How to Solve a System of Equations Using Matrices?

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A step-by-step guide to solving a system of equations using matrices

When solving a system of equations using matrices, we have three matrices \(A, B\), and \(X\), where \(A\) is known as the coefficient matrix, \(B\) is the constant matrix, and \(X\) contains all the variables of the equations, known as a variable matrix. Matrix \(A\) is of the order \(m×n\), while \(B\) is the column matrix of the order \(m×1\). The product of matrix \(A\) and matrix \(X\) reaches matrix \(B\). Hence, \(X\) is a column matrix of order \(n×1\).

The matrices are arranged as:

\(\color{blue}{A. X = B}\)

Let’s understand how to solve a system of equations using a matrix with the help of an example. We have a set of two equations given below. The equations are:

\(\begin{cases}x+y=6 \\ 2x+3y=14\end{cases}\)

Arrange all the coefficients, variables, and constants of the matrix in such a way that whenever we find the product of the matrices, the obtained result should be an equation. Then the matrix equation is, \(AX = B\) where:

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\(A= \begin{bmatrix}1 & 1 \\2 & 3 \end{bmatrix}\)

\(X= \begin{bmatrix}x \\y \end{bmatrix}\)

\(B= \begin{bmatrix}6 \\14 \end{bmatrix}\)

We need to find matrix \(X\), to solve the equations. It can be found by multiplying the inverse of matrix \(A\) with \(B\), which is obtained as \(X=\left(A^{-1}\right)B\).

To find the determinant of matrix \(A\), we follow the following steps:

\(|A|= \begin{bmatrix}1 & 1 \\2 & 3 \end{bmatrix}\)

Therefore, \(|A|= 3\: – 2 = 1\)

\(|A|≠0\), it is possible to find the inverse of matrix \(A\).

Now, by using the formula for finding the inverse of \(2×2\) matrix:

\(A^{-1}= \begin{bmatrix}3 & -1 \\-2 & 1 \end{bmatrix}\)

Now to find the matrix \(X\), we’ll multiply \(A^{-1}\) and \(B\). We get,

\(\begin{bmatrix}3 & -1 \\-2 & 1 \end{bmatrix}\)\(\begin{bmatrix}6 \\14 \end{bmatrix}\)\(=\begin{bmatrix}4\\2 \end{bmatrix}\)

So, the value of matrix \(X\) is, \(\begin{bmatrix}4\\2 \end{bmatrix}\).