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If $A=\left[\begin{array}{cc}-1 & 0 \\ 0 & 2\end{array}\right]$, then $A^3-A^2$ is equal to
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Verified Answer
The correct answer is:
2A
Given that,
Now,
$\begin{aligned}
A^2 & =\left[\begin{array}{cc}
-1 & 0 \\
0 & 2
\end{array}\right]\left[\begin{array}{cc}
-1 & 0 \\
0 & 2
\end{array}\right] \\
& =\left[\begin{array}{ll}
1 & 0 \\
0 & 4
\end{array}\right]
\end{aligned}$
$\begin{aligned} & A^3=A^2 \cdot A=\left[\begin{array}{ll}1 & 0 \\ 0 & 4\end{array}\right]\left[\begin{array}{cc}-1 & 0 \\ 0 & 2\end{array}\right] \\ & =\left[\begin{array}{cc}-1 & 0 \\ 0 & 8\end{array}\right] \\ & \therefore \quad A^3-A^2=\left[\begin{array}{cc}-1 & 0 \\ 0 & 8\end{array}\right]-\left[\begin{array}{ll}1 & 0 \\ 0 & 4\end{array}\right] \\ & \end{aligned}$
Now,
$\begin{aligned}
A^2 & =\left[\begin{array}{cc}
-1 & 0 \\
0 & 2
\end{array}\right]\left[\begin{array}{cc}
-1 & 0 \\
0 & 2
\end{array}\right] \\
& =\left[\begin{array}{ll}
1 & 0 \\
0 & 4
\end{array}\right]
\end{aligned}$
$\begin{aligned} & A^3=A^2 \cdot A=\left[\begin{array}{ll}1 & 0 \\ 0 & 4\end{array}\right]\left[\begin{array}{cc}-1 & 0 \\ 0 & 2\end{array}\right] \\ & =\left[\begin{array}{cc}-1 & 0 \\ 0 & 8\end{array}\right] \\ & \therefore \quad A^3-A^2=\left[\begin{array}{cc}-1 & 0 \\ 0 & 8\end{array}\right]-\left[\begin{array}{ll}1 & 0 \\ 0 & 4\end{array}\right] \\ & \end{aligned}$
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