If P = {pmatrix} 2 & -3 \\ 1 & 1 {pmatrix} , what is P^{-1}

Correct Answer is : [1535−1525]{bmatrix} 1/5 & 3/5 \\ -1/5 & 2/5 {bmatrix}[51−515352]

{bmatrix} -1/5 & -3/5 \\ -1/5 & -2/5 {bmatrix}

{bmatrix} 1/5 & 3/5 \\ 1/5 & 2/5 {bmatrix}

{bmatrix} -1/5 & 3/5 \\ -1/5 & 2/5 {bmatrix}

{bmatrix} 1/5 & 3/5 \\ -1/5 & 2/5 {bmatrix}

Correct Answer is : [1535−1525]{bmatrix} 1/5 & 3/5 \\ -1/5 & 2/5 {bmatrix}[51−515352]

Test Index

JAMB Mathematics 2010 Paper

Question : 37

Total: 49

If P =

\begin{pmatrix} 2 & -3 \\ 1 & 1 \end{pmatrix}

^{-1}

$\begin{bmatrix} -\frac{1}{5} & -\frac{3}{5} \\ -\frac{1}{5} & -\frac{2}{5} \end{bmatrix}$
$\begin{bmatrix} \frac{1}{5} & \frac{3}{5} \\ \frac{1}{5} & \frac{2}{5} \end{bmatrix}$
$\begin{bmatrix} -\frac{1}{5} & \frac{3}{5} \\ -\frac{1}{5} & \frac{2}{5} \end{bmatrix}$
$\begin{bmatrix} \frac{1}{5} & \frac{3}{5} \\ -\frac{1}{5} & \frac{2}{5} \end{bmatrix}$

Solution:

Concept:

The inverse of a

2 \times 2

matrix

A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}

is found using the determinant and the adjugate matrix.

Explanation:

For

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

, compute the determinant

|P| = (2)(1) - (-3)(1) = 2 + 3 = 5

Swap the positions of

2

and

1

, then change the signs of

-3

and

1

to form the adjugate matrix

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

Multiply the adjugate by

\frac{1}{|P|} = \frac{1}{5}

P^{-1} = \frac{1}{5}\begin{bmatrix} 1 & 3 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} \frac{1}{5} & \frac{3}{5} \\ -\frac{1}{5} & \frac{2}{5} \end{bmatrix}

Answer:

The correct option is D:

\begin{bmatrix} \frac{1}{5} & \frac{3}{5} \\ -\frac{1}{5} & \frac{2}{5} \end{bmatrix}

Go to Question:

Prev Question

Next Question