Let 0 neq a in mathbb{Z} and A = begin{bmatri | vedclass

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(D) We have,$\det(A) = \begin{vmatrix} a & a & a-y \ a & a+x & a \ a & a & a \end{vmatrix}$.Applying the column operation $C_1 ightarrow C_1 - C_3

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a rectangular hyperbola

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(D) We have,$\det(A) = \begin{vmatrix} a & a & a-y \ a & a+x & a \ a & a & a \end{vmatrix}$.Applying the column operation $C_1 ightarrow C_1 - C_3$:$\det(A) = \begin{vmatrix} y & a & a-y \ 0 & a+x & a \ 0 & a & a \end{vmatrix}$.Expanding along the first column:$\det(A) = y \cdot \begin{vmatrix} a+x & a \ a & a \end{vmatrix} - 0 + 0$.$\det(A) = y(a(a+x) - a^2) = y(a^2 + ax - a^2) = axy$.Given $\det(A) = 16$,we have $axy = 16$,which implies $xy = \frac{16}{a}$.Since $a$ is a non-zero constant,this equation is of the form $xy = k$,which represents a rectangular hyperbola.

Let $0 \neq a \in \mathbb{Z}$ and $A = \begin{bmatrix} a & a & a-y \\ a & a+x & a \\ a & a & a \end{bmatrix}$ be a matrix. Then,the equation $\det(A) = 16$ represents:

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