Solving a System of Equations Using the Product of Two Matrices

This lesson solves a system of three linear equations in three unknowns using a supplied product of two matrices. We show the product is the identity, identify the inverse we need, and read off the solution.

The given product

We are given two matrices and asked to multiply them:

$P = \begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}, \qquad Q = \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}.$

The system to solve is

$\begin{aligned} x + 3z &= 9, \\ -x + 2y - 2z &= 4, \\ 2x - 3y + 4z &= -3. \end{aligned}$

Computing the product $PQ$

We multiply row by column. For example, the top-left entry is the first row of $P$ times the first column of $Q$:

$1(-2) + (-1)(9) + 2(6) = -2 - 9 + 12 = 1.$

The first row of $P$ against the second and third columns of $Q$ gives $0$ and $0$. Continuing this for every entry:

$PQ = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I.$

Since $PQ = I$, each matrix is the inverse of the other, so

$P^{-1} = Q.$

Writing the system in matrix form

The coefficient matrix of the system is

$A = \begin{bmatrix} 1 & 0 & 3 \\ -1 & 2 & -2 \\ 2 & -3 & 4 \end{bmatrix}, \qquad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \qquad B = \begin{bmatrix} 9 \\ 4 \\ -3 \end{bmatrix},$

so the system is $AX = B$ and therefore $X = A^{-1}B$.

Spotting the transpose

Notice that $A$ is exactly the transpose of $P$:

$A = P^{\mathsf T}.$

Inverse of a transpose

We use the rule that the inverse of a transpose is the transpose of the inverse:

$A^{-1} = \left(P^{\mathsf T}\right)^{-1} = \left(P^{-1}\right)^{\mathsf T} = Q^{\mathsf T}.$

Taking the transpose of $Q$:

$Q^{\mathsf T} = \begin{bmatrix} -2 & 9 & 6 \\ 0 & 2 & 1 \\ 1 & -3 & -2 \end{bmatrix}.$

Multiplying out for the solution

Now $X = A^{-1}B = Q^{\mathsf T}B$:

$\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -2 & 9 & 6 \\ 0 & 2 & 1 \\ 1 & -3 & -2 \end{bmatrix} \begin{bmatrix} 9 \\ 4 \\ -3 \end{bmatrix}.$

Multiplying row by column:

$\begin{aligned} x &= -2(9) + 9(4) + 6(-3) = -18 + 36 - 18 = 0, \\ y &= 0(9) + 2(4) + 1(-3) = 0 + 8 - 3 = 5, \\ z &= 1(9) + (-3)(4) + (-2)(-3) = 9 - 12 + 6 = 3. \end{aligned}$

So the solution is

$x = 0, \quad y = 5, \quad z = 3.$

Key takeaways

• If the product of two square matrices is the identity, then each is the inverse of the other.
• A system $AX = B$ is solved by $X = A^{-1}B$ whenever $A$ is invertible.
• The inverse of a transpose equals the transpose of the inverse, $\left(A^{\mathsf T}\right)^{-1} = \left(A^{-1}\right)^{\mathsf T}$, which lets you reuse an inverse you already know.