Q.The value of $x$, for which the matrix $A = \begin{bmatrix} e^{x-2} & e^{7+x} \\ e^{2+x} & e^{2x+3} \end{bmatrix}$ is singular is

Setting $\det A = 0$ and matching exponents gives $x=8$.

$\det A = e^{x-2}\cdot e^{2x+3} - e^{7+x}\cdot e^{2+x} = e^{(x-2)+(2x+3)} - e^{(7+x)+(2+x)} = e^{3x+1} - e^{2x+9}$.

For $A$ to be singular, $\det A = 0$: $e^{3x+1} = e^{2x+9}$.

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$A$ is singular when $\det A=0$; multiplying the exponential entries and equating exponents gives $x=8$.

$\det A = e^{x-2}\cdot e^{2x+3} - e^{7+x}\cdot e^{2+x}$.

Add exponents in each product: $e^{(x-2)+(2x+3)} - e^{(7+x)+(2+x)} = e^{3x+1} - e^{2x+9}$.

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