If P - begin bmatrix 1 - lalpha - 3 1 - 3 - 3 2 - 4 - 4 end bmatrix is the adjoint of a 3 - 3 matrix A and det A -4 then -A- 4B- 11C- 5D- 7

The correct option is B 11|P| = 2 α 6, now adj A = P ⇒ |adj A| = |P| ⇒ |A|^2 = |P| 16 = 2 α 6 α = 11 ...

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If P = begin ...

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If $P = ⎡ ⎢ ⎣ \begin{matrix} 1 & α & 3 1 & 3 & 3 2 & 4 & 4 \end{matrix} ⎤ ⎥ ⎦$ is the adjoint of a $3 \times 3$ matrix A and det (A) = 4 then $α =$

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If P = $⎛ ⎜ ⎝ \begin{matrix} 1 & \propto & 3 1 & 3 & 3 2 & 4 & 4 \end{matrix} ⎞ ⎟ ⎠$ is the adjoint of a 3x3 matrix A and det(A)=4,then α is equal to

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