JAMB 1994 · UME · Q24

Matrices & determinants

Solve for $x$ and $y$ : $\begin{pmatrix} 1 & 1 \\ 3 & y \end{pmatrix}\begin{pmatrix} x \\ 1 \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \end{pmatrix}$ .

Worked solution (try it first)

Multiply each row of the square matrix by the column.
The first row gives $1 \times x + 1 \times 1 = x + 1$ .
The second row gives $3 \times x + y \times 1 = 3x + y$ .
Match with $\begin{pmatrix} 4 \\ 1 \end{pmatrix}$ : $x + 1 = 4$ , so $x = 3$ .
Then $3(3) + y = 1$ , so $y = 1 - 9 = -8$ .
The answer is $x = 3$ , $y = -8$ , option C.

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