#### please solve rd sharma class 12 chapter adjoint and inverse of a matrix exercise multiple choice questions  question 3 maths textbook solution

Answer: option (d) none of these

Given: $A=\left[\begin{array}{ll} 3 & 4 \\ 2 & 4 \end{array}\right] \& B=\left[\begin{array}{cc} -2 & -2 \\ 0 & -1 \end{array}\right]$

Hint:  add the matrices A and B then find the inverse

Solution:

\begin{aligned} &A+B=\left[\begin{array}{ll} 3 & 4 \\ 2 & 4 \end{array}\right]+\left[\begin{array}{cc} -2 & -2 \\ 0 & -1 \end{array}\right] \\ &=\left[\begin{array}{ll} 1 & 2 \\ 2 & 3 \end{array}\right] \\ &|A+B|=3-4=-1 \end{aligned}

Now,   $(A+B)^{-1}=\frac{\operatorname{adj}(A+B)}{|A+B|}$

\begin{aligned} &C_{11}=(-1)^{1+1} 3=3 \\ &C_{12}=(-1)^{1+2} 2=-2 \\ &C_{21}=(-1)^{2+1} 2=-2 \\ &C_{22}=(-1)^{2+2} 1=1 \\ &\operatorname{Adj}(A+B)=\left[\begin{array}{cc} 3 & -2 \\ -2 & 1 \end{array}\right]^{T}=\left[\begin{array}{cc} 3 & -2 \\ -2 & 1 \end{array}\right] \end{aligned}

Hence,  $(A+B)^{-1}=\frac{\operatorname{adj}(A+B)}{|A+B|}$

\begin{aligned} &=\frac{1}{-1}\left[\begin{array}{cc} 3 & -2 \\ -2 & 1 \end{array}\right] \\ &=-1\left[\begin{array}{cc} 3 & -2 \\ -2 & 1 \end{array}\right] \\ &=\left[\begin{array}{cc} -3 & 2 \\ 2 & -1 \end{array}\right] \end{aligned}

We can clearly see that $(A+B)^{-1}$ exists. So, option (c) is incorrect.

Now we will check the other options.

\begin{aligned} &\mathrm{A}^{-1}=\frac{1}{4}\left[\begin{array}{cc} 4 & -2 \\ -4 & 3 \end{array}\right]^{T}=\frac{1}{4}\left[\begin{array}{cc} 4 & -4 \\ -2 & 3 \end{array}\right] \\ &\mathrm{B}^{-1}=\frac{1}{2}\left[\begin{array}{cc} -1 & 0 \\ 2 & -2 \end{array}\right]^{T}=\frac{1}{2}\left[\begin{array}{cc} -1 & 2 \\ 0 & -2 \end{array}\right] \\ &\mathrm{A}^{-1}+\mathrm{B}^{-1}=\frac{1}{4}\left[\begin{array}{cc} 4 & -2 \\ -4 & 3 \end{array}\right]+\frac{1}{2}\left[\begin{array}{cc} -1 & 2 \\ 0 & -2 \end{array}\right] \\ &=\left[\begin{array}{cc} 1 & -\frac{1}{2} \\ -1 & \frac{3}{4} \end{array}\right]+\left[\begin{array}{rc} -\frac{1}{2} & 0 \\ 1 & -1 \end{array}\right] \\ &=\left[\begin{array}{cc} \frac{1}{2} & \frac{-1}{2} \\ 0 & -\frac{1}{4} \end{array}\right] \\ &\Rightarrow(A+B)^{-1} \neq \mathrm{A}^{-1}+\mathrm{B}^{-1} \end{aligned}

This shows option (b) is incorrect.

It is not skew symmetric matrix and for this we will Show $\left ( \left ( A+B \right )^{-1} \right )^{T}\neq -\left ( A+B \right )^{-1}$

From the above calculation

\begin{aligned} &\text { L.H.S }=\left((A+B)^{-1}\right)^{T}=\left[\begin{array}{cc} -3 & 2 \\ 2 & -1 \end{array}\right]^{T}=\left[\begin{array}{cc} -3 & 2 \\ 2 & -1 \end{array}\right] \\ &\text { R.H.S }=-(A+B)^{-1}=-\left[\begin{array}{cc} -3 & 2 \\ 2 & -1 \end{array}\right]=\left[\begin{array}{cc} 3 & -2 \\ -2 & 1 \end{array}\right] \\ &\text { L.H.S } \neq \text { R.H.S } \end{aligned}

Hence option (d) is correct.