MCQ
If $A=\left[\begin{array}{ll}1 & 1 \\ 8 & 3\end{array}\right]$, then $A^2-4 A=$
  • A
    $\left[\begin{array}{rr}-3 & -3 \\ 60 & 5\end{array}\right]$
  • B
    $\left[\begin{array}{rr}-3 & -3 \\ 8 & 3\end{array}\right]$
  • $\left[\begin{array}{ll}5 & 0 \\ 0 & 5\end{array}\right]$
  • D
    $\left[\begin{array}{cc}8 & 3 \\ 24 & 14\end{array}\right]$

Answer

Correct option: C.
$\left[\begin{array}{ll}5 & 0 \\ 0 & 5\end{array}\right]$
(c) $\left[\begin{array}{ll}5 & 0 \\ 0 & 5\end{array}\right]$
Explanation:
$\begin{array}{l} A ^2-4 A= A \cdot A -4 A \\ =\left[\begin{array}{ll}1 & 1 \\ 8 & 3\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 8 & 3\end{array}\right]-4\left[\begin{array}{ll}1 & 1 \\ 8 & 3\end{array}\right] \\ =\left[\begin{array}{cc}9 & 4 \\ 32 & 17\end{array}\right]-\left[\begin{array}{cc}4 & 4 \\ 32 & 12\end{array}\right] \\ =\left[\begin{array}{ll}5 & 0 \\ 0 & 5\end{array}\right]\end{array}$

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