Question
If $A=\left[\begin{array}{rrr}1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1\end{array}\right]$, find $A^{-1}$ Using $A^{-1}$ solve the system of linear equations $x-2 y=10,2 x-y-z=8,-2 y+z=7$.
✓
Answer
$\begin{array}{l}\text { Here, }| A |=\left|\begin{array}{rrr}1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1\end{array}\right|=1(-1-2)-2(-2-0) \\ =-3+4=1 \neq 0 \\ \Rightarrow A^{-1} \text { exists and } A ^{-1}=\frac{1}{|A|} \text { adj } A \\ \text { Here, } A _{11}=\left|\begin{array}{rr}-1 & -2 \\ -1 & 1\end{array}\right|=-3, A_{12}=-\left|\begin{array}{rr}-2 & -2 \\ 0 & 1\end{array}\right|=2, A_{13}=\left|\begin{array}{rr}-2 & -1 \\ 0 & -1\end{array}\right|=2 ; \\ A _{21}=-\left|\begin{array}{rr}2 & 0 \\ -1 & 1\end{array}\right|=-2, A_{22}=\left|\begin{array}{rr}1 & 0 \\ 0 & 1\end{array}\right|=1, A_{23}=-\left|\begin{array}{rr}1 & 2 \\ 0 & -1\end{array}\right|=1 ; \\ A _{31}=\left|\begin{array}{rr}2 & 0 \\ -1 & -2\end{array}\right|=-4, A_{32}=-\left|\begin{array}{rr}1 & 0 \\ -2 & -2\end{array}\right|=2, A_{33}=\left|\begin{array}{rr}1 & 2 \\ -2 & -1\end{array}\right|=3 \\ \left.\therefore \text { adj } A =\left\lvert\, \begin{array}{rrr}-3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3\end{array}\right.\right]=\left[\begin{array}{rr}-3 & -2 \\ 2 & 1 \\ 2 & -4 \\ 2 & 1 \\ 3\end{array}\right] \\ \therefore A ^{-1}=\frac{1}{|A|} \text { adj } A =\frac{1}{1}\left[\begin{array}{rrr}2 & 1 & 2 \\ 2 & -2 & -4 \\ 2 & 1 & 3\end{array}\right]=\left[\begin{array}{rrr}-3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3\end{array}\right]\end{array}$
The given system of equations can be written as
$
\left[\begin{array}{rrr}
1 & -2 & 0 \\
2 & -1 & -1 \\
0 & -2 & 1
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{r}
10 \\
8 \\
7
\end{array}\right]
$
or $A ^{\prime} X = B$, where $X =\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B =\left[\begin{array}{r}10 \\ 8 \\ 7\end{array}\right]$
Now $\left|A^{\prime}\right|=|A|=1 \neq 0 \Rightarrow\left(A^{\prime}\right)^{-1}$ exists
$\Rightarrow$ the given system has a unique solution $X=\left(A^{\prime}\right)^{-1} B$
$
\begin{array}{l}
\text { or } X=\left(A^{-1}\right)^{\prime} B\left(\because\left(A^{\prime}\right)-1=\left(A^{-1}\right)^{\prime}\right) \\
\Rightarrow X=\left[\begin{array}{rrr}
-3 & -2 & -4 \\
2 & 1 & 2 \\
2 & 1 & 3
\end{array}\right]^{\prime}\left[\begin{array}{r}
10 \\
8 \\
7
\end{array}\right]=\left[\begin{array}{lll}
-3 & 2 & 2 \\
-2 & 1 & 1 \\
-4 & 2 & 3
\end{array}\right]\left[\begin{array}{r}
10 \\
8 \\
7
\end{array}\right] \\
\Rightarrow\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{c}
-30+16+14 \\
-20+8+7 \\
-40+16+21
\end{array}\right]=\left[\begin{array}{r}
0 \\
-5 \\
-3
\end{array}\right] \Rightarrow x=0, y=-5, z=-3
\end{array}
$
Hence, the solution of the given system of equations is $x=0, y=-5, z=-3$.
The given system of equations can be written as
$
\left[\begin{array}{rrr}
1 & -2 & 0 \\
2 & -1 & -1 \\
0 & -2 & 1
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{r}
10 \\
8 \\
7
\end{array}\right]
$
or $A ^{\prime} X = B$, where $X =\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B =\left[\begin{array}{r}10 \\ 8 \\ 7\end{array}\right]$
Now $\left|A^{\prime}\right|=|A|=1 \neq 0 \Rightarrow\left(A^{\prime}\right)^{-1}$ exists
$\Rightarrow$ the given system has a unique solution $X=\left(A^{\prime}\right)^{-1} B$
$
\begin{array}{l}
\text { or } X=\left(A^{-1}\right)^{\prime} B\left(\because\left(A^{\prime}\right)-1=\left(A^{-1}\right)^{\prime}\right) \\
\Rightarrow X=\left[\begin{array}{rrr}
-3 & -2 & -4 \\
2 & 1 & 2 \\
2 & 1 & 3
\end{array}\right]^{\prime}\left[\begin{array}{r}
10 \\
8 \\
7
\end{array}\right]=\left[\begin{array}{lll}
-3 & 2 & 2 \\
-2 & 1 & 1 \\
-4 & 2 & 3
\end{array}\right]\left[\begin{array}{r}
10 \\
8 \\
7
\end{array}\right] \\
\Rightarrow\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{c}
-30+16+14 \\
-20+8+7 \\
-40+16+21
\end{array}\right]=\left[\begin{array}{r}
0 \\
-5 \\
-3
\end{array}\right] \Rightarrow x=0, y=-5, z=-3
\end{array}
$
Hence, the solution of the given system of equations is $x=0, y=-5, z=-3$.
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