Question

Find the inverse of the matrix (if it exists):
\[A=\left[ \begin{matrix}
   1 & 0 & 0 \\
   3 & 3 & 0 \\
   5 & 2 & -1 \\
\end{matrix} \right]\]

Answer 1

Hint: To first find that if the inverse of the matrix exists we need to find the determinant of the matrix and then we need to find the inverse by using the Adjugate Matrix Method. To find the determinant we use the formula:
\[A=\left[ \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & i \\
\end{matrix} \right]\]
Determinant \[=a\left( e\times i-f\times h \right)+b\left( d\times i-f\times g \right)-c\left( d\times h-e\times g \right)\]
And for inverse of the matrix we use the formula:
\[{{A}^{\left( -1 \right)}}=\dfrac{1}{\left| A \right|}\times \left( \begin{matrix}
   {{C}_{1,1}} & {{C}_{2,1}} & {{C}_{3,1}} \\
   {{C}_{1,2}} & {{C}_{2,2}} & {{C}_{3,2}} \\
   {{C}_{1,3}} & {{C}_{2,3}} & {{C}_{3,3}} \\
\end{matrix} \right)\]
where \[n\] is the positional value from \[0,0\] to \[2,2\] .

Complete step-by-step answer:
We use the triangle’s law as
\[{{C}_{1,1}}={{\left( -1 \right)}^{1+1}}\left[ \begin{matrix}
   1 & 0 & 0 \\
   3 & 3 & 0 \\
   5 & 2 & -1 \\
\end{matrix} \right]=1\times \left( 3\times -1+0\times 2 \right)=-3\]
\[{{C}_{1,1}}={{\left( -1 \right)}^{1+2}}\left[ \begin{matrix}
   1 & 0 & 0 \\
   3 & 3 & 0 \\
   5 & 2 & -1 \\
\end{matrix} \right]=-1\times \left( 3\times \left( -1 \right)-0\times 5 \right)=3\]
\[{{C}_{1,1}}={{\left( -1 \right)}^{1+3}}\left[ \begin{matrix}
   1 & 0 & 0 \\
   3 & 3 & 0 \\
   5 & 2 & -1 \\
\end{matrix} \right]=1\times \left( 3\times 2-3\times 5 \right)=-9\]
\[{{C}_{1,1}}={{\left( -1 \right)}^{2+1}}\left[ \begin{matrix}
   1 & 0 & 0 \\
   3 & 3 & 0 \\
   5 & 2 & -1 \\
\end{matrix} \right]=-1\times \left( 0\times \left( -1 \right)-0\times 2 \right)=0\]
\[{{C}_{1,1}}={{\left( -1 \right)}^{2+2}}\left[ \begin{matrix}
   1 & 0 & 0 \\
   3 & 3 & 0 \\
   5 & 2 & -1 \\
\end{matrix} \right]=-1\times \left( 1\times \left( -1 \right)-0\times 5 \right)=-1\]
\[{{C}_{1,1}}={{\left( -1 \right)}^{2+3}}\left[ \begin{matrix}
   1 & 0 & 0 \\
   3 & 3 & 0 \\
   5 & 2 & -1 \\
\end{matrix} \right]=-1\times \left( 1\times 2-0\times 5 \right)\]
\[{{C}_{1,1}}={{\left( -1 \right)}^{3+1}}\left[ \begin{matrix}
   1 & 0 & 0 \\
   3 & 3 & 0 \\
   5 & 2 & -1 \\
\end{matrix} \right]=-1\times \left( 0\times 0-0\times 3 \right)=0\]
\[{{C}_{1,1}}={{\left( -1 \right)}^{3+2}}\left[ \begin{matrix}
   1 & 0 & 0 \\
   3 & 3 & 0 \\
   5 & 2 & -1 \\
\end{matrix} \right]=-1\times \left( 1\times 0-0\times 3 \right)=0\]
\[{{C}_{1,1}}={{\left( -1 \right)}^{3+3}}\left[ \begin{matrix}
   1 & 0 & 0 \\
   3 & 3 & 0 \\
   5 & 2 & -1 \\
\end{matrix} \right]=1\times \left( 1\times 3-0\times 3 \right)=3\]
Now placing the values in the formula we get:
\[{{A}^{\left( -1 \right)}}=\dfrac{1}{\left| A \right|}\times \left( \begin{matrix}
   {{C}_{1,1}} & {{C}_{2,1}} & {{C}_{3,1}} \\
   {{C}_{1,2}} & {{C}_{2,2}} & {{C}_{3,2}} \\
   {{C}_{1,3}} & {{C}_{2,3}} & {{C}_{3,3}} \\
\end{matrix} \right)\]
\[{{A}^{\left( -1 \right)}}=\dfrac{1}{\left| -3 \right|}\times \left( \begin{matrix}
   -3 & 0 & 0 \\
   3 & -1 & 0 \\
   -9 & -2 & 3 \\
\end{matrix} \right)\]
\[{{A}^{\left( -1 \right)}}=\left( \begin{matrix}
   \dfrac{-3}{-3} & 0 & 0 \\
   \dfrac{3}{-3} & \dfrac{-1}{-3} & 0 \\
   \dfrac{-9}{-3} & \dfrac{-2}{-3} & \dfrac{3}{-3} \\
\end{matrix} \right)=\left( \begin{matrix}
   1 & 0 & 0 \\
   -1 & \dfrac{1}{3} & 0 \\
   3 & \dfrac{2}{3} & -1 \\
\end{matrix} \right)\]
Hence, the inverse of the matrix is
 \[{{A}^{-1}}=\left( \begin{matrix}
   1 & 0 & 0 \\
   -1 & \dfrac{1}{3} & 0 \\
   3 & \dfrac{2}{3} & -1 \\
\end{matrix} \right)\]

Note: Students may go wrong while finding the cofactor matrix and then arrange them in the matrix according to the value of \[{{C}_{n,n}}\] as \[\left( \begin{matrix}
   {{C}_{1,1}} & {{C}_{2,1}} & {{C}_{3,1}} \\
   {{C}_{1,2}} & {{C}_{2,2}} & {{C}_{3,2}} \\
   {{C}_{1,3}} & {{C}_{2,3}} & {{C}_{3,3}} \\
\end{matrix} \right)\] and similarly the determinant is moded hence, the determinant will not change sign.