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Algebra 2 Matrices (introductory)

Solving systems using matrices

20 practice questions 0 video lessons Theory + worked examples

Solving Systems Using Matrices

California Algebra 2 • Standard A-REI.9 • Matrices

Solving Systems Using 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 a system of linear equations as a matrix equation and solve it using the inverse of the coefficient matrix.

A linear system can be written \(AX=B\) and solved with the inverse, \(X=A^{-1}B\), when the coefficient matrix has a nonzero determinant.

California Algebra 2 › Matrices › Solving Systems Using Matrices  —  Standard A-REI.9

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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.
A system as a matrix equation A linear system is written AX = B and solved with the inverse matrix. 2 1 1 -1 x y = 5 1 A X = B, then X = A⁻¹B
The system as \(AX=B\).
Solving A X = B Solving A X = B Solving A X = B write the system as A X = B A: coefficients, B: constants solve: X = A⁻¹ B needs det(A) ≠ 0
Solving with the inverse matrix.

Matrix solution:

\[AX=B\ \Rightarrow\ X=A^{-1}B,\qquad \det A\neq0\]
write the system as A X equals B and solve with the inverse of A
Technology finds \(A^{-1}\) for larger systems.

How to solve with matrices

  1. Write the system as \(AX=B\).
  2. Check \(\det A\neq0\).
  3. Find \(A^{-1}\).
  4. 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.

\(\begin{bmatrix}2&1\\1&-1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}\)\(=\)\(\begin{bmatrix}5\\1\end{bmatrix}\)
the system as a matrix equation
Example 2 — Solve with the inverse
Solve the system from Example 1.
Solution

Use \(X=A^{-1}B\) (or substitution).

\(x\)\(=\)\(2\)
\(y\)\(=\)\(1\)
x equals 2 and y equals 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.

yes, the determinant is negative 3, so it 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.

if the determinant is zero there is 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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