The inverse of matrix N = {vmatrix} 2 & 3 \\ 1 & 4 {vmatrix}^{-1} is

Correct Answer is : 15⋅∣4−3−12∣1/5 · {vmatrix} 4 & -3 \\ -1 & 2 {vmatrix}51⋅4−1−32

1/5 · {vmatrix} 2 & 1 \\ 3 & 4 {vmatrix}

1/5 · {vmatrix} 4 & -3 \\ -1 & 2 {vmatrix}

1/5 {vmatrix} 2 & -1 \\ -3 & 4 {vmatrix}

1/5 {vmatrix} 4 & 3 \\ 1 & 2 {vmatrix}

Correct Answer is : 15⋅∣4−3−12∣1/5 · {vmatrix} 4 & -3 \\ -1 & 2 {vmatrix}51⋅4−1−32

Test Index

JAMB Mathematics 2011 Paper

Question : 23

Total: 46

\begin{vmatrix} 2 & 3 \\ 1 & 4 \end{vmatrix}^{-1}

Solution:

Concept:

For a

2 \times 2

matrix

M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}

, the inverse is given by

M^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}

Explanation:

Let

N = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}

First, calculate the determinant of

N

|N| = (2 \times 4) - (3 \times 1) = 8 - 3 = 5

Now form the adjugate by interchanging the entries on the main diagonal and changing the signs of the other two entries:

\text{adj}(N) = \begin{pmatrix} 4 & -3 \\ -1 & 2 \end{pmatrix}

Thus, the inverse is

N^{-1} = \frac{1}{|N|} \text{adj}(N) = \frac{1}{5} \begin{pmatrix} 4 & -3 \\ -1 & 2 \end{pmatrix}

Answer:

N^{-1} = \frac{1}{5} \begin{pmatrix} 4 & -3 \\ -1 & 2 \end{pmatrix}

, which is Option B.

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