WAEC - Further Mathematics (2009 - No. 21)
If \(P = \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix}\), find \((P^{2} + P)\).
\(\begin{vmatrix} 4 & 3 \\ 6 & 1 \end{vmatrix}\)
\(\begin{vmatrix} 4 & 3 \\ 6 & 4 \end{vmatrix}\)
\(\begin{vmatrix} 2 & 2 \\ 6 & 2 \end{vmatrix}\)
\(\begin{vmatrix} 3 & 2 \\ 6 & 4 \end{vmatrix}\)
Explanation
\( P^{2} = \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix}\)
\(\begin{vmatrix} 1 \times 1 + 1 \times 2 & 1 \times 1 + 1 \times 1 \\ 2 \times 1 + 1 \times 2 & 2 \times 1 + 1 \times 1 \end{vmatrix}\)
= \(\begin{vmatrix} 3 & 2 \\ 4 & 3 \end{vmatrix} + \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix}\)
= \(\begin{vmatrix} 4 & 3 \\ 6 & 4 \end{vmatrix}\)
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