Question

For the matrix A = \left[ {\begin{array}{*{20}{c}} a&b; \\\ c&{\dfrac{{1 + bc}}{a}} \end{array}} \right], show that $a{A^{ - 1}} = \left( {{a^2} + bc + 1} \right)I - aA$.

Answer 1

First, we will use the formula of the inverse of the matrix A = \left[ {\begin{array}{*{20}{c}} a&b; \\\ c&d; \end{array}} \right] by, {A^{ - 1}} = \dfrac{1}{{\left| A \right|}}\left[ {\begin{array}{*{20}{c}} d&{ - b} \\\ { - c}&a; \end{array}} \right], where $\left| A \right|$ is the determinant$A$. Then we will find the value of $a$, $b$, $c$ and $d$ from the given matrix $A$ and then substitute them in the formula of inverse of matrix and multiply it with $a$for left sides of the equation. Then we will substitute the value of A and I in the right hand side of the equation to show the required result.

Complete step-by-step answer:
We are given that the matrix is A = \left[ {\begin{array}{*{20}{c}} a&b; \\\ c&{\dfrac{{1 + bc}}{a}} \end{array}} \right].
We know that the inverse of the matrix A = \left[ {\begin{array}{*{20}{c}} h&i; \\\ j&k; \end{array}} \right] by using the formula, {A^{ - 1}} = \dfrac{1}{{\left| A \right|}}\left[ {\begin{array}{*{20}{c}} k&{ - i} \\\ { - j}&h; \end{array}} \right], where $\left| A \right|$ is the determinant of $A$.
Finding the value of $h$, $i$, $k$ and $l$ from the given matrix $A$, we get
$\Rightarrow h = a$
$\Rightarrow i = b$
$\Rightarrow j = c$
$\Rightarrow l = \dfrac{{1 + bc}}{a}$
Then we will compute the value of determinant of $A$ using the above values, we get

\Rightarrow \left| A \right| = \left| {\begin{array}{*{20}{c}} a&b; \\\ c&{\dfrac{{1 + bc}}{a}} \end{array}} \right| \\\ \Rightarrow \left| A \right| = a\left( {\dfrac{{1 + bc}}{a}} \right) - bc \\\ \Rightarrow \left| A \right| = 1 + bc - bc \\\ \Rightarrow \left| A \right| = 1 \\\

Substituting the above values of the determinant of A, $h$, $i$, $j$ and $k$ in the formula of inverse of matrix, we get

\Rightarrow {A^{ - 1}} = \dfrac{1}{1}\left[ {\begin{array}{*{20}{c}} {\dfrac{{1 + bc}}{a}}&{ - b} \\\ { - c}&a; \end{array}} \right] \\\ \Rightarrow {A^{ - 1}} = \left[ {\begin{array}{*{20}{c}} {\dfrac{{1 + bc}}{a}}&{ - b} \\\ { - c}&a; \end{array}} \right] \\\

Multiplying the above equation by $a$ on both sides, we get

{\dfrac{{1 + bc}}{a}}&b; \\\ c&a; \end{array}} \right]$$ Using the property in the above equation of matrix if a matrix multiplies a constant, then every element of the matrix will multiply it, we get

\Rightarrow a{A^{ - 1}} = \left[ {\begin{array}{{20}{c}}
{a\left( {\dfrac{{1 + bc}}{a}} \right)}&{ab} \\
{ac}&{{a^2}}
\end{array}} \right] \\
\Rightarrow a{A^{ - 1}} = \left[ {\begin{array}{{20}{c}}
{1 + bc}&{ab} \\
{ac}&{{a^2}}
\end{array}} \right]{\text{ ......eq.(1)}} \\

Now we will substitute the value of matrix A and identity matrix in the equation$$\left( {{a^2} + bc + 1} \right)I - aA$$, we get

\Rightarrow \left( {{a^2} + bc + 1} \right)\left[ {\begin{array}{{20}{c}}
1&0 \\
0&1
\end{array}} \right] - a\left[ {\begin{array}{{20}{c}}
a&b; \\
c&{\dfrac{{1 + bc}}{a}}
\end{array}} \right] \\
\Rightarrow \left[ {\begin{array}{{20}{c}}
{\left( {{a^2} + bc + 1} \right)}&0 \\
0&{\left( {{a^2} + bc + 1} \right)}
\end{array}} \right] - \left[ {\begin{array}{{20}{c}}
{{a^2}}&{ab} \\
{ac}&{a\left( {\dfrac{{1 + bc}}{a}} \right)}
\end{array}} \right] \\
\Rightarrow \left[ {\begin{array}{{20}{c}}
{\left( {{a^2} + bc + 1} \right)}&0 \\
0&{\left( {{a^2} + bc + 1} \right)}
\end{array}} \right] - \left[ {\begin{array}{{20}{c}}
{{a^2}}&{ab} \\
{ac}&{1 + bc}
\end{array}} \right] \\
\Rightarrow \left[ {\begin{array}{{20}{c}}
{{a^2} + bc + 1 - {a^2}}&{0 - ab} \\
{0 - ac}&{{a^2} + bc + 1 - bc - 1}
\end{array}} \right] \\
\Rightarrow \left[ {\begin{array}{{20}{c}}
{bc + 1}&{ - ab} \\
{ - ac}&{{a^2}}
\end{array}} \right]{\text{ ......eq.(2)}} \\

From equation (1) and equation (2), we get $$a{A^{ - 1}} = \left( {{a^2} + bc + 1} \right)I - aA$$ Hence, proved. **Note:** In these types of questions, the key concept is to find the inverse by putting the values in the formula of inverse. Students should know that the matrix $$\left[ {\begin{array}{*{20}{c}} d&{ - b} \\\ { - c}&a; \end{array}} \right]$$ is the adjoint matrix of $$A$$. We can remember this matrix to save some time in the $$2 \times 2$$ matrix but we have to compute the value of $$adjA$$ in the matrix with more than 2 rows and 2 columns. When students know the formula of inverse, the solution is very simple and easy. We need to know that an identity matrix I is a $$2 \times 2$$ matrix $$\left[ {\begin{array}{*{20}{c}} 1&0 \\\ 0&1 \end{array}} \right]$$.