If A = begin{bmatrix} 1/3 & 2 0 & 2x - 3 end | vedclass

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(B) Given $A = \begin{bmatrix} 1/3 & 2 \ 0 & 2x - 3 \end{bmatrix}$ and $B = \begin{bmatrix} 3 & 6 \ 0 & -1 \end{bmatrix}$.We are given that $AB = I$

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(B) Given $A = \begin{bmatrix} 1/3 & 2 \ 0 & 2x - 3 \end{bmatrix}$ and $B = \begin{bmatrix} 3 & 6 \ 0 & -1 \end{bmatrix}$.We are given that $AB = I$,where $I$ is the identity matrix $\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$.Calculating the product $AB$:$AB = \begin{bmatrix} 1/3 & 2 \ 0 & 2x - 3 \end{bmatrix} \begin{bmatrix} 3 & 6 \ 0 & -1 \end{bmatrix} = \begin{bmatrix} (1/3 	imes 3) + (2 	imes 0) & (1/3 	imes 6) + (2 	imes -1) \ (0 	imes 3) + ((2x - 3) 	imes 0) & (0 	imes 6) + ((2x - 3) 	imes -1) \end{bmatrix}$$AB = \begin{bmatrix} 1 + 0 & 2 - 2 \ 0 + 0 & 0 - (2x - 3) \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 3 - 2x \end{bmatrix}$.Since $AB = I$,we have $\begin{bmatrix} 1 & 0 \ 0 & 3 - 2x \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$.Comparing the elements,we get $3 - 2x = 1$.$2x = 3 - 1 = 2$.$x = 1$.

If $A = \begin{bmatrix} 1/3 & 2 \\ 0 & 2x - 3 \end{bmatrix}$,$B = \begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix}$ and $AB = I$,then $x =$

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