Given P = {bmatrix}1 & 2\\2 & 3{bmatrix}, find P² - 4P - I where I is the identity matrix

Correct Answer is : [0000]{bmatrix}0 & 0\\0 & 0{bmatrix}[0000]

{bmatrix}-1 & 0\\0 & -1{bmatrix}

Correct Answer is : [0000]{bmatrix}0 & 0\\0 & 0{bmatrix}[0000]

Test Index

JAMB Mathematics 2024 Paper

Question : 43

Total: 50

Given P =

\begin{bmatrix}1 & 2\\2 & 3\end{bmatrix}

^2

- 4P - I where I is the identity matrix

Solution:

Concept:

Matrix multiplication and scalar subtraction are performed element by element.

Here, compute

P^2

, then scale

P

4

, and subtract the identity matrix

I

Explanation:

Given

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

and

I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

First, find

P^2

by multiplying

P

by itself.

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

So,

P^2 = \begin{bmatrix} 5 & 8 \\ 8 & 13 \end{bmatrix}

Next, compute

4P

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

Now subtract:

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

This equals

I

Therefore,

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

Answer:

The result is

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

, which is option B.

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