Solving matrix equations and systems | Year 11

01 / LEARN

Understand the idea.

Build the coefficient matrix

Choose a fixed variable order, such as $x$ then $y$ , in every equation. Put their coefficients in the matching columns of $A$ , the unknowns in a column $X$ , and the constants in a column $b$ to obtain $A X = b$ .

Keep a zero coefficient when a variable is missing. The matrix positions must still represent the same variables in each row. Expanding $A X$ back into equations is a useful check that the model matches the original system.

Undo on the left

When $A$ is invertible, multiply $A X = b$ on the left by $A^{- 1}$ . The adjacent factors $A^{- 1} A$ become $I$ , giving $X = A^{- 1} b$ . Check the resulting values in every original equation.

The side of multiplication matters. For a matrix unknown, $A X = B$ gives $X = A^{- 1} B$ , whereas $X A = B$ gives $X = B A^{- 1}$ . There is no general division operation that lets you ignore the order.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Solve a pair

Solve $2 x + y = 7$ and $3 x + 2 y = 12$ .

Worked solution

1
Write the coefficient model in variable order.
$(\begin{array}{cccccccccc} 2 & 1 \\ 3 & 2 \end{array}) (\begin{array}{cccccccccc} x \\ y \end{array}) = (\begin{array}{cccccccccc} 7 \\ 12 \end{array})$
Both rows use the order $x, y$ , so each row reproduces one original equation.
2
Calculate the inverse of the coefficient matrix.
$det A = 2 (2) - 1 (3) = 1, A^{- 1} = (\begin{array}{cccccccccc} 2 & - 1 \\ - 3 & 2 \end{array})$
The non-zero determinant permits the inverse formula.
3
Multiply the constant column on the left.
$X = A^{- 1} b = (\begin{array}{cccccccccc} 2 (7) - 12 \\ - 3 (7) + 2 (12) \end{array}) = (\begin{array}{cccccccccc} 2 \\ 3 \end{array})$
Left multiplication cancels the coefficient matrix in $A X = b$ .
4
Check both original equations.
$\begin{aligned} 2 (2) + 3 & = 7 \\ 3 (2) + 2 (3) & = 12 \end{aligned}$
The pair $x = 2, y = 3$ satisfies both constraints.

Answer

$x = 2$ , $y = 3$ .

EXAMPLE 02

Solve for a whole matrix

Solve $(\begin{array}{cccccccccc} 1 & 1 \\ 0 & 1 \end{array}) X = (\begin{array}{cccccccccc} 3 & 5 \\ 2 & 4 \end{array})$ .

Worked solution

1
Find the inverse of the left-hand factor.
$A = (\begin{array}{cccccccccc} 1 & 1 \\ 0 & 1 \end{array}), det A = 1, A^{- 1} = (\begin{array}{cccccccccc} 1 & - 1 \\ 0 & 1 \end{array})$
The determinant is non-zero, and the inverse undoes the upper-right mixing.
2
Multiply both sides on the left.
$X = A^{- 1} B = (\begin{array}{cccccccccc} 1 & - 1 \\ 0 & 1 \end{array}) (\begin{array}{cccccccccc} 3 & 5 \\ 2 & 4 \end{array})$
The inverse must sit beside $A$ to produce $A^{- 1} A = I$ .
3
Evaluate each row-column product.
$X = (\begin{array}{cccccccccc} 3 - 2 & 5 - 4 \\ 2 & 4 \end{array}) = (\begin{array}{cccccccccc} 1 & 1 \\ 2 & 4 \end{array})$
The first output row subtracts the second row of $B$ from its first; the second row is retained.
4
Verify the original matrix equation.
$\begin{aligned} A X & = (\begin{array}{cccccccccc} 1 & 1 \\ 0 & 1 \end{array}) (\begin{array}{cccccccccc} 1 & 1 \\ 2 & 4 \end{array}) \\ = (\begin{array}{cccccccccc} 1 + 2 & 1 + 4 \\ 2 & 4 \end{array}) \\ = (\begin{array}{cccccccccc} 3 & 5 \\ 2 & 4 \end{array}) \end{aligned}$
The product recovers every entry of the specified right-hand matrix.

Answer

$X = (\begin{array}{cccccccccc} 1 & 1 \\ 2 & 4 \end{array})$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

If $X A = B$ and A is invertible, what is X?

Show answer & reasoning

$X = B A^{- 1}$ .

Right-multiply $X A = B$ by $A^{- 1}$ : $X (A A^{- 1}) = B A^{- 1}$ , so $X = B A^{- 1}$ . Multiplying on the left would not place the inverse beside $A$ .

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Write $3 x - y = 4$ , $2 x + 5 y = 1$ as one matrix equation.
2 marks
Solve $x + y = 5$ , $x - y = 1$ .
2 marks
Solve $2 x + 3 y = 1$ , $x + 2 y = 2$ .
2 marks
Solve $(\begin{array}{cccccccccc} 2 & 0 \\ 0 & 3 \end{array}) X = (\begin{array}{cccccccccc} 8 \\ 9 \end{array})$ .
2 marks
Solve $X (\begin{array}{cccccccccc} 1 & 2 \\ 0 & 1 \end{array}) = (\begin{array}{cccccccccc} 3 & 7 \\ 1 & 4 \end{array})$ .
2 marks
A ticket model gives $2 a + b = 17$ and $a + 2 b = 19$ . Find $a$ and $b$ and check both equations.
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Write $3 x - y = 4$ , $2 x + 5 y = 1$ as one matrix equation.
2 marks
Keep $x$ first and $y$ second in both rows.
$(\begin{array}{cccccccccc} 3 & - 1 \\ 2 & 5 \end{array}) (\begin{array}{cccccccccc} x \\ y \end{array}) = (\begin{array}{cccccccccc} 4 \\ 1 \end{array})$ .
Solve $x + y = 5$ , $x - y = 1$ .
2 marks
Add the equations to get $2 x = 6$ .
Then $y = 5 - 3 = 2$ ; the matrix method gives the same result.
$x = 3$ , $y = 2$ .
Solve $2 x + 3 y = 1$ , $x + 2 y = 2$ .
2 marks
The determinant is 1.
The inverse $(\begin{array}{cccccccccc} 2 & - 3 \\ - 1 & 2 \end{array})$ applied to ${(1, 2)}^{T}$ gives ${(- 4, 3)}^{T}$ .
$x = - 4$ , $y = 3$ .
Solve $(\begin{array}{cccccccccc} 2 & 0 \\ 0 & 3 \end{array}) X = (\begin{array}{cccccccccc} 8 \\ 9 \end{array})$ .
2 marks
Undo the two diagonal scales.
$X = (\begin{array}{cccccccccc} 4 \\ 3 \end{array})$ .
Solve $X (\begin{array}{cccccccccc} 1 & 2 \\ 0 & 1 \end{array}) = (\begin{array}{cccccccccc} 3 & 7 \\ 1 & 4 \end{array})$ .
2 marks
Multiply the right-hand matrix on the right by $(\begin{array}{cccccccccc} 1 & - 2 \\ 0 & 1 \end{array})$ .
$X = (\begin{array}{cccccccccc} 3 & 1 \\ 1 & 2 \end{array})$ .
A ticket model gives $2 a + b = 17$ and $a + 2 b = 19$ . Find $a$ and $b$ and check both equations.
2 marks
Eliminate to obtain $3 a = 15$ .
Then $b = 17 - 10 = 7$ ; the second equation gives $5 + 14 = 19$ .
$a = 5$ , $b = 7$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics: matrices and transformations; original lessons follow the approved school-sized chapter structure.

Official curriculum source ↗

In this lesson

Understand the idea.

Build the coefficient matrix

Undo on the left

See the method in action.

Solve a pair

Write the coefficient model in variable order.

Calculate the inverse of the coefficient matrix.

Multiply the constant column on the left.

Check both original equations.

Solve for a whole matrix

Find the inverse of the left-hand factor.

Multiply both sides on the left.

Evaluate each row-column product.

Verify the original matrix equation.

Try it yourself.

Check your understanding

Keep practising