Let A be a $3 \times 3$ matrix such that
$A^T \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 5 \\ 2 \\ 2 \end{bmatrix}$, $A^T \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 1 \\ 1 \end{bmatrix}$, $A \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 4 \\ 4 \end{bmatrix}$ and $A \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 3 \\ 1 \end{bmatrix}$.
If $\text{det}(A) = 1$, then $\text{det}(\text{adj}(A^2 + A))$ is equal to:
The problem asks for the determinant of the adjugate of the matrix $A^2 + A$, given certain properties of matrix $A$ and its determinant.
Matrix A Determination
First, let's determine the matrix $A$. Let $A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$.
- From $A \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 3 \\ 1 \end{bmatrix}$, the third column of $A$ is $\begin{bmatrix} 1 \\ 3 \\ 1 \end{bmatrix}$. So, $a_{13}=1, a_{23}=3, a_{33}=1$.
- From $A^T \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 1 \\ 1 \end{bmatrix}$, the third column of $A^T$ is $\begin{bmatrix} 3 \\ 1 \\ 1 \end{bmatrix}$. This implies the third row of $A$ is $\begin{bmatrix} 3 & 1 & 1 \end{bmatrix}$. So, $a_{31}=3, a_{32}=1, a_{33}=1$. (This is consistent with the previous point for $a_{33}$).
- The matrix A now looks like: $A = \begin{bmatrix} a_{11} & a_{12} & 1 \\ a_{21} & a_{22} & 3 \\ 3 & 1 & 1 \end{bmatrix}$.
- Using $A \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 4 \\ 4 \end{bmatrix}$: $\begin{bmatrix} a_{11} & a_{12} & 1 \\ a_{21} & a_{22} & 3 \\ 3 & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} a_{11}+1 \\ a_{21}+3 \\ 3+1 \end{bmatrix} = \begin{bmatrix} 3 \\ 4 \\ 4 \end{bmatrix}$. This gives $a_{11}+1=3 \implies a_{11}=2$ and $a_{21}+3=4 \implies a_{21}=1$.
- Using $A^T \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 5 \\ 2 \\ 2 \end{bmatrix}$: $A^T = \begin{bmatrix} 2 & 1 & 3 \\ a_{12} & a_{22} & 1 \\ 1 & 3 & 1 \end{bmatrix}$. $\begin{bmatrix} 2 & 1 & 3 \\ a_{12} & a_{22} & 1 \\ 1 & 3 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 2+3 \\ a_{12}+1 \\ 1+1 \end{bmatrix} = \begin{bmatrix} 5 \\ 2 \\ 2 \end{bmatrix}$. This gives $a_{12}+1=2 \implies a_{12}=1$.
- The matrix A is $A = \begin{bmatrix} 2 & 1 & 1 \\ 1 & a_{22} & 3 \\ 3 & 1 & 1 \end{bmatrix}$.
- We are given $\text{det}(A) = 1$. Calculate the determinant of A in terms of $a_{22}$: $\text{det}(A) = 2(a_{22} \times 1 - 3 \times 1) - 1(1 \times 1 - 3 \times 3) + 1(1 \times 1 - a_{22} \times 3)$ $\text{det}(A) = 2(a_{22}-3) - 1(1-9) + 1(1-3a_{22})$ $\text{det}(A) = 2a_{22}-6 + 8 + 1-3a_{22} = 3 - a_{22}$.
- Setting $\text{det}(A)=1$: $1 = 3 - a_{22} \implies a_{22} = 2$.
- So, $A = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix}$.
Calculate $A^2 + A$
First, calculate $A^2$:
$A^2 = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix} \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 4+1+3 & 2+2+1 & 2+3+1 \\ 2+2+9 & 1+4+3 & 1+6+3 \\ 6+1+3 & 3+2+1 & 3+3+1 \end{bmatrix} = \begin{bmatrix} 8 & 5 & 6 \\ 13 & 8 & 10 \\ 10 & 6 & 7 \end{bmatrix}$
Now, calculate $A^2 + A$:
$A^2 + A = \begin{bmatrix} 8 & 5 & 6 \\ 13 & 8 & 10 \\ 10 & 6 & 7 \end{bmatrix} + \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 10 & 6 & 7 \\ 14 & 10 & 13 \\ 13 & 7 & 8 \end{bmatrix}$
Calculate Determinant of $A^2 + A$
Let $M = A^2 + A = \begin{bmatrix} 10 & 6 & 7 \\ 14 & 10 & 13 \\ 13 & 7 & 8 \end{bmatrix}$.
Calculate the determinant of $M$:
$\text{det}(M) = 10 \begin{vmatrix} 10 & 13 \\ 7 & 8 \end{vmatrix} - 6 \begin{vmatrix} 14 & 13 \\ 13 & 8 \end{vmatrix} + 7 \begin{vmatrix} 14 & 10 \\ 13 & 7 \end{vmatrix}$
$\text{det}(M) = 10(10 \times 8 - 13 \times 7) - 6(14 \times 8 - 13 \times 13) + 7(14 \times 7 - 10 \times 13)$
$\text{det}(M) = 10(80 - 91) - 6(112 - 169) + 7(98 - 130)$
$\text{det}(M) = 10(-11) - 6(-57) + 7(-32)$
$\text{det}(M) = -110 + 342 - 224 = 342 - 334 = 8$.
Calculate Determinant of Adjugate Matrix
We need to find $\text{det}(\text{adj}(A^2 + A)) = \text{det}(\text{adj}(M))$.
For an $n \times n$ matrix $M$, the determinant of its adjugate is given by the formula $\text{det}(\text{adj}(M)) = (\text{det}(M))^{n-1}$.
Here, $M$ is a $3 \times 3$ matrix, so $n=3$.
$\text{det}(\text{adj}(M)) = (\text{det}(M))^{3-1} = (\text{det}(M))^2$.
Substituting $\text{det}(M) = 8$:
$\text{det}(\text{adj}(A^2 + A)) = (8)^2 = 64$.
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