Question

If $A^{-1} = \left[\begin{array}{cc}2 & -3 \\ -1 & 2\end{array}\right]$ and $B^{-1} = \left[\begin{array}{cc}1 & 0 \\ -3 & 1\end{array}\right]$, then $(AB)^{-1} =$

• A. $\left[\begin{array}{cc}2 & 7 \\ 3 & -1\end{array}\right]$
• B. $\left[\begin{array}{cc}2 & -7 \\ -3 & 11\end{array}\right]$
• C. $\left[\begin{array}{cc}2 & -3 \\ -7 & 11\end{array}\right]$
• D. $\left[\begin{array}{cc}2 & 3 \\ 7 & -11\end{array}\right]$

Hint:

The reversal law, $(AB)^{-1} = B^{-1}A^{-1}$ (and similarly for transposes: $(AB)^T = B^T A^T$), is a fundamental property of matrices. Multiplying them in the wrong order ($A^{-1}B^{-1}$) will lead you straight to a trap option!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are provided with the inverse matrices $A^{-1}$ and $B^{-1}$ and asked to compute the inverse of the matrix product $AB$, denoted as $(AB)^{-1}$.

Step 2: Key Formula or Approach:
The reversal law for the inverse of a matrix product states that $(AB)^{-1} = B^{-1}A^{-1}$. Note the order of multiplication is reversed!

Step 3: Detailed Explanation:
Set up the matrix multiplication using the reversal law:
$$(AB)^{-1} = B^{-1}A^{-1} = \left[\begin{array}{cc}1 & 0 \\ -3 & 1\end{array}\right] \left[\begin{array}{cc}2 & -3 \\ -1 & 2\end{array}\right]$$ Perform the standard row-by-column matrix multiplication:
Element in row 1, column 1: $(1)(2) + (0)(-1) = 2 + 0 = 2$
Element in row 1, column 2: $(1)(-3) + (0)(2) = -3 + 0 = -3$
Element in row 2, column 1: $(-3)(2) + (1)(-1) = -6 - 1 = -7$
Element in row 2, column 2: $(-3)(-3) + (1)(2) = 9 + 2 = 11$
Construct the resulting matrix from these elements:
$$(AB)^{-1} = \left[\begin{array}{cc}2 & -3 \\ -7 & 11\end{array}\right]$$

Step 4: Final Answer:
The required matrix is $\left[\begin{array}{cc}2 & -3 \\ -7 & 11\end{array}\right]$, matching option (C).