Matrices & Determinants
General
Grade 12
Question:
If $[1 \quad x \quad 2] \begin{bmatrix} 2 & 3 & 1 \\ 0 & 4 & 2 \\ 0 & 3 & 2 \end{bmatrix} \begin{bmatrix} x \\ 1 \\ -1 \end{bmatrix} = O$, then the value of $x$ is
Step-by-Step Solution
Key Concept: General
The LHS of the equation is $$[1 \quad x \quad 2] \begin{bmatrix} 2 & 3 & 1 \\ 0 & 4 & 2 \\ 0 & 3 & 2 \end{bmatrix} \begin{bmatrix} x \\ 1 \\ -1 \end{bmatrix} = [2 \quad 4x + 9 \quad 2x + 5] \begin{bmatrix} x \\ 1 \\ -1 \end{bmatrix}$$ $$= [2x + 4x + 9 - 2x - 5] = [4x + 4]$$ Thus $4x + 4 = 0 \Rightarrow x = -1$.
Correct Answer: -1