Question

If \( A = \begin{bmatrix} 2 & 1 5 & 3 \end{bmatrix} \), then \( A^{-1} = \) ____.

• A. \( \begin{bmatrix} 3 & -1 -5 & 2 \end{bmatrix} \)
• B. \( \begin{bmatrix} 3 & 1 -5 & 2 \end{bmatrix} \)
• C. \( \begin{bmatrix} 3 & 1 5 & 2 \end{bmatrix} \)
• D. None

Hint:

When the determinant $|A| = 1$, the inverse of the matrix is simply its adjoint. Always check the determinant first; if it's 1, you can write the answer by inspection!

Answer 1

The Correct Option is A

Step 1: Let \( A^{-1} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), so that \( A \cdot A^{-1} = I \) gives \( \begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix}\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \).

Step 2: Multiplying out the first column gives two equations, \( 2a + c = 1 \) and \( 5a + 3c = 0 \). From the second, \( c = -\frac{5a}{3} \); substituting into the first gives \( 2a - \frac{5a}{3} = 1 \), so \( \frac{a}{3} = 1 \), giving \( a = 3 \) and \( c = -5 \).

Step 3: Multiplying out the second column gives \( 2b + d = 0 \) and \( 5b + 3d = 1 \). From the first, \( d = -2b \); substituting gives \( 5b - 6b = 1 \), so \( -b = 1 \), giving \( b = -1 \) and \( d = 2 \).

Assembling the four entries gives the inverse matrix directly: \[ \boxed{A^{-1} = \begin{bmatrix} 3 & -1 \\ -5 & 2 \end{bmatrix}} \]