Let alpha beta neq 0 and A = begin{bmatrix} b | vedclass

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(D) We know that if $B$ is the matrix of cofactors of $A$,then $AB = \operatorname{det}(A)I$,where $I$ is the identity matrix. Thus,$\operatorname{det

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$216$

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(D) We know that if $B$ is the matrix of cofactors of $A$,then $AB = \operatorname{det}(A)I$,where $I$ is the identity matrix. Thus,$\operatorname{det}(AB) = \operatorname{det}(A) \cdot \operatorname{det}(B)$.Since $B = \operatorname{adj}(A)$,we have $\operatorname{det}(B) = \operatorname{det}(\operatorname{adj}(A)) = (\operatorname{det}(A))^{n-1}$,where $n=3$ is the order of the matrix.Therefore,$\operatorname{det}(AB) = \operatorname{det}(A) \cdot (\operatorname{det}(A))^{3-1} = (\operatorname{det}(A))^3$.To find $\operatorname{det}(A)$,we use the cofactor $B_{21} = -\alpha$. The cofactor of $A_{21}$ is $(-1)^{2+1} \begin{vmatrix} \alpha & 3 \ \alpha & 2\alpha \end{vmatrix} = -(2\alpha^2 - 3\alpha) = 3\alpha - 2\alpha^2$.Given $B_{21} = -\alpha$,we have $3\alpha - 2\alpha^2 = -\alpha$,which implies $2\alpha^2 - 4\alpha = 0$. Since $\alpha 
eq 0$,we get $\alpha = 2$.Using $B_{12} = -9$,the cofactor of $A_{12}$ is $(-1)^{1+2} \begin{vmatrix} \alpha & \beta \ -\beta & 2\alpha \end{vmatrix} = -(2\alpha^2 + \beta^2) = -9$. Substituting $\alpha = 2$,we get $-(8 + \beta^2) = -9$,so $\beta^2 = 1$. Since $\beta 
eq 0$,$\beta = 1$ or $-1$.Using $B_{22} = 7$,the cofactor of $A_{22}$ is $(-1)^{2+2} \begin{vmatrix} \beta & 3 \ -\beta & 2\alpha \end{vmatrix} = 2\alpha\beta + 3\beta = 7$. Substituting $\alpha = 2$,we get $4\beta + 3\beta = 7$,so $7\beta = 7$,which gives $\beta = 1$.Now,$A = \begin{bmatrix} 1 & 2 & 3 \ 2 & 2 & 1 \ -1 & 2 & 4 \end{bmatrix}$.$\operatorname{det}(A) = 1(8-2) - 2(8+1) + 3(4+2) = 6 - 18 + 18 = 6$.Thus,$\operatorname{det}(AB) = (\operatorname{det}(A))^3 = 6^3 = 216$.

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