Question
If $A=\left[\begin{array}{ll}2 & x \\ 0 & 1\end{array}\right]$ and $B=\left[\begin{array}{cc}4 & 36 \\ 0 & 1\end{array}\right]$, find the value of $x$, given that $A^2-B$
✓
Answer
Given
$
\begin{aligned}
& A ^2=\left[\begin{array}{ll}
2 & x \\
0 & 1
\end{array}\right]\left[\begin{array}{ll}
2 & x \\
0 & 1
\end{array}\right] \\
& A ^2=\left[\begin{array}{ll}
2 & x \\
0 & 1
\end{array}\right]\left[\begin{array}{ll}
2 & x \\
0 & 1
\end{array}\right] \\
& =\left[\begin{array}{cc}
4+0 & 2 x+x \\
0+0 & 0+1
\end{array}\right] \\
& =\left[\begin{array}{cc}
4 & 3 x \\
0 & 1
\end{array}\right] \\
& \because A ^2= B
\end{aligned}
$
$
\therefore\left[\begin{array}{cc}
4 & 3 x \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
4 & 36 \\
0 & 1
\end{array}\right]
$
Corresponding the corresponding elements
$
\begin{aligned}
& 3 x =36 \\
& \Rightarrow x =12
\end{aligned}
$
Hence $x=12$.
$
\begin{aligned}
& A ^2=\left[\begin{array}{ll}
2 & x \\
0 & 1
\end{array}\right]\left[\begin{array}{ll}
2 & x \\
0 & 1
\end{array}\right] \\
& A ^2=\left[\begin{array}{ll}
2 & x \\
0 & 1
\end{array}\right]\left[\begin{array}{ll}
2 & x \\
0 & 1
\end{array}\right] \\
& =\left[\begin{array}{cc}
4+0 & 2 x+x \\
0+0 & 0+1
\end{array}\right] \\
& =\left[\begin{array}{cc}
4 & 3 x \\
0 & 1
\end{array}\right] \\
& \because A ^2= B
\end{aligned}
$
$
\therefore\left[\begin{array}{cc}
4 & 3 x \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
4 & 36 \\
0 & 1
\end{array}\right]
$
Corresponding the corresponding elements
$
\begin{aligned}
& 3 x =36 \\
& \Rightarrow x =12
\end{aligned}
$
Hence $x=12$.
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