Matrices & Determinants
Matrices and Determinants
Allen Star Batch
Grade 12
Question:
$A = \begin{pmatrix} -3 & -1 & 2 \\ 3 & 1 & -1 \\ 4 & 2 & 5 \end{pmatrix}$, $A \begin{pmatrix} x_1 \\ y_1 \\ z_1 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$, $A \begin{pmatrix} x_2 \\ y_2 \\ z_2 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$, $A \begin{pmatrix} x_3 \\ y_3 \\ z_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$, $B = \begin{pmatrix} x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \\ z_1 & z_2 & z_3 \end{pmatrix}$, then:
Trace $(B) = -8$
$|AdjB| = 4$
$|AdjB| = \frac{1}{4}$
Sum of all elements of $B$ is $-10$
Step-by-Step Solution
Key Concept: Recognize that the three column vectors of B form A^(-1) when AB equals the 3×3 identity matrix. Use the relationship |adj(B)| = |B|^(n-1) for an n×n matrix and compute trace(B) = sum of diagonal elements of A^(-1).
Given $AB = I$, we find $B = A^{-1} = \frac{1}{2}\begin{pmatrix}7 & 9 & -1 \\ -19 & -23 & 3 \\ 2 & 2 & 0\end{pmatrix}$. To verify, compute $|\text{adj}(B)| = |B|^2 = \frac{1}{4}$ since $|A| = 2$, confirming the inverse formula and that $A$ is invertible.
Correct Answer: 1,3,4