Algebra 2
Matrices (introductory)
Solving systems using matrices
20 practice questions
0 video lessons
Theory + worked examples
Theory
A linear system can be written as a matrix equation:
\[AX=B\quad\Longrightarrow\quad X=A^{-1}B,\]
where \(A\) holds the coefficients, \(X\) the variables, and \(B\) the constants.
An inverse exists only when \(\det A\neq0\). If it is \(0\), there is no unique solution.
The system as \(AX=B\).
Solving with the inverse matrix.
Matrix solution:
\[AX=B\ \Rightarrow\ X=A^{-1}B,\qquad \det A\neq0\]
Technology finds \(A^{-1}\) for larger systems.
How to solve with matrices
- Write the system as \(AX=B\).
- Check \(\det A\neq0\).
- Find \(A^{-1}\).
- Compute \(X=A^{-1}B\).
Example 1 — Write the matrix equation
Write \(2x+y=5,\ x-y=1\) as \(AX=B\).
Solution
Coefficients, variables, constants.
| \(\begin{bmatrix}2&1\\1&-1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}\) | \(=\) | \(\begin{bmatrix}5\\1\end{bmatrix}\) |
Example 2 — Solve with the inverse
Solve the system from Example 1.
Solution
Use \(X=A^{-1}B\) (or substitution).
| \(x\) | \(=\) | \(2\) |
| \(y\) | \(=\) | \(1\) |
Example 3 — Determinant check
Is \(A=\begin{bmatrix}2&1\\1&-1\end{bmatrix}\) invertible?
Solution
Compute \(\det A=ad-bc\).
| \(\det A\) | \(=\) | \((2)(-1)-(1)(1)=-3\neq0\) |
Nonzero, so \(A\) is invertible.
Example 4 — When no inverse exists
What if \(\det A=0\)?
Solution
Then \(A^{-1}\) does not exist, and the system has no unique solution.
Common pitfalls
Order matters: \(X=A^{-1}B\), not \(BA^{-1}\).
Check the determinant before inverting.
\(\det A=0\) means no unique solution.
Frequently asked questions
How do you write a system as a matrix equation?
\(AX=B\), with \(A\) the coefficients, \(X\) the variables, \(B\) the constants.
How do you solve \(AX=B\)?
Multiply by the inverse: \(X=A^{-1}B\).
When does the inverse exist?
When the determinant of \(A\) is nonzero.
What if the determinant is zero?
There is no unique solution.
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