Find the inverse of the matrix A= a & b c & 1+bc a ans show that … - Vidyadip

Question

Find the inverse of the matrix $\text{A}=\begin{bmatrix}\text{a} & \text{b} \\ \text{c} & \frac{1+\text{bc}}{\text{a}} \end{bmatrix}$ ans show that $aA^{-1} = (a^2 + bc + 1) I - aA$.

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Answer

$\text{A}=\begin{bmatrix}\text{a} & \text{b} \\ \text{c} & \frac{1+\text{bc}}{\text{a}} \end{bmatrix}$
$\Rightarrow\ |\text{A}|=(1+\text{bc})-\text{bc}=1\neq0$
$\text{A}^{-1}=\frac{1}{|\text{A}|}\text{ adj A}=\frac{1}{1}\text{A}=\begin{bmatrix}\frac{1+\text{bc}}{\text{a}} & -\text{b} \\ -\text{c} & \text{a} \end{bmatrix}$
Now $aA^{-1} = (a^2 + bc + 1)I - aA$
$\text{L.H.S}:\text{aA}^{-1}=\text{a}\begin{bmatrix}\frac{1+\text{bc}}{\text{a}} & -\text{b} \\ -\text{c} & \text{a} \end{bmatrix}$
$=\begin{bmatrix} 1+\text{bc} & -\text{ab} \\ -\text{ac} & \text{a}^2 \end{bmatrix}$
$\text{R.H.S}:(\text{a}^2+\text{bc}+1)\text{I}-\text{aA}$
$=\begin{bmatrix}\text{a}^2+\text{bc}+1 & 0 \\0 & \text{a}^2+\text{bc}+1 \end{bmatrix}-\begin{bmatrix}\text{a}^2 & \text{ab} \\ \text{ac} & 1+\text{bc} \end{bmatrix}$
$=\begin{bmatrix}1+\text{bc} & -\text{ab} \\ -\text{ac} & \text{a}^2 \end{bmatrix}$
Since, L.H.S = R.H.S
Hence proved.

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