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Given P = \(\begin{bmatrix}1 & 2\\2 & 3\end{bmatrix}\), find P\(^2\) - 4P - I where...

Mathematics
JAMB 2024

Given P = \(\begin{bmatrix}1 & 2\\2 & 3\end{bmatrix}\), find P\(^2\) - 4P - I where I is the identity matrix

  • A. \(\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}\)
  • B. \(\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}\)
  • C. \(\begin{bmatrix}-1 & 0\\0 & -1\end{bmatrix}\)
  • D. \(\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}\)
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Correct Answer: Option B
Explanation

Given the matrix 

\(P = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix},\)

we want to find \( P^2 - 4P - I \), where \( I \) is the identity matrix:

\(I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.\)

Calculate {\( P^2 \)}

\(P^2 = P \times P = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix} \times \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix} = \begin{bmatrix} 5 & 8 \\ 8 & 13 \end{bmatrix}.\)

Calculate {\( 4P \)}

\(4P = 4 \times \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix} = \begin{bmatrix} 4 & 8 \\ 8 & 12 \end{bmatrix}\)

Calculate {\( P^2 - 4P \)}

\(P^2 - 4P = \begin{bmatrix} 5 & 8 \\ 8 & 13 \end{bmatrix} - \begin{bmatrix} 4 & 8 \\ 8 & 12 \end{bmatrix} = \begin{bmatrix} 5 - 4 & 8 - 8 \\ 8 - 8 & 13 - 12 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)

Calculate {\( P^2 - 4P - I \)}

\(P^2 - 4P - I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)

Thus, the result of \( P^2 - 4P - I \) is 

\(\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)

There is an explanation video available below.


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Explanation Video

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