Determinants

Solving Systems of Linear Equations

Understand

A system such as \(a_1x + b_1y + c_1z = d_1\), … can be written as \(AX = B\), where \(A\) is the coefficient matrix, \(X\) the column of unknowns and \(B\) the column of constants.

Deciding the type of solution

If \(|A| \ne 0\): the system is consistent with the unique solution \(X = A^{-1}B\).
If \(|A| = 0\): compute \((\operatorname{adj}A)B\).
- If \((\operatorname{adj}A)B \ne O\): the system is inconsistent (no solution).
- If \((\operatorname{adj}A)B = O\): the system may have infinitely many solutions or none; examine the equations further.

Method

Write \(A\), \(X\), \(B\).
Find \(|A|\); check it is non-zero.
Find \(\operatorname{adj}A\) and \(A^{-1}\).
Compute \(X = A^{-1}B\) and check by substitution.

Key Concepts

Unique solution ⇔ \(|A| \ne 0\).
\(X = A^{-1}B\) – note the order.

Formula Bank

Matrix method

\[AX = B \;\Rightarrow\; X = A^{-1}B \quad (|A| \ne 0)\]

Note the order: \(A^{-1}\) multiplies \(B\) from the left.

Key Points

Verify by substitution

After solving AX = B, substitute the values back into one or two original equations. It takes 30 seconds and catches most errors.

Consistency test (AX = B)

\(|A|\neq0\): unique solution \(X=A^{-1}B\).
\(|A|=0\) and \((\operatorname{adj}A)B\neq O\): no solution (inconsistent).
\(|A|=0\) and \((\operatorname{adj}A)B=O\): infinitely many solutions or none – check further.

Common Mistakes

Writing X = BA⁻¹

Multiplying the constants column on the wrong side.

Correct: from \(AX = B\), pre-multiply by \(A^{-1}\): \(X = A^{-1}B\). \(BA^{-1}\) is not even defined for a 3 × 1 column \(B\).

Solved Examples

Solving a 3 × 3 system

Solve \(x + y + z = 4\), \(x - y + z = 2\), \(2x + y - z = 1\) by the matrix method.

\(A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 2 & 1 & -1 \end{bmatrix}\), \(B = \begin{bmatrix} 4 \\ 2 \\ 1 \end{bmatrix}\); \(|A| = 6\).
\(\operatorname{adj}A = \begin{bmatrix} 0 & 2 & 2 \\ 3 & -3 & 0 \\ 3 & 1 & -2 \end{bmatrix}\).
\(X = \frac{1}{6}\operatorname{adj}A\,B = \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}\).
Check in the first equation: 1 + 1 + 2 = 4. ✓

Final Answer: \(x = 1,\ y = 1,\ z = 2\)

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Topic Summary

\(AX = B\). \(|A| \ne 0\) ⇒ unique \(X = A^{-1}B\). \(|A| = 0\) and \((\operatorname{adj}A)B \ne O\) ⇒ no solution. Always verify by substitution.

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