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: Understanding the Concept
From \(AB = C\), \(B = A^{-1}C\). The inverse of a product reverses the order: \(B^{-1} = C^{-1}A\).

Step 2: Find C inverse
\(|C| = 7 \times 6 - 3 \times 0 = 42\). So \(C^{-1} = \dfrac{1}{42}\begin{bmatrix} 6 & -3 \\ 0 & 7 \end{bmatrix}\).

Step 3: Multiply
\[ B^{-1} = \frac{1}{42}\begin{bmatrix} 6 & -3 \\ 0 & 7 \end{bmatrix}\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \frac{1}{42}\begin{bmatrix} 21 & 0 \\ -7 & 14 \end{bmatrix} = \frac16\begin{bmatrix} 3 & 0 \\ -1 & 2 \end{bmatrix} \]
This is option (B). Option (A) forgets to divide the 7s out, (C) and (D) keep entries of C.

Final Answer:
The inverse is \(\frac16\begin{bmatrix} 3 & 0 \\ -1 & 2 \end{bmatrix}\), option (B). \[ \boxed{\frac{1}{6}\begin{bmatrix} 3 & 0 \\ -1 & 2 \end{bmatrix}} \]