Question

If $A^{-1}=\left[\begin{array}{ccc}3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 5 & 5\end{array}\right]$, then $A=$

• A. $\left[\begin{array}{ccc}-5 & 20 & -2 \\ -1 & 3 & 0 \\ 3 & -11 & 1\end{array}\right]$
• B. $\left[\begin{array}{ccc}-5 & 20 & 2 \\ -1 & 3 & 0 \\ 3 & 11 & 1\end{array}\right]$
• C. $\left[\begin{array}{ccc}-5 & 20 & 2 \\ 1 & 3 & 0 \\ 3 & 11 & -1\end{array}\right]$
• D. $\left[\begin{array}{ccc}-5 & 20 & -2 \\ 1 & 3 & 0 \\ 3 & 11 & 1\end{array}\right]$

Hint:

To save time on a $3 \times 3$ matrix inversion in a multiple-choice setting, calculating just the first column of the inverse (which comes from the cofactors of the first row) is often enough to eliminate incorrect options.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given the inverse of a matrix, $A^{-1}$, and we need to find the original matrix $A$. By properties of inverse matrices, $(A^{-1})^{-1} = A$.

Step 2: Key Formula or Approach:
To find the inverse of $A^{-1}$, we use the formula $(A^{-1})^{-1} = \frac{1}{|A^{-1}|} \text{Adj}(A^{-1})$.

Step 3: Detailed Explanation:
Let's first calculate the determinant of $A^{-1}$:
$$|A^{-1}| = 3(5 - 10) - 2(5 - 4) + 6(5 - 2)$$ $$|A^{-1}| = 3(-5) - 2(1) + 6(3)$$ $$|A^{-1}| = -15 - 2 + 18 = 1$$ Since $|A^{-1}| = 1$, the matrix $A$ is simply the adjoint of $A^{-1}$.
Now, calculate the cofactors of the elements of $A^{-1}$:
$C_{11} = (5 - 10) = -5$
$C_{12} = -(5 - 4) = -1$
$C_{13} = (5 - 2) = 3$
$C_{21} = -(10 - 30) = 20$
$C_{22} = (15 - 12) = 3$
$C_{23} = -(15 - 4) = -11$
$C_{31} = (4 - 6) = -2$
$C_{32} = -(6 - 6) = 0$
$C_{33} = (3 - 2) = 1$
The adjoint matrix is the transpose of the cofactor matrix. So, we place these cofactors into columns:
$$\text{Adj}(A^{-1}) = \left[\begin{array}{ccc}-5 & 20 & -2 \\ -1 & 3 & 0 \\ 3 & -11 & 1\end{array}\right]$$ Since the determinant is 1, $A = \text{Adj}(A^{-1})$.

Step 4: Final Answer:
The matrix $A$ is $\left[\begin{array}{ccc}-5 & 20 & -2 \\ -1 & 3 & 0 \\ 3 & -11 & 1\end{array}\right]$, matching option (A).