Question

Find the inverse of the matrix:
\[ \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix} \]

• A. $\dfrac{1}{5} \begin{pmatrix} 4 & -3 \\ -1 & 2 \end{pmatrix}$
• B. $\begin{pmatrix} 4 & 3 \\ -1 & 2 \end{pmatrix}$
• C. $\begin{pmatrix} 2 & -3 \\ -1 & 4 \end{pmatrix}$
• D. $\begin{pmatrix} -4 & 3 \\ 1 & -2 \end{pmatrix}$

Hint:

For a 2x2 matrix, the inverse is found using the determinant and adjugate. Always verify by checking if $A \cdot A^{-1} = I$.

Answer 1

The Correct Option is A

Step 1: Calculate the Determinant

First, compute the determinant of matrix \( A \):

\[ \det(A) = (2)(4) - (3)(1) = 8 - 3 = 5. \]

Since the determinant is not zero (\(\det(A) = 5 \neq 0\)), the matrix \( A \) is invertible.

Step 2: Apply the Inverse Formula for a 2x2 Matrix

The inverse of a 2x2 matrix

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

is given by:

\[ A^{-1} = \frac{1}{\det(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}. \]

Applying this to our matrix:

\[ A^{-1} = \frac{1}{5} \begin{pmatrix} 4 & -3 \\ -1 & 2 \end{pmatrix}. \]

Step 3: Write the Final Inverse Matrix

Multiply each element of the matrix by \(\frac{1}{5}\):

\[ A^{-1} = \begin{pmatrix} \frac{4}{5} & -\frac{3}{5} \\ -\frac{1}{5} & \frac{2}{5} \end{pmatrix}. \]

Final Answer

\[ \boxed{\begin{pmatrix} \dfrac{4}{5} & -\dfrac{3}{5} \\ -\dfrac{1}{5} & \dfrac{2}{5} \end{pmatrix}} \]

Alternatively, it can also be expressed as:

\[ \boxed{\dfrac{1}{5} \begin{pmatrix} 4 & -3 \\ -1 & 2 \end{pmatrix}} \]