Find the inverses of the following matrices by the transformation…

Question

Find the inverses of the following matrices by the transformation method:$\left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]$

✓

Answer

Let $A =\left(\begin{array}{rrr}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right)$
Then $| A |=\left|\begin{array}{rrr}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right|$
$
\begin{aligned}
& =2(3-0)-0(15-0)-1(5-0) \\
& =6-0-5=1 \neq 0
\end{aligned}
$
$\therefore A ^{-1}$ exists.
We write $AA ^{-1}= I$
$
\therefore\left(\begin{array}{rrr}
2 & 0 & -1 \\
5 & 1 & 0 \\
0 & 1 & 3
\end{array}\right) A ^{-1}=\left(\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right)
$
By $3 R_1$, we get
$
\left(\begin{array}{rrr}
6 & 0 & -3 \\
5 & 1 & 0 \\
0 & 1 & 3
\end{array}\right) A ^{-1}=\left(\begin{array}{lll}
3 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right)
$
By $R_1-R_2$, we get
$
\left[\begin{array}{rrr}
1 & -1 & -3 \\
5 & 1 & 0 \\
0 & 1 & 3
\end{array}\right] A ^{-1}=\left(\begin{array}{rrr}
3 & -1 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]
$
By $R_2-5 R_1$, we get
$
\left[\begin{array}{rrr}
1 & -1 & -3 \\
0 & 6 & 15 \\
0 & 1 & 3
\end{array}\right] A ^{-1}=\left[\begin{array}{rrr}
3 & -1 & 0 \\
-15 & 6 & 0 \\
0 & 0 & 1
\end{array}\right]
$
By $R_2-5 R_3$, we get
$
\left(\begin{array}{rrr}
1 & -1 & -3 \\
0 & 1 & 0 \\
0 & 1 & 3
\end{array}\right] A^{-1}=\left[\begin{array}{rrr}
3 & -1 & 0 \\
-15 & 6 & -5 \\
0 & 0 & 1
\end{array}\right]
$
By $R_1+R_2$ and $R_3-R_2$, we get
$
\left(\begin{array}{rrr}
1 & 0 & -3 \\
0 & 1 & 0 \\
0 & 0 & 3
\end{array}\right) A^{-1}=\left(\begin{array}{rrr}
-12 & 5 & -5 \\
-15 & 6 & -5 \\
15 & -6 & 6
\end{array}\right)
$
By $\left(\frac{1}{3}\right) R_3$, we get
$
\left(\begin{array}{rrr}
1 & 0 & -3 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right) A^{-1}=\left(\begin{array}{rrr}
-12 & 5 & -5 \\
-15 & 6 & -5 \\
5 & -2 & 2
\end{array}\right)
$
By $R_1+3 R_3$, we get
$
\begin{aligned}
& {\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] A ^{-1}=\left[\begin{array}{rrr}
3 & -1 & 1 \\
-15 & 6 & -5 \\
5 & -2 & 2
\end{array}\right]} \\
& \therefore A ^{-1}=\left[\begin{array}{rrr}
3 & -1 & 1 \\
-15 & 6 & -5 \\
5 & -2 & 2
\end{array}\right]
\end{aligned}
$