Q.Use product $\begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix} \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}$ to solve the system of equations $x - y + 2z = 1$, $2y - 3z = 1$, $3x - 2y + 4z = 2$.

Concept: Inverse Matrix Method – If $AB = I$, then $B = A^{-1}$, and the system $A\mathbf{x} = \mathbf{b}$ has solution $\mathbf{x} = B\mathbf{b}$.

First, compute the given product:

So the second matrix is the inverse of the coefficient matrix $A = \begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}$. …

The core idea here is the Inverse Matrix Method. If you have a system written as $A\mathbf{x} = \mathbf{b}$, and you know $A^{-1}$, then the solution is simply $\mathbf{x} = A^{-1}\mathbf{b}$. The problem has cleverly given you the product of two matrices — if that product equals the identity matrix $I$, then the second matrix is $A^{-1}$. Let's check that first, then use it.

1. Compute the product to confirm it's the identity matrix.

Let $A = \begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}$.

Multiply $A \times B$:

• Row 1 × Column 1: $(1)(-2) + (-1)(9) + (2)(6) = -2 - 9 + 12 = 1$

• Row 1 × Column 2: $(1)(0) + (-1)(2) + (2)(1) = 0 - 2 + 2 = 0$

• Row 1 × Column 3: $(1)(1) + (-1)(-3) + (2)(-2) = 1 + 3 - 4 = 0$

• Row 2 × Column 1: $(0)(-2) + (2)(9) + (-3)(6) = 0 + 18 - 18 = 0$

• Row 2 × Column 2: $(0)(0) + (2)(2) + (-3)(1) = 0 + 4 - 3 = 1$

• Row 2 × Column 3: $(0)(1) + (2)(-3) + (-3)(-2) = 0 - 6 + 6 = 0$

• Row 3 × Column 1: $(3)(-2) + (-2)(9) + (4)(6) = -6 - 18 + 24 = 0$

• Row 3 × Column 2: $(3)(0) + (-2)(2) + (4)(1) = 0 - 4 + 4 = 0$

• Row 3 × Column 3: $(3)(1) + (-2)(-3) + (4)(-2) = 3 + 6 - 8 = 1$

So $A \times B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I$.

2. Write the system in matrix form.

The given equations:

This is $A\mathbf{x} = \mathbf{b}$ where:

3. Apply the inverse matrix method.

If $A\mathbf{x} = \mathbf{b}$, multiply both sides on the left by $A^{-1}$: …

Method: Recognizing a Given Matrix Product as the Inverse, Then Solving the System

This method uses a matrix product the problem hands you directly — if that product turns out to be the identity matrix, the second matrix IS the inverse of the first, and the system can be solved immediately without computing cofactors at all.

Steps

Step 1: Multiply the two given matrices

Carry out the full matrix multiplication, row by column, exactly as given — don't assume the result in advance.

Step 2: Check whether the product is the identity matrix

If every diagonal entry comes out $1$ and every off-diagonal entry comes out $0$, the product is $I$.

Step 3: Conclude which matrix is the inverse

If $A\cdot B=I$ for square matrices of the same size, then $B=A^{-1}$ (and also $A=B^{-1}$) — this is a direct consequence of the definition of an inverse, no separate adjoint computation needed.

Step 4: Identify the coefficient matrix $A$ and constant column $B$ from the given system

Match the system of equations to the matrix $A$ whose inverse you just found — write the system as $A\mathbf{x}=\mathbf{b}$. …

Common Mistakes

Mistake 1: Assuming which matrix is $A^{-1}$ without checking the product order

Why it's wrong: the problem gives $A\times B$; concluding $B=A^{-1}$ is only valid once $A\times B$ is actually confirmed to equal $I$ — assuming it without computing the product risks using the wrong matrix as the inverse. Correct approach: always compute the given product exactly as written and confirm it equals the identity matrix before using either matrix as an "inverse."

Mistake 2: Multiplying $\mathbf{b}$ and $A^{-1}$ in the wrong order …