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 vlaue of $x$ given that $A^2=B$
✓
Answer
$A^2=\left[\begin{array}{ll}2 & x \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}2 & x \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}4+0 & 2 x+x \\ 0+0 & 0+1\end{array}\right]=\left[\begin{array}{cc}4 & 3 x \\ 0 & 1\end{array}\right]$
Given $A^2=B$
$\left[\begin{array}{cc}4 & 3 x \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}4 & 36 \\ 0 & 1\end{array}\right]$
Comparing the two matrices we get
$3 x=36 \Rightarrow x=12$
Given $A^2=B$
$\left[\begin{array}{cc}4 & 3 x \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}4 & 36 \\ 0 & 1\end{array}\right]$
Comparing the two matrices we get
$3 x=36 \Rightarrow x=12$
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