For the matrix $A=\left[\begin{array}{lll}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{array}\right] \cdot…
For the matrix $A=\left[\begin{array}{lll}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{array}\right] \cdot A^{-1}=$
- $A$
- $A^2$
- $A^3$
- $A^4$
Solution
Given,
$
\begin{aligned}
& \text { Given, } A=\left[\begin{array}{lll}
3 & -3 & 4 \\
2 & -3 & 4 \\
0 & -1 & 1
\end{array}\right] \\
& \text { Now, } \quad|A|=\left[\begin{array}{lll}
3 & -3 & 4 \\
2 & -3 & 4 \\
0 & -1 & 1
\end{array} \mid\right. \\
& =3(-3+4)+3(2-0)+4(-2-0)=3+6-8=1 \neq 0
\end{aligned}
$
Here,
$
\begin{aligned}
& A_{11}=1, A_{12}=-2, A_{13}=-2 \\
& A_{21}=-1, A_{22}=3, A_{23}=3 \\
& A_{31}=0, A_{32}=-4, A_{33}=-3
\end{aligned}
$
$\operatorname{adj}(A)=\left[\begin{array}{ccc}1 & -2 & -2 \\ -1 & 3 & 3 \\ 0 & -4 & -3\end{array}\right]^1=\left[\begin{array}{ccc}1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3\end{array}\right]$
Now,
$
\begin{aligned}
A^{-1} & =\frac{\operatorname{adj}(A)}{|A|}=\left[\begin{array}{ccc}
1 & -1 & 0 \\
-2 & 3 & -4 \\
-2 & 3 & -3
\end{array}\right] \\
A^2 & =A \cdot A=\left[\begin{array}{lll}
3 & -3 & 4 \\
2 & -3 & 4 \\
0 & -1 & 1
\end{array}\right]\left[\begin{array}{lll}
3 & -3 & 4 \\
2 & -3 & 4 \\
0 & -1 & 1
\end{array}\right] \\
A^2 & =\left[\begin{array}{ccc}
3 & -4 & 4 \\
0 & -1 & 0 \\
-2 & 2 & -3
\end{array}\right]
\end{aligned}
$
$
\begin{aligned}
\therefore A^3=A^2 \cdot A & =\left[\begin{array}{ccc}
3 & -4 & 4 \\
0 & -1 & 0 \\
-2 & 2 & -3
\end{array}\right]\left[\begin{array}{lll}
3 & -3 & 4 \\
2 & -3 & 4 \\
0 & -1 & 1
\end{array}\right] \\
A^3 & =\left[\begin{array}{ccc}
1 & -1 & 0 \\
-2 & 3 & -4 \\
-2 & 3 & -3
\end{array}\right]
\end{aligned}
$
From Eqs. (i) and (ii), we get $A^{-1}=A^3$
Asked in: AP EAMCET 2018 (22 Apr Shift 1)
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