Question on: JAMB Mathematics - 2013
Find the inverse \(\begin{vmatrix} 5 & 3 \ 6 & 4 \end{vmatrix}\)
A
\(\begin{vmatrix} 2 & -\frac{3}{2} \ -3 & -\frac{5}{2} \end{vmatrix}\)
B
\(\begin{vmatrix} 2 & -\frac{3}{2} \ -3 & \frac{5}{2} \end{vmatrix}\)
C
\(\begin{vmatrix} 2 & \frac{3}{2} \ -3 & \frac{5}{2} \end{vmatrix}\)
D
\(\begin{vmatrix} 2 & \frac{3}{2} \ -3 & \frac{5}{2} \end{vmatrix}\)
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Correct Option: B
Let A = \(\begin{pmatrix} 5 & 3 \ 6 & 4 \end{pmatrix}\)
Then |A| = \(\begin{pmatrix} 5 & 3 \ 6 & 4 \end{pmatrix}\) = 20 - 18 = 2
Hence A-1 = \(\frac{1}{|A|}\begin{pmatrix} 4 & -3 \ -6 & 5 \end{pmatrix}\)
= \(\frac{1}{2}\begin{pmatrix} 4 & -3 \ -6 & 5 \end{pmatrix}\)
= \(\begin{pmatrix} 4 \times 1/2 & -3 \times 1/2 \ -6 \times 1/2 & 5 \times 1/2 \end{pmatrix}\)
= \(\begin{vmatrix} 2 & -\frac{3}{2} \ -3 & \frac{5}{2} \end{vmatrix}\)
Then |A| = \(\begin{pmatrix} 5 & 3 \ 6 & 4 \end{pmatrix}\) = 20 - 18 = 2
Hence A-1 = \(\frac{1}{|A|}\begin{pmatrix} 4 & -3 \ -6 & 5 \end{pmatrix}\)
= \(\frac{1}{2}\begin{pmatrix} 4 & -3 \ -6 & 5 \end{pmatrix}\)
= \(\begin{pmatrix} 4 \times 1/2 & -3 \times 1/2 \ -6 \times 1/2 & 5 \times 1/2 \end{pmatrix}\)
= \(\begin{vmatrix} 2 & -\frac{3}{2} \ -3 & \frac{5}{2} \end{vmatrix}\)
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