MCQ
If $A=\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right]$, then $A^2=p A$, then the value of $p$ is
- A2
- ✓4
- C$-2$
- D$-4$
✓
Answer
Correct option: B.
4
$
A=\left[\begin{array}{cc}
2 & -2 \\
-2 & 2
\end{array}\right]
$
and $A^2=p A$
$
\begin{aligned}
& A^2=A \times A=\left[\begin{array}{cc}
2 & -2 \\
-2 & 2
\end{array}\right] \times\left[\begin{array}{cc}
2 & -2 \\
-2 & 2
\end{array}\right] \\
& =\left[\begin{array}{cc}
4+4 & -4-4 \\
-4-4 & 4+4
\end{array}\right] \\
& =\left[\begin{array}{cc}
8 & -8 \\
-8 & 8
\end{array}\right] \\
& PA = p \left[\begin{array}{cc}
2 & -2 \\
-2 & 2
\end{array}\right] \\
& =\left[\begin{array}{cc}
2 p & -2 p \\
-2 p & 2 p
\end{array}\right] \\
& \because A ^2= pA \\
& \therefore\left[\begin{array}{cc}
8 & -8 \\
8 & 8
\end{array}\right]=\left[\begin{array}{cc}
2 p & -2 p \\
-2 p & 2 p
\end{array}\right]
\end{aligned}
$
Comparing, we get
$
\begin{aligned}
& 8=2 p \\
& \Rightarrow p=4
\end{aligned}
$
A=\left[\begin{array}{cc}
2 & -2 \\
-2 & 2
\end{array}\right]
$
and $A^2=p A$
$
\begin{aligned}
& A^2=A \times A=\left[\begin{array}{cc}
2 & -2 \\
-2 & 2
\end{array}\right] \times\left[\begin{array}{cc}
2 & -2 \\
-2 & 2
\end{array}\right] \\
& =\left[\begin{array}{cc}
4+4 & -4-4 \\
-4-4 & 4+4
\end{array}\right] \\
& =\left[\begin{array}{cc}
8 & -8 \\
-8 & 8
\end{array}\right] \\
& PA = p \left[\begin{array}{cc}
2 & -2 \\
-2 & 2
\end{array}\right] \\
& =\left[\begin{array}{cc}
2 p & -2 p \\
-2 p & 2 p
\end{array}\right] \\
& \because A ^2= pA \\
& \therefore\left[\begin{array}{cc}
8 & -8 \\
8 & 8
\end{array}\right]=\left[\begin{array}{cc}
2 p & -2 p \\
-2 p & 2 p
\end{array}\right]
\end{aligned}
$
Comparing, we get
$
\begin{aligned}
& 8=2 p \\
& \Rightarrow p=4
\end{aligned}
$
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