Question

If a,b,c are non-zero real numbers, then the inverse of the matrix $A=\left[\begin{array}{c}a\\ 0\\ 0\end{array}\begin{array}{c}0\\ b\\ 0\end{array}\begin{array}{c}0\\ 0\\ c\end{array}\right]$ is

• A. $\left[\begin{array}{c}{a}^{-1}\\ 0\\ 0\end{array}\begin{array}{c}0\\ b^{-1}\\ 0\end{array}\begin{array}{c}0\\ 0\\ c^{-1}\end{array}\right]$
• B. $abc\left[\begin{array}{c}{a}^{-1}\\ 0\\ 0\end{array}\begin{array}{c}0\\ b^{-1}\\ 0\end{array}\begin{array}{c}0\\ 0\\ c^{-1}\end{array}\right]$
• C. $\frac{1}{abc}\left[\begin{array}{c}1\\ 0\\ 0\end{array}\begin{array}{c}0\\ 1\\ 0\end{array}\begin{array}{c}0\\ 0\\ 1\end{array}\right]$
• D. $\frac{1}{abc}\left[\begin{array}{c}a\\ 0\\ 0\end{array}\begin{array}{c}0\\ b\\ 0\end{array}\begin{array}{c}0\\ 0\\ c\end{array}\right]$

Answer 1

The correct option is A $\left[\begin{array}{c}{a}^{-1}\\ 0\\ 0\end{array}\begin{array}{c}0\\ b^{-1}\\ 0\end{array}\begin{array}{c}0\\ 0\\ c^{-1}\end{array}\right]$

Consider

$A=\left|\begin{array}{ccc}a& 0& 0\\ 0& b& 0\\ 0& 0& c\end{array}\right|$

Firstly we find the determinant of $A$ as shown below:

$\left|A\right|=a[\left(b\right)\left(c\right)-\left(0\right)\left(0\right)]-0[\left(0\right)\left(c\right)-\left(0\right)\left(0\right)]+0[\left(0\right)\left(0\right)-\left(0\right)\left(b\right)]$

$\left|A\right|=a\left[bc\right]-0+0$

$\left|A\right|=abc$$\ne$$0$

Therefore, $A^{-1}$ exists.

Now to find $\text{adjoint}ofA$, we calculate the cofactors of A, so let Cij be cofactor of aij in A. therefore the cofactors are as shown below:

$C_{11}=\left|\begin{array}{cc}b& 0\\ 0& c\end{array}\right|=b\left(c\right)-0\left(0\right)=bc$

$\Rightarrow$$C_{11}=bc$

$C_{12}=\left|\begin{array}{cc}0& 0\\ 0& c\end{array}\right|=0\left(c\right)-0\left(0\right)=0$

$\Rightarrow$$C_{12}=0$

$C_{13}=\left|\begin{array}{cc}0& b\\ 0& 0\end{array}\right|=0\left(0\right)-0\left(b\right)=0$

$\Rightarrow$$C_{13}=0$

$C_{21}=\left|\begin{array}{cc}0& 0\\ 0& c\end{array}\right|=0\left(c\right)-0\left(0\right)=0$

$\Rightarrow$$C_{21}=0$

$C_{22}=\left|\begin{array}{cc}a& 0\\ 0& c\end{array}\right|=a\left(c\right)-0\left(0\right)=ac$

$\Rightarrow$$C_{22}=ac$

$C_{23}=\left|\begin{array}{cc}a& 0\\ 0& 0\end{array}\right|=a\left(0\right)-0\left(0\right)=0$

$\Rightarrow$$C_{23}=0$

$C_{31}=\left|\begin{array}{cc}0& 0\\ b& 0\end{array}\right|=0\left(0\right)-b\left(0\right)=0$

$\Rightarrow$$C_{31}=0$

$C_{32}=\left|\begin{array}{cc}a& 0\\ 0& 0\end{array}\right|=a\left(0\right)-0\left(0\right)=0$

$\Rightarrow$ $C_{32}=0$

$C_{33}=\left|\begin{array}{cc}a& 0\\ 0& b\end{array}\right|=a\left(b\right)-0\left(0\right)=ab$

$\Rightarrow$$C_{33}=ab$

Hence the adjoint of A is as follows:

$\text{adj}A={\left[\begin{array}{ccc}{C}_{11}& C_{12}& C_{13}\\ C_{21}& C_{22}& C_{23}\\ C_{31}& C_{32}& C_{33}\end{array}\right]}^T={\left[\begin{array}{ccc}bc& 0& 0\\ 0& ac& 0\\ 0& 0& ab\end{array}\right]}^T$

$\Rightarrow$$\text{adj}A=\left[\begin{array}{ccc}bc& 0& 0\\ 0& ac& 0\\ 0& 0& ab\end{array}\right]$

Now we find the inverse of A:

$A^{-1}=\frac{1}{\left|A\right|}\cdot \text{adj}.A$

$A^{-1}=\frac{1}{abc}\cdot \left[\begin{array}{ccc}bc& 0& 0\\ 0& ac& 0\\ 0& 0& ab\end{array}\right]$

$A^{-1}=\left[\begin{array}{ccc}\frac{bc}{abc}& 0& 0\\ 0& \frac{ac}{abc}& 0\\ 0& 0& \frac{ab}{abc}\end{array}\right]$

$A^{-1}=\left[\begin{array}{ccc}\frac{1}{a}& 0& 0\\ 0& \frac{1}{b}& 0\\ 0& 0& \frac{1}{c}\end{array}\right]$

$A^{-1}=\left[\begin{array}{ccc}{a}^{-1}& 0& 0\\ 0& b^{-1}& 0\\ 0& 0& c^{-1}\end{array}\right]$