If \(P = \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix}\), find \((P^{2} + P)\).
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option B
Explanation:
\( P^{2} = \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix}\)
\(\begin{vmatrix} 1 \times 1 + 1 \times 2 & 1 \times 1 + 1 \times 1 \\ 2 \times 1 + 1 \times 2 & 2 \times 1 + 1 \times 1 \end{vmatrix}\)
= \(\begin{vmatrix} 3 & 2 \\ 4 & 3 \end{vmatrix} + \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix}\)
= \(\begin{vmatrix} 4 & 3 \\ 6 & 4 \end{vmatrix}\)
\( P^{2} = \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix}\)
\(\begin{vmatrix} 1 \times 1 + 1 \times 2 & 1 \times 1 + 1 \times 1 \\ 2 \times 1 + 1 \times 2 & 2 \times 1 + 1 \times 1 \end{vmatrix}\)
= \(\begin{vmatrix} 3 & 2 \\ 4 & 3 \end{vmatrix} + \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix}\)
= \(\begin{vmatrix} 4 & 3 \\ 6 & 4 \end{vmatrix}\)