & -2 \\ 3 & 1 \\ \end{matrix} \right]$

 

$\left| B \right|$ = $\left| \begin{matrix} 1 & 2 \\ -3 & -5 \\ \end{matrix} \right|$

$\left| B \right|$ = $1\times \left( -5 \right)-2\times \left( -3 \right)$

$\left| B \right|$ = $-5+6$

$\left| B \right|$ = $1\ne 0$

Since B is a non-singular matrix therefore solution is possible

 

${{\text{B}}^{-1}}$ = $\frac{\left[ \begin{matrix} -5 & -2 \\ 3 & 1 \\ \end{matrix} \right]}{1}$

= $\left[ \begin{matrix} -5 & -2 \\ 3 & 1 \\ \end{matrix} \right]$

${{\text{B}}^{-1}}$ = $\left[ \begin{matrix} -5 & -2 \\ 3 & 1 \\ \end{matrix} \right]$

 

(iii) C = $\left[ \begin{matrix} -2 & 6 \\ 3 & -9 \\ \end{matrix} \right]$

The multiplicative inverse of matrix C is calculated as:

${{\text{C}}^{-1}}$ = $\frac{Adj~\text{C}}{\left| \text{C} \right|}$

 

$\text{C}$ = $\left[ \begin{matrix} -2 & 6 \\ 3 & -9 \\ \end{matrix} \right]$

$Adj~\text{C}$ = $\left[ \begin{matrix} -9 & -6 \\ -3 & -2 \\ \end{matrix} \right]$

 

$\left| C \right|$ = $\left| \begin{matrix} -2 & 6 \\ 3 & -9 \\ \end{matrix} \right|$

$\left| C \right|$ = $\left( -2 \right)\times \left( -9 \right)-6\times 3$

$\left| C \right|$ = $18-18$

$\left| C \right|$ = $0$

Since C is a singular matrix therefore solution is not possible.

 

Therefore, ${{\text{C}}^{-1}}$ does not exist

 

(iv) D =$\left[ \begin{matrix} \frac{1}{2} & \frac{3}{4} \\ 1 & 2 \\ \end{matrix} \right]$

The multiplicative inverse of matrix D is calculated as:

${{\text{D}}^{-1}}$ = $\frac{Adj~\text{D}}{\left| \text{D} \right|}$

 

$\text{D}$ =$\left[ \begin{matrix} \frac{1}{2} & \frac{3}{4} \\ 1 & 2 \\ \end{matrix} \right]$

$Adj\text{ }\!\!~\!\!\text{ D}$ = $\left[ \begin{matrix} 2 & \frac{-3}{4} \\ -1 & \frac{1}{2} \\ \end{matrix} \right]$

 

$\left| D \right|$ = $\left| \begin{matrix} \frac{1}{2} & \frac{3}{4} \\ 1 & 2 \\ \end{matrix} \right|$

$\left| D \right|$ = $\frac{1}{2}\times 2-\frac{3}{4}\times 1$

$\left| D \right|$ = $1-\frac{3}{4}$

$\left| D \right|$ = $\frac{4-3}{4}$

$\left| D \right|$ = $\frac{1}{4}\ne 0$

Since D is a non-singular matrix therefore solution is possible.

 

${{\text{D}}^{-1}}$ = $\frac{\left[ \begin{matrix} 2 & \frac{-3}{4} \\ -1 & \frac{1}{2} \\ \end{matrix} \right]}{\frac{1}{4}}$

= $4\times \left[ \begin{matrix} 2 & \frac{-3}{4} \\ -1 & \frac{1}{2} \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 2~\times 4 & \frac{-3}{4}~\times 4 \\ -1~\times 4 & \frac{1}{2}~\times 4 \\ \end{matrix} \right]$

${{\text{D}}^{-1}}$ = $\left[ \begin{matrix} 8 & -3 \\ -4 & 2 \\ \end{matrix} \right]$

Solution:

(i) $\text{A}\left( \text{Adj }\!\!~\!\!\text{ A} \right)\text{= }\!\!~\!\!\text{ }\left( \text{Adj }\!\!~\!\!\text{ A} \right)\text{ }\!\!~\!\!\text{ A=}\left( \text{det }\!\!~\!\!\text{ A} \right)\text{ }\!\!~\!\!\text{ I}$

Matrix A is given as

A = $\left[ \begin{matrix} 1 & 2 \\ 4 & 6 \\ \end{matrix} \right]$

 

Adj A = $\left[ \begin{matrix} 6 & -2 \\ -4 & 1 \\ \end{matrix} \right]$

 

det A = $\left| \text{A} \right|$

= $\left| \begin{matrix} 1 & 2 \\ 4 & 6 \\ \end{matrix} \right|$

= $1\times 6-2\times 4$

= $6-8$

$\left| \text{A} \right|$ = $-2$

 

Now

$\text{A}\left( Adj\text{ }\!\!~\!\!\text{ A} \right)$

= $\left[ \begin{matrix} 1 & 2 \\ 4 & 6 \\ \end{matrix} \right]\times \left[ \begin{matrix} 6 & -2 \\ -4 & 1 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1~\times 6+2~\times \left( -4 \right) & 1~\times \left( -2 \right)+2~\times 1 \\ 4~\times 6+6~\times \left( -4 \right) & 4\times \left( -2 \right)+6~\times 1 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 6-8 & -2+2 \\ 24-24 & -8+6 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -2 & 0 \\ 0 & -2 \\ \end{matrix} \right] \quad\quad\quad\quad\quad...~(1)$ 

 

$\left( Adj\text{ }\!\!~\!\!\text{ A} \right)\text{ }\!\!~\!\!\text{ A}$

= $\left[ \begin{matrix} 6 & -2 \\ -4 & 1 \\ \end{matrix} \right]\times \left[ \begin{matrix} 1 & 2 \\ 4 & 6 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 6~\times 1+\left( -2 \right)~\times 4 & 6~\times 2+\left( -2 \right)~\times 6 \\ -4~\times 1+1~\times 4 & -4~\times 2+1~\times 6 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 6-8 & 12-12 \\ -4+4 & -8+6 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -2 & 0 \\ 0 & -2 \\ \end{matrix} \right] \quad\quad\quad\quad\quad...~(2)$

 

$\left( \text{det }\!\!~\!\!\text{ A} \right)\text{ }\!\!~\!\!\text{ I}$

= $~\left( -2 \right)\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -2 & 0 \\ 0 & -2 \\ \end{matrix} \right] \quad\quad\quad\quad\quad...~(3)$

 

From $(1)$, $(2)$ and $(3)$ it is clear that:

$\text{A}\left( \text{Adj }\!\!~\!\!\text{ A} \right)\text{= }\!\!~\!\!\text{ }\left( \text{Adj }\!\!~\!\!\text{ A} \right)\text{ }\!\!~\!\!\text{ A=}\left( \text{det }\!\!~\!\!\text{ A} \right)\text{ }\!\!~\!\!\text{ I}$

 

 

(ii) $\text{B}{{\text{B}}^{-1}}=I=~{{\text{B}}^{-1}}\text{B}$

The multiplicative inverse of matrix B is calculated as:

${{\text{B}}^{-1}}$ = $\frac{Adj~\text{B}}{\left| \text{B} \right|}$

 

B = $\left[ \begin{matrix} 3 & -1 \\ 2 & -2 \\ \end{matrix} \right]$

$\text{Adj }\!\!~\!\!\text{ B}$ = $\left[ \begin{matrix} -2 & 1 \\ -2 & 3 \\ \end{matrix} \right]$

 

$\left| \text{B} \right|$ = $\left| \begin{matrix} 3 & -1 \\ 2 & -2 \\ \end{matrix} \right|$

$\left| \text{B} \right|$ = $3\times \left( -2 \right)-\left( -1 \right)\times 2$

$\left| \text{B} \right|$ = $-6+2$

$\left| \text{B} \right|$ = $-4\ne 0$

 

Since B is a non-singular matrix therefore ${{\text{B}}^{-1}}$ is possible

 

${{\text{B}}^{-1}}$ = $\frac{\left[ \begin{matrix} -2 & 1 \\ -2 & 3 \\ \end{matrix} \right]}{-4}$

= $\left[ \begin{matrix} \frac{-2}{-4} & \frac{1}{-4} \\ \frac{-2}{-4} & \frac{3}{-4} \\ \end{matrix} \right]$

${{\text{B}}^{-1}}$ = $\left[ \begin{matrix} \frac{1}{2} & -\frac{1}{4} \\ \frac{1}{2} & -\frac{3}{4} \\ \end{matrix} \right]$

 

Now

$\text{B}{{\text{B}}^{-1}}$

= $\left[ \begin{matrix} 3 & -1 \\ 2 & -2 \\ \end{matrix} \right]\times \left[ \begin{matrix} \frac{1}{2} & \frac{-1}{4} \\ \frac{1}{2} & \frac{-3}{4} \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 3\times \left( \frac{1}{2} \right)+\left( -1 \right)\times \left( \frac{1}{2} \right) & 3~\times \left( -\frac{1}{4} \right)+\left( -1 \right)\times \left( -\frac{3}{4} \right) \\ 2~\times \left( \frac{1}{2} \right)+\left( -2 \right)\times \left( \frac{1}{2} \right) & 2\times \left( -\frac{1}{4} \right)+\left( -2 \right)\times \left( -\frac{3}{4} \right) \\ \end{matrix} \right]$

= $\left[ \begin{matrix} \frac{3}{2}-\left( \frac{1}{2} \right) & -\frac{3}{4}+\frac{3}{4} \\ 1-1 & -\frac{1}{2}+\frac{3}{2} \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]$ = $I \quad\quad\quad\quad\quad...~(1)$

 

${{\text{B}}^{-1}}\text{B}$

= $\left[ \begin{matrix} \frac{1}{2} & -\frac{1}{4} \\ \frac{1}{2} & -\frac{3}{4} \\ \end{matrix} \right]~\times \text{ }\!\!~\!\!\text{ }\left[ \begin{matrix} 3 & -1 \\ 2 & -2 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} \frac{1}{2}\times 3+\left( -\frac{1}{4} \right)\times 2 & \frac{1}{2}\times \left( -1 \right)+\left( -\frac{1}{4} \right)\times \left( -2 \right) \\ \frac{1}{2}\times 3+\left( -\frac{3}{4} \right)\times 2 & \frac{1}{2}\times \left( -1 \right)+\left( -\frac{3}{4} \right)\times \left( -2 \right) \\ \end{matrix} \right]$

= $\left[ \begin{matrix} \frac{3}{2}-\frac{1}{2} & -\frac{1}{2}+\frac{1}{2} \\ \frac{3}{2}-\frac{3}{2} & -\frac{1}{2}+\frac{3}{2} \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]$ = $I \quad\quad\quad\quad\quad...~(2)$

 

From $(1)$ and $(2)$, it is clear that:

$\text{B}{{\text{B}}^{-1}}=I=~{{\text{B}}^{-1}}\text{B}$

Solution:

Two matrices A and B are the multiplicative inverse of each other if their product is the identity matrix.

AB = $I$ = $\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]$

 

(i) $\left[ \begin{matrix} 3 & 5 \\ 4 & 7 \\ \end{matrix} \right]$ and $\left[ \begin{matrix} 7 & -5 \\ -4 & 3 \\ \end{matrix} \right]$

Let A = $\left[ \begin{matrix} 3 & 5 \\ 4 & 7 \\ \end{matrix} \right]$ and B = $\left[ \begin{matrix} 7 & -5 \\ -4 & 3 \\ \end{matrix} \right]$

 

AB = $\left[ \begin{matrix} 3 & 5 \\ 4 & 7 \\ \end{matrix} \right]\left[ \begin{matrix} 7 & -5 \\ -4 & 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 3\times 7+5~\times \left( -4 \right) & 3\times \left( -5 \right)+5~\times 3 \\ 4\times 7+7~\times \left( -4 \right) & 4\times \left( -5 \right)+7~\times 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 21-20 & -15+15 \\ 28-28 & -20+21 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]$ = $I$

 

Hence the given matrices are multiplicative inverses of each other.

 

(ii) $\left[ \begin{matrix} 1 & 2 \\ 2 & 3 \\ \end{matrix} \right]$ and $\left[ \begin{matrix} -3 & 2 \\ 2 & -1 \\ \end{matrix} \right]$

Let A = $\left[ \begin{matrix} 1 & 2 \\ 2 & 3 \\ \end{matrix} \right]$ and B = $\left[ \begin{matrix} -3 & 2 \\ 2 & -1 \\ \end{matrix} \right]$

 

AB = $\left[ \begin{matrix} 1 & 2 \\ 2 & 3 \\ \end{matrix} \right]\left[ \begin{matrix} -3 & 2 \\ 2 & -1 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1\times \left( -3 \right)+2~\times 2 & 1\times 2+2~\times \left( -1 \right) \\ 2\times \left( -3 \right)+3~\times 2 & 2\times 2+3\times \left( -1 \right) \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -3+4 & 2-2 \\ -6+6 & 4-3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]$ = $I$

 

Hence the given matrices are multiplicative inverses of each other.

Solution:

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

Solving Left Hand Side: ${{\left( \text{AB} \right)}^{-1}}$

$(AB)$ = $\left( \left[ \begin{matrix} 4 & 0 \\ -1 & 2 \\ \end{matrix} \right]\text{ }\!\!~\!\!\text{ }\times \text{ }\!\!~\!\!\text{ }\left[ \begin{matrix} -4 & -2 \\ 1 & -1 \\ \end{matrix} \right] \right)$

= $\left[ \begin{matrix} 4\times \left( -4 \right)+0\text{ }\!\!~\!\!\text{ }\times 1 & 4\times \left( -2 \right)+0\text{ }\!\!~\!\!\text{ }\times \left( -1 \right) \\ -1\times \left( -4 \right)+2\text{ }\!\!~\!\!\text{ }\times 1 & \left( -1 \right)\times \left( -2 \right)+2\text{ }\!\!~\!\!\text{ }\times \left( -1 \right) \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -16+0 & -8+0 \\ 4+2 & 2-2 \\ \end{matrix} \right]$

$(AB)$ = $\left[ \begin{matrix} -16 & -8 \\ 6 & 0 \\ \end{matrix} \right]$

 

The multiplicative inverse of $(AB)$ is calculated as:

${{\left( \text{AB} \right)}^{-1}}$= $\frac{Adj~\left( \text{AB} \right)}{\left| \text{AB} \right|}$

 

$(AB)$ = $\left[ \begin{matrix} -16 & -8 \\ 6 & 0 \\ \end{matrix} \right]$

$Adj\text{ }\!\!~\!\!\text{ }\left( \text{AB} \right)$ = $\left[ \begin{matrix} 0 & 8 \\ -6 & -16 \\ \end{matrix} \right]$

 

$\left| \text{AB} \right|$ = $\left| \begin{matrix} -16 & -8 \\ 6 & 0 \\ \end{matrix} \right|$

= $\left( -16 \right)\times 0-\left( -8 \right)\times 6$

= $0+48$

$\left|\text{AB} \right|$ = $48\ne 0$

Since $(AB)$ is non-singular therefore ${{\left( \text{AB} \right)}^{-1}}$ is possible

 

${{\left( \text{AB} \right)}^{-1}}$ = $\frac{\left[ \begin{matrix} 0 & 8 \\ -6 & -16 \\ \end{matrix} \right]}{48}$

= $\left[ \begin{matrix} \frac{0}{48} & \frac{8}{48} \\ \frac{-6}{48} & \frac{-16}{48} \\ \end{matrix} \right]$

${{\left( \text{AB} \right)}^{-1}}$ = $\left[ \begin{matrix} 0 & \frac{1}{6} \\ -\frac{1}{8} & -\frac{1}{3} \\ \end{matrix} \right] \quad\quad\quad\quad\quad ...~(1)$

 

Solving Right Hand Side: ${{\text{B}}^{-1}}{{\text{A}}^{-1}}$

The multiplicative inverse of matrix B is calculated as:

${{\text{B}}^{-1}}$ = $\frac{Adj~B}{\left| \text{B} \right|}$

 

B = $\left[ \begin{matrix} -4 & -2 \\ 1 & -1 \\ \end{matrix} \right]$

$\text{Adj }\!\!~\!\!\text{ B}$ = $\left[ \begin{matrix} -1 & 2 \\ -1 & -4 \\ \end{matrix} \right]$

 

$\left| \text{B} \right|$ = $\left| \begin{matrix} -4 & -2 \\ 1 & -1 \\ \end{matrix} \right|$

= $\left( -4 \right)\times \left( -1 \right)-\left( -2 \right)\times 1$

= $4+2$

$\left| \text{B} \right|$ = $6\ne 0$

Since B is non-singular therefore ${{\text{B}}^{-1}}$ is possible

 

${{\text{B}}^{-1}}$ = $\frac{\left[ \begin{matrix} -1 & 2 \\ -1 & -4 \\ \end{matrix} \right]}{6}$

= $\left[ \begin{matrix} \frac{-1}{6} & \frac{2}{6} \\ \frac{-1}{6} & \frac{-4}{6} \\ \end{matrix} \right]$

${{\text{B}}^{-1}}$ = $\left[ \begin{matrix} \frac{-1}{6} & \frac{1}{3} \\ \frac{-1}{6} & \frac{-2}{3} \\ \end{matrix} \right]$ 

 

Similarly, the multiplicative inverse of matrix A is calculated as:
${{\text{A}}^{-1}}$ = $\frac{Adj~\text{A}}{\left| \text{A} \right|}$

 

A = $\left[ \begin{matrix} 4 & 0 \\ -1 & 2 \\ \end{matrix} \right]$

$\text{Adj }\!\!~\!\!\text{ A}$ = $\left[ \begin{matrix} 2 & -0 \\ 1 & 4 \\ \end{matrix} \right]$

 

$\left| \text{A} \right|$ = $\left| \begin{matrix} 4 & 0 \\ -1 & 2 \\ \end{matrix} \right|$

= $4\times 2-0\times \left( -1 \right)$

= $8+0$

$\left| \text{A} \right|$ = $8\ne 0$

Since A is non-singular therefore ${{\text{A}}^{-1}}$ is possible

 

${{\text{A}}^{-1}}$ = $\frac{\left[ \begin{matrix} 2 & -0 \\ 1 & 4 \\ \end{matrix} \right]}{8}$

= $\left[ \begin{matrix} \frac{2}{8} & \frac{0}{8} \\ \frac{1}{8} & \frac{4}{8} \\ \end{matrix} \right]$

${{\text{A}}^{-1}}$ = $\left[ \begin{matrix} \frac{1}{4} & 0 \\ \frac{1}{8} & \frac{1}{2} \\ \end{matrix} \right]$ 

 

Now

${{\text{B}}^{-1}}{{\text{A}}^{-1}}$ = $\left[ \begin{matrix} \frac{-1}{6} & \frac{1}{3} \\ \frac{-1}{6} & \frac{-2}{3} \\ \end{matrix} \right]\times \left[ \begin{matrix} \frac{1}{4} & 0 \\ \frac{1}{8} & \frac{1}{2} \\ \end{matrix} \right]$

= \[\left[ \begin{matrix} \left( \frac{-1}{6}\times \frac{1}{4} \right)+\left( \frac{1}{3}\times \frac{1}{8} \right) & \left( \frac{-1}{6}\times 0 \right)+\left( \frac{1}{3}\times \frac{1}{2} \right) \\ \left( \frac{-1}{6}\times \frac{1}{4} \right)+\left( \frac{-2}{3}\times \frac{1}{8} \right) & \left( \frac{-1}{6}\times 0 \right)+\left( \frac{-2}{3}\times \frac{1}{2} \right) \\ \end{matrix} \right]\]

= $\left[ \begin{matrix} -\frac{1}{24}+\frac{1}{24} & 0+\frac{1}{6} \\ -\frac{1}{24}-\frac{2}{24} & 0-\frac{2}{6} \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 0 & \frac{1}{6} \\ \frac{-1-2}{24} & -\frac{2}{6} \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 0 & \frac{1}{6} \\ \frac{-3}{24} & -\frac{2}{6} \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 0 & \frac{1}{6} \\ -\frac{1}{8} & -\frac{1}{3} \\ \end{matrix} \right] \quad\quad\quad\quad\quad ...~(2)$

 

From $(1)$ and $(2)$, it is clear that:

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

 

 

(ii) ${{\left( \text{DA} \right)}^{-1}}$ = ${{\text{A}}^{-1}}{{\text{D}}^{-1}}$

Solving Left Hand Side: ${{\left( \text{DA} \right)}^{-1}}$

$\left( \text{DA} \right)$ = $\left[ \begin{matrix} 3 & 1 \\ -2 & 2 \\ \end{matrix} \right]\text{ }\!\!~\!\!\text{ }\times \text{ }\!\!~\!\!\text{ }\left[ \begin{matrix} 4 & 0 \\ -1 & 2 \\ \end{matrix} \right]\text{ }\!\!~\!\!\text{ }$

= $\left[ \begin{matrix} 3\times 4+1\text{ }\!\!~\!\!\text{ }\times \left( -1 \right) & 3\times 0+1\text{ }\!\!~\!\!\text{ }\times 2 \\ -2\times 4+2\text{ }\!\!~\!\!\text{ }\times \left( -1 \right) & -2\times 0+2\text{ }\!\!~\!\!\text{ }\times 2 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 12-1 & 0+2 \\ -8-2 & 0+4 \\ \end{matrix} \right]$

$\left( \text{DA} \right)$ = $\left[ \begin{matrix} 11 & 2 \\ -10 & 4 \\ \end{matrix} \right]$

 

The multiplicative inverse of $\left( \text{DA} \right)$ is calculated as:

${{\left( \text{DA} \right)}^{-1}}$= $\frac{Adj\text{ }\!\!~\!\!\text{ }\left( \text{DA} \right)}{\left| \text{DA} \right|}$

 

$\left( \text{DA} \right)$ = $\left[ \begin{matrix} 11 & 2 \\ -10 & 4 \\ \end{matrix} \right]$

$\text{Adj }\!\!~\!\!\text{ }\left( \text{DA} \right)$ = $\left[ \begin{matrix} 4 & -2 \\ 10 & 11 \\ \end{matrix} \right]$

 

$\left| \text{DA} \right|$ = $\left| \begin{matrix} 11 & 2 \\ -10 & 4 \\ \end{matrix} \right|$

= $11\times 4-2\times \left( -10 \right)$

= $44+20$

$\left| \text{DA} \right|$ = $64\ne 0$

Since $\left( \text{DA} \right)$ is non-singular therefore ${{\left( \text{DA} \right)}^{-1}}$ is possible

 

${{\left( \text{DA} \right)}^{-1}}$ = $\frac{\left[ \begin{matrix} 4 & -2 \\ 10 & 11 \\ \end{matrix} \right]}{64}$

= $\left[ \begin{matrix} \frac{4}{64} & \frac{-2}{64} \\ \frac{10}{64} & \frac{11}{64} \\ \end{matrix} \right]$

= $\left[ \begin{matrix} \frac{1}{16} & \frac{-1}{32} \\ \frac{5}{32} & \frac{11}{64} \\ \end{matrix} \right]$

${{\left( \text{DA} \right)}^{-1}}$ = $\left[ \begin{matrix} \frac{1}{16} & \frac{-1}{32} \\ \frac{5}{32} & \frac{11}{64} \\ \end{matrix} \right] \quad\quad\quad\quad\quad ...~(1)$

 

Solving Right Hand Side: ${{\text{A}}^{-1}}{{\text{D}}^{-1}}$

The multiplicative inverse of matrix A is calculated as:

${{\text{A}}^{-1}}$ = $\frac{Adj~\text{A}}{\left| \text{A} \right|}$

 

A = $\left[ \begin{matrix} 4 & 0 \\ -1 & 2 \\ \end{matrix} \right]$

$\text{Adj }\!\!~\!\!\text{ A}$ = $\left[ \begin{matrix} 2 & -0 \\ 1 & 4 \\ \end{matrix} \right]$

 

$\left| \text{A} \right|$ = $\left| \begin{matrix} 4 & 0 \\ -1 & 2 \\ \end{matrix} \right|$

= $4\times 2-0\times \left( -1 \right)$

= $8+0$

$\left| \text{A} \right|$ = $8\ne 0$

Since A is non-singular therefore ${{\text{A}}^{-1}}$ is possible

 

${{\text{A}}^{-1}}$ = $\frac{\left[ \begin{matrix} 2 & -0 \\ 1 & 4 \\ \end{matrix} \right]}{8}$

= $\left[ \begin{matrix} \frac{2}{8} & \frac{0}{8} \\ \frac{1}{8} & \frac{4}{8} \\ \end{matrix} \right]$

${{\text{A}}^{-1}}$ = $\left[ \begin{matrix} \frac{1}{4} & 0 \\ \frac{1}{8} & \frac{1}{2} \\ \end{matrix} \right]$

 

Similarly, the multiplicative inverse of matrix D is calculated as:

${{\text{D}}^{-1}}$ = $\frac{Adj~\text{D}}{\left| \text{D} \right|}$

 

D = $\left[ \begin{matrix} 3 & 1 \\ -2 & 2 \\ \end{matrix} \right]$

$\text{Adj }\!\!~\!\!\text{ D}$ = $\left[ \begin{matrix} 2 & -1 \\ 2 & 3 \\ \end{matrix} \right]$

 

$\left| \text{D} \right|$ = $\left| \begin{matrix} 3 & 1 \\ -2 & 2 \\ \end{matrix} \right|$

= $3\times 2-1\times \left( -2 \right)$

= $6+2$

$\left| \text{D} \right|$ = \[8\ne 0\]

Since D is non-singular therefore ${{\text{D}}^{-1}}$ is possible

 

${{\text{D}}^{-1}}$ = \[\frac{\left[ \begin{matrix} 2 & -1 \\ 2 & 3 \\ \end{matrix} \right]}{8}\]

= $\left[ \begin{matrix} \frac{2}{8} & -\frac{1}{8} \\ \frac{2}{8} & \frac{3}{8} \\ \end{matrix} \right]$

${{\text{D}}^{-1}}$ = $\left[ \begin{matrix} \frac{1}{4} & -\frac{1}{8} \\ \frac{1}{4} & \frac{3}{8} \\ \end{matrix} \right]$

 

Now

${{\text{A}}^{-1}}{{\text{D}}^{-1}}$ = $\left[ \begin{matrix} \frac{1}{4} & 0 \\ \frac{1}{8} & \frac{1}{2} \\ \end{matrix} \right]\times \left[ \begin{matrix} \frac{1}{4} & \frac{-1}{8} \\ \frac{1}{4} & \frac{3}{8} \\ \end{matrix} \right]$

= $\left[ \begin{matrix} \left( \frac{1}{4}\times \frac{1}{4} \right)+\left( 0\times \frac{1}{4} \right) & \left( \frac{1}{4}\times \frac{-1}{8} \right)+\left( 0\times \frac{3}{8} \right) \\ \left( \frac{1}{8}\times \frac{1}{4} \right)+\left( \frac{1}{2}\times \frac{1}{4} \right) & \left( \frac{1}{8}\times \frac{-1}{8} \right)+\left( \frac{1}{2}\times \frac{3}{8} \right) \\ \end{matrix} \right]$

= $\left[ \begin{matrix} \frac{1}{16}+0 & -\frac{1}{32}+0 \\ \frac{1}{32}+\frac{1}{8} & -\frac{1}{64}+\frac{3}{16} \\ \end{matrix} \right]$

= $\left[ \begin{matrix} \frac{1}{16} & -\frac{1}{32} \\ \frac{1+4}{32} & \frac{-1+12}{64} \\ \end{matrix} \right]$

${{\text{A}}^{-1}}{{\text{D}}^{-1}}$ = $\left[ \begin{matrix} \frac{1}{16} & -\frac{1}{32} \\ \frac{5}{32} & \frac{11}{64} \\ \end{matrix} \right]\quad\quad\quad\quad\quad ...~(2)$ 

 

From $(1)$ and $(2)$, it is clear that:

${{\left( \text{DA} \right)}^{-1}}$ = ${{\text{A}}^{-1}}{{\text{D}}^{-1}}$