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MATHEMATICS · Single correct · CBSE · Mathematics · Determinants · Determinants
Problem Let A = [ x 3 2 1 y 4 2 2 z ] A=\begin{bmatrix}x&3&2\\1&y&4\\2&2&z\end{bmatrix} A = x 1 2 3 y 2 2 4 z . If x y z = 2 λ xyz=2\lambda x y z = 2 λ and 8 x + 4 y + 3 z = λ + 28 8x+4y+3z=\lambda+28 8 x + 4 y + 3 z = λ + 28 , then adj ( A ) A \operatorname{adj}(A)A adj ( A ) A equals
(a) [ λ + 1 0 0 0 λ + 1 0 0 0 λ + 1 ] \begin{bmatrix}\lambda+1&0&0\\0&\lambda+1&0\\0&0&\lambda+1\end{bmatrix} λ + 1 0 0 0 λ + 1 0 0 0 λ + 1
(b) [ λ 0 0 0 λ 0 0 0 λ ] \begin{bmatrix}\lambda&0&0\\0&\lambda&0\\0&0&\lambda\end{bmatrix} λ 0 0 0 λ 0 0 0 λ
(c) [ λ 2 0 0 0 λ 2 0 0 0 λ 2 ] \begin{bmatrix}\lambda^2&0&0\\0&\lambda^2&0\\0&0&\lambda^2\end{bmatrix} λ 2 0 0 0 λ 2 0 0 0 λ 2
(d) [ λ + 2 0 0 0 λ + 2 0 0 0 λ + 2 ] \begin{bmatrix}\lambda+2&0&0\\0&\lambda+2&0\\0&0&\lambda+2\end{bmatrix} λ + 2 0 0 0 λ + 2 0 0 0 λ + 2
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