Q.If $A^{-1} = \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix}$, find $(AB)^{-1}$.

Concept: Inverse of a Product — $(AB)^{-1} = B^{-1}A^{-1}$.

We are given $A^{-1}$ directly. We first need $B^{-1}$, then multiply in reverse order.

Step 1: Find $B^{-1}$.

For $B = \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix}$, compute $\det(B)$:

Step 2: Adjugate and inverse.

Cofactor matrix:

Transpose to get adjugate, then $B^{-1} = \frac{1}{\det(B)} \cdot \text{adj}(B)$:

Step 3: Multiply $B^{-1}A^{-1}$.

Compute:

• Row 1: $(9 -30 +30,\; -3 +12 -12,\; 3 -10 +12) = (9, -3, 5)$
• Row 2: $(3 -15 +10,\; -1 +6 -4,\; 1 -5 +4) = (-2, 1, 0)$
• Row 3: $(6 -30 +25,\; -2 +12 -10,\; 2 -10 +10) = (1, 0, 2)$

Thus:

The key idea here is one of the most elegant properties of matrix inverses: when you multiply two matrices and then invert the result, the inverses come back in reverse order. That is, $(AB)^{-1} = B^{-1}A^{-1}$. This isn't just a trick — it follows from the definition of an inverse. If you multiply $AB$ by $B^{-1}A^{-1}$, you get $A(BB^{-1})A^{-1} = AIA^{-1} = AA^{-1} = I$. So the reverse order is forced by the way matrix multiplication works.

We are given $A^{-1}$ directly, so we only need to find $B^{-1}$ and then multiply.

1. Find $B^{-1}$. We have $B = \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix}$. To find its inverse, we can use the formula $B^{-1} = \frac{1}{\det(B)} \operatorname{adj}(B)$. First compute the determinant:

Since $\det(B) = 1$, the inverse is simply the adjugate (transpose of the cofactor matrix). That saves us from dividing.

Now compute the cofactor matrix. For each entry $(i,j)$, the cofactor is $(-1)^{i+j}$ times the determinant of the submatrix after removing row $i$ and column $j$.

• $C_{11} = + \begin{vmatrix} 3 & 0 \\ -2 & 1 \end{vmatrix} = 3$
• $C_{12} = - \begin{vmatrix} -1 & 0 \\ 0 & 1 \end{vmatrix} = -(-1) = 1$
• $C_{13} = + \begin{vmatrix} -1 & 3 \\ 0 & -2 \end{vmatrix} = 2$
• $C_{21} = - \begin{vmatrix} 2 & -2 \\ -2 & 1 \end{vmatrix} = -(2\cdot 1 - (-2)(-2)) = -(2 - 4) = 2$
• $C_{22} = + \begin{vmatrix} 1 & -2 \\ 0 & 1 \end{vmatrix} = 1$
• $C_{23} = - \begin{vmatrix} 1 & 2 \\ 0 & -2 \end{vmatrix} = -(-2) = 2$
• $C_{31} = + \begin{vmatrix} 2 & -2 \\ 3 & 0 \end{vmatrix} = 0 - (-6) = 6$
• $C_{32} = - \begin{vmatrix} 1 & -2 \\ -1 & 0 \end{vmatrix} = -(0 - 2) = 2$
• $C_{33} = + \begin{vmatrix} 1 & 2 \\ -1 & 3 \end{vmatrix} = 3 - (-2) = 5$

So the cofactor matrix is:

The adjugate is its transpose:

Since $\det(B)=1$, we have $B^{-1} = \operatorname{adj}(B)$.

2. Multiply $B^{-1}$ by $A^{-1}$. We need $(AB)^{-1} = B^{-1}A^{-1}$. So compute:

Let's do this step by step. The entry in row $i$, column $j$ is the dot product of row $i$ of $B^{-1}$ with column $j$ of $A^{-1}$.

Row 1:

• Col 1: $3\cdot 3 + 2\cdot(-15) + 6\cdot 5 = 9 - 30 + 30 = 9$
• Col 2: $3\cdot(-1) + 2\cdot 6 + 6\cdot(-2) = -3 + 12 - 12 = -3$
• Col 3: $3\cdot 1 + 2\cdot(-5) + 6\cdot 2 = 3 - 10 + 12 = 5$

Row 2:

• Col 1: $1\cdot 3 + 1\cdot(-15) + 2\cdot 5 = 3 - 15 + 10 = -2$
• Col 2: $1\cdot(-1) + 1\cdot 6 + 2\cdot(-2) = -1 + 6 - 4 = 1$
• Col 3: $1\cdot 1 + 1\cdot(-5) + 2\cdot 2 = 1 - 5 + 4 = 0$

Row 3:

• Col 1: $2\cdot 3 + 2\cdot(-15) + 5\cdot 5 = 6 - 30 + 25 = 1$
• Col 2: $2\cdot(-1) + 2\cdot 6 + 5\cdot(-2) = -2 + 12 - 10 = 0$
• Col 3: $2\cdot 1 + 2\cdot(-5) + 5\cdot 2 = 2 - 10 + 10 = 2$

So the product is:

Method: Finding $(AB)^{-1}$ Using the Reversal Rule

When a question gives one matrix's inverse directly and asks for the inverse of a PRODUCT involving it, use the reversal identity instead of ever computing $AB$ itself.

Steps

Step 1: Recall the identity $(AB)^{-1} = B^{-1}A^{-1}$

The order of the two inverses is reversed compared to the original product — this comes directly from undoing $AB$ in the opposite order it was built (the "socks and shoes" idea).

Step 2: Identify which inverse is already given and which you must compute

Read the question carefully to see which matrix's inverse is handed to you directly, and compute the inverse of the other matrix using the standard adjoint method ($\det$, cofactors, transpose, divide).

Step 3: Multiply in the reversed order — never the original order

Once both inverses are known, compute $B^{-1}A^{-1}$ (not $A^{-1}B^{-1}$) entry by entry, taking the dot product of each row of $B^{-1}$ with each column of $A^{-1}$.

Step 4: Sanity-check by matching the dimensions and, if time allows, spot-checking one entry

Since matrix multiplication order matters, a quick way to catch a mistake is to verify that multiplying your final answer by $AB$ (formed directly) gives the identity — even checking just one entry can reveal an order mistake.

Common Mistakes

Mistake 1: Multiplying in the original order, $A^{-1}B^{-1}$, instead of the reversed order

Why it's wrong: $(AB)^{-1}=A^{-1}B^{-1}$ is false in general because matrix multiplication does not commute — only $(AB)^{-1}=B^{-1}A^{-1}$ actually undoes $AB$. Correct approach: always write the reversal rule down first, then substitute — never multiply the two given/computed inverses in the same order they were named in the question.

Mistake 2: Sign errors while computing the "missing" matrix's inverse via cofactors

Why it's wrong: since this method needs a full adjoint computation for one of the two matrices, the usual $(-1)^{i+j}$ sign slip risk applies and will corrupt the final product even if the multiplication step is done correctly. Correct approach: verify each cofactor's sign against the checkerboard pattern before transposing into the adjoint.

Mistake 3: Arithmetic slips in the final $3\times3$ matrix multiplication

Why it's wrong: computing $B^{-1}A^{-1}$ requires nine separate dot-product entries, and a single misread number anywhere silently changes one entry of the final answer. Correct approach: compute one full row at a time, writing out each dot product explicitly rather than doing it mentally.