Find the inverse of {pmatrix}4 & 2 \\ -3 & -2{pmatrix}

Correct Answer is : (11−1.5−2){pmatrix}1 & 1 \\ -1.5 & -2{pmatrix}(1−1.51−2)

{pmatrix}1 & 1 \\ -1.5 & -2{pmatrix}

{pmatrix}1 & -1 \\ 1.5 & -2{pmatrix}

{pmatrix}-2 & 1 \\ 1.5 & 1{pmatrix}

{pmatrix}-2 & -1 \\ 1.5 & 1{pmatrix}

Correct Answer is : (11−1.5−2){pmatrix}1 & 1 \\ -1.5 & -2{pmatrix}(1−1.51−2)

Test Index

WAEC Further Mathematics 2021 Paper

Question : 13

Total: 38

Find the inverse of

\begin{pmatrix}4 & 2 \\ -3 & -2\end{pmatrix}

Solution:

Concept:

For a

2 \times 2

matrix

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

, the inverse is given by

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

provided the determinant

ad-bc \neq 0

Explanation:

Let

A=\begin{pmatrix}4 & 2 \\ -3 & -2\end{pmatrix}

First, find the determinant:

|A| = 4(-2) - 2(-3) = -8 + 6 = -2

The adjugate matrix is formed by swapping the diagonal elements and changing the signs of the off-diagonal elements:

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

Now multiply the adjugate by

\frac{1}{|A|}

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

This gives:

A^{-1} = \begin{pmatrix}1 & 1 \\ -1.5 & -2\end{pmatrix}

Answer:

Option A:

A^{-1} = \begin{pmatrix}1 & 1 \\ -1.5 & -2\end{pmatrix}

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