Question

The inverse of the matrix $\left[ \begin{matrix}
   2 & 0 & -1 \\
   5 & 1 & 0 \\
   0 & 1 & 3 \\
\end{matrix} \right]$ is
(a) $\left[ \begin{matrix}
   3 & 1 & 1 \\
   -15 & 6 & -5 \\
   5 & -2 & 2 \\
\end{matrix} \right]$
(b) $\left[ \begin{matrix}
   3 & -1 & 1 \\
   -15 & 6 & 5 \\
   5 & -2 & 2 \\
\end{matrix} \right]$
(c) $\left[ \begin{matrix}
   3 & -1 & 1 \\
   -15 & 6 & -5 \\
   5 & 2 & 2 \\
\end{matrix} \right]$
(d) None of these

Answer 1

Hint:We have a formula from which we can find the inverse of any matrix provided, the determinant of that matrix is not equal to $0$. The formula to find the inverse of the matrix $A$ is ${{A}^{-1}}=\dfrac{1}{\left| A \right|}adj\left( A \right)$ where $\left| A \right|$ is the determinant and $adj\left( A \right)$ is the adjoint of matrix $A$.

Complete step-by-step answer:
Before proceeding with the question, we must know all the formulas that will be required to solve this question.
The inverse of a matrix $A$ is given by the formula,
${{A}^{-1}}=\dfrac{1}{\left| A \right|}adj\left( A \right)..............\left( 1 \right)$
Here $\left| A \right|$ is the determinant and $adj\left( A \right)$ is the adjoint of matrix $A$.
The determinant of a matrix $A=\left( \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & i \\
\end{matrix} \right)$ is given by the formula,
$\left| A \right|=a\left( ei-fh \right)-b\left( di-fg \right)+c\left( dh-eg \right).................\left( 2 \right)$
To find the adjoint of the matrix, we will first find the cofactor matrix which is given by,
$\left( \begin{matrix}
   ei-fh & fg-di & dh-eg \\
   ch-bi & ai-cg & bg-ah \\
   bf-ce & cd-af & ae-bd \\
\end{matrix} \right).......................\left( 3 \right)$
The adjoint of the matrix can be found by taking the transpose of the cofactor matrix. So, the transpose of the matrix is given by,
$adj\left( A \right)=\left( \begin{matrix}
   ei-fh & ch-bi & bf-ce \\
   fg-di & ai-cg & cd-af \\
   dh-eg & bg-ah & ae-bd \\
\end{matrix} \right)$
In the question, we are given a matrix $\left[ \begin{matrix}
   2 & 0 & -1 \\
   5 & 1 & 0 \\
   0 & 1 & 3 \\
\end{matrix} \right]$. We have to find the inverse of this matrix.
Using formula $\left( 2 \right)$, the determinant of this matrix is,
$\begin{align}
  & \left| A \right|=2\left( 1.3-0.1 \right)-0\left( 5.3-0.0 \right)-1\left( 5.1-1.0 \right) \\
 & \Rightarrow \left| A \right|=2\left( 3 \right)-0\left( 15 \right)-1\left( 5 \right) \\
 & \Rightarrow \left| A \right|=6-5 \\
 & \Rightarrow \left| A \right|=1.............\left( 4 \right) \\
\end{align}$
The adjoint of the matrix can be found by using the steps shown in the above paragraph,
We have a matrix $\left[ \begin{matrix}
   2 & 0 & -1 \\
   5 & 1 & 0 \\
   0 & 1 & 3 \\
\end{matrix} \right]$. Using formula (3), the cofactor matrix for this matrix will be,
$\begin{align}
  & \left[ \begin{matrix}
   1.3-0.1 & 0.0-5.3 & 5.1-1.0 \\
   -1.1-0.3 & 2.3-\left( -1 \right).0 & 0.0-2.1 \\
   0.0-\left( -1 \right).1 & \left( -1 \right).5-2.0 & 2.1-0.5 \\
\end{matrix} \right] \\
 & \Rightarrow \left[ \begin{matrix}
   3 & -15 & 5 \\
   -1 & 6 & -2 \\
   1 & -5 & 2 \\
\end{matrix} \right] \\
\end{align}$
The adjoint of the matrix can be found by taking the transpose of the matrix and will be equal to,
$adj\left( A \right)=\left[ \begin{matrix}
   3 & -1 & 1 \\
   -15 & 6 & -5 \\
   5 & -2 & 2 \\
\end{matrix} \right]$
Since we have got the determinant and the adjoint of the matrix, we can now find it’s inverse. Using formula $\left( 1 \right)$, we get,
$\begin{align}
  & {{A}^{-1}}=\dfrac{1}{1}\left[ \begin{matrix}
   3 & -1 & 1 \\
   -15 & 6 & -5 \\
   5 & -2 & 2 \\
\end{matrix} \right] \\
 & \Rightarrow {{A}^{-1}}=\left[ \begin{matrix}
   3 & -1 & 1 \\
   -15 & 6 & -5 \\
   5 & -2 & 2 \\
\end{matrix} \right] \\
\end{align}$
Since none of the options are matching, hence the answer is option (d).

Note: There is a possibility that one may commit a mistake while finding the adjoint of the matrix. It is a very common mistake that one does not take the transpose of the cofactor matrix while finding the adjoint of the matrix and this leads us to an incorrect answer.