If $A=\left[\begin{array}{cc}-1 & 0 \\ 0 & 2\end{array}\right]$, then $A^3-A^2$ is equal to

• Correct answer
  A

  2A
• B

  2I
• C

  A
• D

  I

Solution

Given that,
Now, $\begin{aligned} A^2 & =\left[\begin{array}{cc} -1 & 0 \\ 0 & 2 \end{array}\right]\left[\begin{array}{cc} -1 & 0 \\ 0 & 2 \end{array}\right] \\ & =\left[\begin{array}{ll} 1 & 0 \\ 0 & 4 \end{array}\right] \end{aligned}$ $\begin{aligned} & A^3=A^2 \cdot A=\left[\begin{array}{ll}1 & 0 \\ 0 & 4\end{array}\right]\left[\begin{array}{cc}-1 & 0 \\ 0 & 2\end{array}\right] \\ & =\left[\begin{array}{cc}-1 & 0 \\ 0 & 8\end{array}\right] \\ & \therefore \quad A^3-A^2=\left[\begin{array}{cc}-1 & 0 \\ 0 & 8\end{array}\right]-\left[\begin{array}{ll}1 & 0 \\ 0 & 4\end{array}\right] \\ & \end{aligned}$

Hence, the correct answer is option: 'A'.