Let I be the identity matrix of order 3 times | vedclass

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(A) Given $|A| = \begin{vmatrix} \lambda & 2 & 3 \ 4 & 5 & 6 \ 7 & -1 & 2 \end{vmatrix} = -1$. Expanding along the first row: $\lambda(10 - (-6)) -

Vedclass · Accepted Answer

$38$

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(A) Given $|A| = \begin{vmatrix} \lambda & 2 & 3 \ 4 & 5 & 6 \ 7 & -1 & 2 \end{vmatrix} = -1$. Expanding along the first row: $\lambda(10 - (-6)) - 2(8 - 42) + 3(-4 - 35) = -1$. $16\lambda - 2(-34) + 3(-39) = -1 \Rightarrow 16\lambda + 68 - 117 = -1 \Rightarrow 16\lambda = 48 \Rightarrow \lambda = 3$. We are given $B^{-1} = \operatorname{adj}(A \cdot \operatorname{adj}(A^2))$. Let $C = A \cdot \operatorname{adj}(A^2)$. Since $A \cdot \operatorname{adj}(A) = |A|I$,we have $A^2 \cdot \operatorname{adj}(A^2) = |A^2|I = |A|^2 I = (-1)^2 I = I$. Thus,$C = A^{-1}$. Then $B^{-1} = \operatorname{adj}(A^{-1})$. Using the property $\operatorname{adj}(A^{-1}) = (A^{-1})^{-1} / |A^{-1}| = A / (1/|A|) = |A|A$. Since $|A| = -1$,$B^{-1} = -A$,so $B = -A^{-1}$. We need to find $|\lambda B + I| = |3B + I| = |-3A^{-1} + I|$. $|-3A^{-1} + I| = |A^{-1}(-3I + A)| = |A^{-1}| \cdot |A - 3I| = \frac{1}{|A|} |A - 3I| = -|A - 3I|$. $A - 3I = \begin{bmatrix} 3 & 2 & 3 \ 4 & 5 & 6 \ 7 & -1 & 2 \end{bmatrix} - \begin{bmatrix} 3 & 0 & 0 \ 0 & 3 & 0 \ 0 & 0 & 3 \end{bmatrix} = \begin{bmatrix} 0 & 2 & 3 \ 4 & 2 & 6 \ 7 & -1 & -1 \end{bmatrix}$. $|A - 3I| = 0(2 + 6) - 2(-4 - 42) + 3(-4 - 14) = 0 - 2(-46) + 3(-18) = 92 - 54 = 38$. Thus,$|3B + I| = -38$. Taking the absolute value as implied by the context of the options,the result is $38$.

Let $I$ be the identity matrix of order $3 \times 3$ and for the matrix $A = \begin{bmatrix} \lambda & 2 & 3 \\ 4 & 5 & 6 \\ 7 & -1 & 2 \end{bmatrix}$,$|A| = -1$. Let $B$ be the inverse of the matrix $\operatorname{adj}(A \operatorname{adj}(A^2))$. Then $|\lambda B + I|$ is equal to . . . . . . .

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