MCQ
If matrix $A=\left[\begin{array}{ll}2 & 2 \\ 0 & 2\end{array}\right]$ and $A^2=\left[\begin{array}{ll}4 & x \\ 0 & 4\end{array}\right]$, then thevalue of $x$ is:
- A2
- B4
- ✓8
- D10
✓
Answer
Correct option: C.
8
(c) 8
Explanation:
Given, matrix $A =\left[\begin{array}{ll}2 & 2 \\ 0 & 2\end{array}\right]$.
Then $A ^2=\left[\begin{array}{lll}2 & 2 \\ 0 & 2\end{array}\right]\left[\begin{array}{ll}2 & 2 \\ 0 & 2\end{array}\right]-\left[\begin{array}{ll}4 & 8 \\ 0 & 4\end{array}\right]$
On comparing it with $A ^2=\left[\begin{array}{ll}4 & x \\ 0 & 4\end{array}\right]$, we get that
$x=8$.
Explanation:
Given, matrix $A =\left[\begin{array}{ll}2 & 2 \\ 0 & 2\end{array}\right]$.
Then $A ^2=\left[\begin{array}{lll}2 & 2 \\ 0 & 2\end{array}\right]\left[\begin{array}{ll}2 & 2 \\ 0 & 2\end{array}\right]-\left[\begin{array}{ll}4 & 8 \\ 0 & 4\end{array}\right]$
On comparing it with $A ^2=\left[\begin{array}{ll}4 & x \\ 0 & 4\end{array}\right]$, we get that
$x=8$.
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