Matrices & Determinants Question - None

Question

Given that matrix B is the inverse of a $3 \times 3$ matrix A, where:

$$B = \begin{pmatrix} 5 & 2a & 1 \ 0 & 2 & 1 \ a & 3 & -1 \end{pmatrix}$$

If $\det(A) + 1 = 0$, find the value(s) of $a$.

Accepted Answer

Step 1: Since B is the inverse of A, we have $A \cdot B = I$

Step 2: We know that $\det(A) \cdot \det(B) = \det(I) = 1$

Step 3: Given $\det(A) + 1 = 0$, we get $\det(A) = -1$

Step 4: Therefore: $\det(B) = \frac{1}{\det(A)} = \frac{1}{-1} = -1$

Step 5: Calculate $\det(B)$:

$\det(B) = 5(-2 - 3) - 2a(0 - a) + 1(0 - 2a) = -1$

$-25 + 2a^2 - 2a = -1$

$2a^2 - 2a - 24 = 0$

$a^2 - a - 12 = 0$

$(a - 4)(a + 3) = 0$

∴ $a = 4$ or $a = -3$

Question: