Solving linear systems with matrices
Solving Linear Systems with Matrices
Solving Linear Systems with Matrices is a topic in Matrices in the California Common Core State Standards. It is aligned to Standard A-REI.9, which requires students to represent and solve a system of linear equations using matrices.
A linear system can be written \(AX=B\) and solved by \(X=A^{-1}B\) when the coefficient matrix is invertible, or by row-reducing an augmented matrix.
Theory
A linear system can be packaged as a single matrix equation:
where \(A\) holds the coefficients, \(X\) the variables, and \(B\) the constants. If \(A\) is invertible, multiply both sides by \(A^{-1}\):
Alternatively, form the augmented matrix \([A\,|\,B]\) and solve by row reduction.
The matrix-equation method:
How to solve a system with matrices
- Write the system as \(AX=B\).
- Check \(\det A\neq 0\).
- Compute \(A^{-1}\) and multiply: \(X=A^{-1}B\).
- Verify the solution in the original equations.
Coefficients form \(A\), variables \(X\), constants \(B\).
\(\det A=2(3)-1(1)=5\), so \(A^{-1}=\dfrac{1}{5}\begin{bmatrix}3&-1\\-1&2\end{bmatrix}\).
So \(x=1,\ y=3\).
Substitute into both equations.
| \(2(1)+3\) | \(=\) | \(5\ \checkmark\) |
| \(1+3(3)\) | \(=\) | \(10\ \checkmark\) |
\(A\) has no inverse, so \(X=A^{-1}B\) fails; the system has either no solution or infinitely many.
| \(\det A=0\) | \(\Rightarrow\) | \(\text{no unique solution}\) |
Common pitfalls
Frequently asked questions
How do you write a system as a matrix equation?
Put the coefficients in \(A\), the variables in \(X\), and the constants in \(B\), giving \(AX=B\).
How do you solve AX = B with an inverse?
If \(A\) is invertible, \(X=A^{-1}B\). Compute the inverse and multiply.
What if the determinant is zero?
Then \(A\) has no inverse and the system has either no solution or infinitely many.
What is an augmented matrix?
The coefficient matrix with the constants attached as an extra column, \([A\,|\,B]\), used for row reduction.