Determinant of Adjoint Matrix Sum

To find $\det(\text{adj}(A^2 + A))$, we use the properties of determinants and matrix algebra for $3 \times 3$ matrices.

Step 1: Simplify the expression for $\det(\text{adj}(A^2 + A))$

For any $3 \times 3$ matrix $M$, the determinant of its adjoint is given by: $\det(\text{adj}(M)) = (\det(M))^{3-1} = (\det(M))^2$

Let $M = A^2 + A = A(A + I)$. Using the multiplicative property of determinants: $\det(M) = \det(A(A + I)) = \det(A) \cdot \det(A + I)$

Given that $\det(A) = 1$: $\det(A^2 + A) = 1 \cdot \det(A + I) = \det(A + I)$

Therefore: $\det(\text{adj}(A^2 + A)) = (\det(A + I))^2$

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Step 2: Construct Matrix $A$ from the given conditions

Let the columns of $A$ be $C_1, C_2, C_3$, so $A = \begin{bmatrix} C_1 & C_2 & C_3 \end{bmatrix}$.

From the given matrix-vector products:

1. $A \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} = C_1 + C_3 = \begin{bmatrix} 3 \\ 4 \\ 4 \end{bmatrix} \implies C_1 = \begin{bmatrix} 3 \\ 4 \\ 4 \end{bmatrix} - C_3$
2. $A \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} = C_2 + C_3 = \begin{bmatrix} 1 \\ 3 \\ 1 \end{bmatrix} \implies C_2 = \begin{bmatrix} 1 \\ 3 \\ 1 \end{bmatrix} - C_3$

From the transposed relations $A^T \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 5 \\ 2 \\ 2 \end{bmatrix}$ and $A^T \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 1 \\ 1 \end{bmatrix}$, the entries of $C_3 = \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix}$ satisfy $c_1 = 2 - c_3$ and $c_2 = 1 - c_3$.

Setting $c_3 = x$, we have: $C_3 = \begin{bmatrix} 2-x \\ 1-x \\ x \end{bmatrix}$

Substituting $C_3$ back into the expressions for $C_1$ and $C_2$: $C_1 = \begin{bmatrix} 3 - (2-x) \\ 4 - (1-x) \\ 4 - x \end{bmatrix} = \begin{bmatrix} 1+x \\ 3+x \\ 4-x \end{bmatrix}$

$C_2 = \begin{bmatrix} 1 - (2-x) \\ 3 - (1-x) \\ 1 - x \end{bmatrix} = \begin{bmatrix} -1+x \\ 2+x \\ 1-x \end{bmatrix}$

Thus, the matrix $A$ is:

$$A = \begin{bmatrix} 1+x & -1+x & 2-x \\ 3+x & 2+x & 1-x \\ 4-x & 1-x & x \end{bmatrix}$$

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Step 3: Determine $\det(A)$ and $\det(A + I)$

Using column operations $C_1 \to C_1 + C_3$ and $C_2 \to C_2 + C_3$ on matrix $A$:

$$\det(A) = \begin{vmatrix} 3 & 1 & 2-x \\ 4 & 3 & 1-x \\ 4 & 1 & x \end{vmatrix}$$

Expanding along the third column: $\det(A) = 12x - 15$

Since $\det(A) = 1$: $12x - 15 = 1 \implies x = \frac{4}{3}$

Now, constructing $A + I$:

$$A + I = \begin{bmatrix} 2+x & -1+x & 2-x \\ 3+x & 3+x & 1-x \\ 4-x & 1-x & 1+x \end{bmatrix}$$

Applying column operations $C_1 \to C_1 + C_3$ and $C_2 \to C_2 + C_3$:

$$\det(A + I) = \begin{vmatrix} 4 & 1 & 2-x \\ 4 & 4 & 1-x \\ 5 & 2 & 1+x \end{vmatrix}$$

Evaluating the determinant gives: $\det(A + I) = -8$

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Step 4: Compute $\det(\text{adj}(A^2 + A))$

Substitute $\det(A + I) = -8$ into our simplified formula: $\det(\text{adj}(A^2 + A)) = (\det(A + I))^2 = (-8)^2 = 64$