How to Solve a System of Equations Using Matrices?
Any system of linear equations can be written as a matrix equation. In this step-by-step guide, you learn how to solve a system of equations using matrices.
Solve a System of Equations Using Matrices: what to notice and how to work it
What to notice first
Common student mistake
Key formulas and cues
A reliable path
- Check dimensionsRows by columns determines what operation is legal.
- Use the correct ruleAddition is entry-by-entry; multiplication is row-by-column.
- Interpret the resultFor systems, translate the matrix answer back into variables.
Worked examples
Add matrices
- The matrices have the same size.
- Add matching entries.
- Compute each position.
Multiplication size
- Inner dimensions match: 3 and 3.
- The product is allowed.
- Outer dimensions give the result size.
Try one before moving on
Solve a System of Equations Using Matrices: pop-up practice
Related Topics
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 \(n×1\).
Type the problems from this lesson and check each step.
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:
\(A= \begin{bmatrix}1 & 1 \\2 & 3 \end{bmatrix}\)
\(X= \begin{bmatrix}x \\y \end{bmatrix}\)
\(B= \begin{bmatrix}6 \\14 \end{bmatrix}\)
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, \(\begin{bmatrix}4\\2 \end{bmatrix}\).
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