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Question:

If $A=\left[\begin{array}{cc}\frac{1}{3} & 2 \\ 0 & 2 x-3\end{array}\right], B=\left[\begin{array}{cc}3 & 6 \\ 0 & -1\end{array}\right]$ and $A B=l$, then $x=$______

Solution:

The given matrices are $A=\left[\begin{array}{cc}\frac{1}{3} & 2 \\ 0 & 2 x-3\end{array}\right]$ and $B=\left[\begin{array}{cc}3 & 6 \\ 0 & -1\end{array}\right]$

$A B=I$              (Given)

$\Rightarrow\left[\begin{array}{cc}\frac{1}{3} & 2 \\ 0 & 2 x-3\end{array}\right]\left[\begin{array}{cc}3 & 6 \\ 0 & -1\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}1 & 0 \\ 0 & 3-2 x\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]$

$\Rightarrow 3-2 x=1$

$\Rightarrow 2 x=2$

$\Rightarrow x=1$

Thus, the value of $x$ is 1 .

If $A=\left[\begin{array}{cc}\frac{1}{3} & 2 \\ 0 & 2 x-3\end{array}\right], B=\left[\begin{array}{cc}3 & 6 \\ 0 & -1\end{array}\right]$ and $A B=I$, then $x=$ 1

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