Question

If \(A = [\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}]\), \(C = [\begin{array}{cc}7 & 3 \\ 0 & 6\end{array}]\) and \(AB = C\), then the inverse of matrix B is

• A. \(\frac{1}{42}[\begin{array}{cc}3 & 0 \\ -1 & 2\end{array}]\)
• B. \(\frac{1}{6}[\begin{array}{cc}3 & 0 \\ -1 & 2\end{array}]\)
• C. \(\frac{1}{42}[\begin{array}{cc}6 & 3 \\ -1 & 2\end{array}]\)
• D. \(\frac{1}{6}[\begin{array}{cc}7 & 3 \\ -1 & 3\end{array}]\)

Hint:

Since AB = C, B inverse = C inverse times A.

Answer 1

The Correct Option is B

Step 1: Shortcut
Taking the inverse of both sides of $B = A^{-1}C$ gives $B^{-1} = C^{-1}A$.

Step 2: Adjoint of C
$\operatorname{adj}C = \begin{bmatrix} 6 & -3 \\ 0 & 7 \end{bmatrix}$ and $|C| = 42$.

Step 3: Product
$\operatorname{adj}C \cdot A = \begin{bmatrix} 21 & 0 \\ -7 & 14 \end{bmatrix}$. Dividing by 42 and simplifying by 7 gives $\frac16\begin{bmatrix} 3 & 0 \\ -1 & 2 \end{bmatrix}$. Option (B).

Final Answer:
Option (B). \[ \boxed{\frac{1}{6}\begin{bmatrix} 3 & 0 \\ -1 & 2 \end{bmatrix}} \]