Matrices and Systems of Linear Equations

Three equations in three unknowns can be solved by elimination, but the matrix method scales, and it answers a question elimination hides: whether a unique solution exists at all.

Matrix operations

Addition, subtraction and scalar multiplication work element by element, and need matrices of the same order. Multiplication takes a row from the left and a column from the right:

$$(m \times n) \times (n \times p) = (m \times p)$$

so two $3\times3$ matrices multiply to give another $3\times3$.

The identity matrix

$$I = \begin{pmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{pmatrix}$$

leaves any matrix unchanged: $AI = IA = A$. It is the matrix equivalent of the number 1, and the definition of an inverse is

$$AA^{-1} = A^{-1}A = I$$

The determinant of a 3 × 3 matrix

Expand along the first row, alternating signs:

$$\det\begin{pmatrix} a&b&c \\ d&e&f \\ g&h&i \end{pmatrix} = a\begin{vmatrix} e&f \\ h&i \end{vmatrix} - b\begin{vmatrix} d&f \\ g&i \end{vmatrix} + c\begin{vmatrix} d&e \\ g&h \end{vmatrix}$$

Each $2\times2$ determinant is the minor — what remains after deleting the row and column of that element.

Find $\det\begin{pmatrix} 2&1&3 \\ 1&0&2 \\ 4&1&8 \end{pmatrix}$.

$$= 2(0 \times 8 - 2 \times 1) - 1(1 \times 8 - 2 \times 4) + 3(1 \times 1 - 0 \times 4)$$ $$= 2(-2) - 1(0) + 3(1) = -4 + 0 + 3 = -1$$

And $\det\begin{pmatrix} 1&2&3 \\ 0&1&4 \\ 5&6&0 \end{pmatrix}$:

$$= 1(0 - 24) - 2(0 - 20) + 3(0 - 5) = -24 + 40 - 15 = 1$$

The inverse of a 3 × 3 matrix

$$A^{-1} = \frac{1}{\det A}\,\operatorname{adj}(A)$$

where the adjoint is built in three steps:

1. Replace each element by its minor.
2. Apply the sign pattern to get the cofactor matrix.
3. Transpose it.

For $A = \begin{pmatrix} 1&2&3 \\ 0&1&4 \\ 5&6&0 \end{pmatrix}$, with $\det A = 1$:

The cofactor matrix is

$$\begin{pmatrix} -24 & 20 & -5 \\ 18 & -15 & 4 \\ 5 & -4 & 1 \end{pmatrix}$$

Transposing gives the adjoint, and dividing by $\det A = 1$ leaves

$$A^{-1} = \begin{pmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{pmatrix}$$

Check by multiplying: $AA^{-1} = I$ ✓

Solving systems with matrices

$$\begin{aligned} x + y + z &= 6 \\ 2x - y + z &= 3 \\ x + 2y - z &= 2 \end{aligned}$$

becomes $A\mathbf{x} = \mathbf{b}$:

$$\begin{pmatrix} 1&1&1 \\ 2&-1&1 \\ 1&2&-1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 6 \\ 3 \\ 2 \end{pmatrix}$$

Multiplying on the left by $A^{-1}$:

$$\mathbf{x} = A^{-1}\mathbf{b}$$

The solution is $x = 1$, $y = 2$, $z = 3$.

Check in all three equations: $1+2+3 = 6$ ✓, $2-2+3 = 3$ ✓, $1+4-3 = 2$ ✓

Row reduction

For a single system, row reduction is quicker than inverting. Write the augmented matrix and use row operations to reach a triangle of zeros:

$$\left(\begin{array}{ccc|c} 1&1&1&6 \\ 2&-1&1&3 \\ 1&2&-1&2 \end{array}\right)$$

The three legal operations are: swap two rows, multiply a row by a non-zero constant, and add a multiple of one row to another. None of them changes the solution.

$R_2 - 2R_1$ and $R_3 - R_1$ give

$$\left(\begin{array}{ccc|c} 1&1&1&6 \\ 0&-3&-1&-9 \\ 0&1&-2&-4 \end{array}\right)$$

Then $3R_3 + R_2$ clears the last column:

$$\left(\begin{array}{ccc|c} 1&1&1&6 \\ 0&-3&-1&-9 \\ 0&0&-7&-21 \end{array}\right)$$

The last row reads $-7z = -21$, so $z = 3$. Back-substituting into row 2: $-3y - 3 = -9$, so $y = 2$. Then row 1 gives $x = 1$ ✓

Consistency of a system

A zero determinant means the equations are not independent — one is a combination of the others. Whether that leaves infinitely many solutions or none depends on the constants.

$$x + y = 3 \qquad 2x + 2y = 6$$

The second is just twice the first, so any point on the line satisfies both: infinitely many solutions.

$$x + y = 3 \qquad 2x + 2y = 7$$

The left sides are still proportional, but $2(x+y)$ must be 6, not 7. The equations contradict each other: no solution.