Solving linear equations with matrices

Every­thing in this course comes together here. A sys­tem of lin­ear equa­tions can be writ­ten as one matrix equa­tion $AX = B$, and then solved in three ways: by the inverse matrix, by Cramer's rule using deter­mi­nants, or by row oper­a­tions (the Gauss-Jor­dan method). The deter­mi­nant of $A$ also tells you in advance whether the sys­tem has exactly one solu­tion. This les­son works through each method on the same sys­tems so that you can com­pare them.

Writ­ing a sys­tem as a matrix equa­tion#

Take the sys­tem

$$2x + 3y = 8, \qquad x - y = -1$$

Col­lect the coef­fi­cients, the unknowns and the con­stants into matri­ces:

$$A = \begin{bmatrix} 2 & 3 \\ 1 & -1 \end{bmatrix}, \qquad X = \begin{bmatrix} x \\ y \end{bmatrix}, \qquad B = \begin{bmatrix} 8 \\ -1 \end{bmatrix}$$

Mul­ti­ply­ing out, $AX = \begin{bmatrix} 2x + 3y \\ x - y \end{bmatrix}$, so the sys­tem is exactly the sin­gle equa­tion $AX = B$. Here $A$ is called the coef­fi­cient matrix.

Method 1: the inverse matrix#

If $|A| \neq 0$, mul­ti­ply both sides of $AX = B$ on the left by $A^{-1}$:

$$A^{-1}AX = A^{-1}B \implies IX = A^{-1}B \implies X = A^{-1}B$$

The sys­tem then has exactly one solu­tion. (The mul­ti­pli­ca­tion must be on the left of both sides, because matrix mul­ti­pli­ca­tion is not com­mu­ta­tive.)

Exam­ple 1#

Solve $2x + 3y = 8$, $x - y = -1$ by the inverse matrix.

$|A| = -2 - 3 = -5 \neq 0$, so

$$\displaystyle A^{-1} = \frac{1}{-5}\begin{bmatrix} -1 & -3 \\ -1 & 2 \end{bmatrix} = \frac{1}{5}\begin{bmatrix} 1 & 3 \\ 1 & -2 \end{bmatrix}$$

$$\displaystyle X = A^{-1}B = \frac{1}{5}\begin{bmatrix} 1 & 3 \\ 1 & -2 \end{bmatrix}\begin{bmatrix} 8 \\ -1 \end{bmatrix} = \frac{1}{5}\begin{bmatrix} 8 - 3 \\ 8 + 2 \end{bmatrix} = \frac{1}{5}\begin{bmatrix} 5 \\ 10 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$$

So $x = 1$, $y = 2$. Check: $2 + 6 = 8$ and $1 - 2 = -1$.

Exam­ple 2#

Solve by the inverse matrix:

$$x + 2y + z = 8, \qquad 2x - y + z = 3, \qquad x + y - z = 0$$

Here $A = \begin{bmatrix} 1 & 2 & 1 \\ 2 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix}$ and $B = \begin{bmatrix} 8 \\ 3 \\ 0 \end{bmatrix}$.

$$|A| = 1(1 - 1) - 2(-2 - 1) + 1(2 + 1) = 0 + 6 + 3 = 9$$

Work­ing out the nine cofac­tors and trans­pos­ing gives

$$\text{adj}\,A = \begin{bmatrix} 0 & 3 & 3 \\ 3 & -2 & 1 \\ 3 & 1 & -5 \end{bmatrix}$$

$$\displaystyle X = \frac{1}{9}\begin{bmatrix} 0 & 3 & 3 \\ 3 & -2 & 1 \\ 3 & 1 & -5 \end{bmatrix}\begin{bmatrix} 8 \\ 3 \\ 0 \end{bmatrix} = \frac{1}{9}\begin{bmatrix} 0 + 9 + 0 \\ 24 - 6 + 0 \\ 24 + 3 + 0 \end{bmatrix} = \frac{1}{9}\begin{bmatrix} 9 \\ 18 \\ 27 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$$

So $x = 1$, $y = 2$, $z = 3$. Check in the first equa­tion: $1 + 4 + 3 = 8$.

Method 2: Cramer's rule#

Let $\Delta = |A|$, and let $\Delta_1$, $\Delta_2$, $\Delta_3$ be the deter­mi­nants obtained by replac­ing the first, sec­ond and third columns of $A$ by the col­umn of con­stants $B$. If $\Delta \neq 0$,

$$\displaystyle x = \frac{\Delta_1}{\Delta}, \qquad y = \frac{\Delta_2}{\Delta}, \qquad z = \frac{\Delta_3}{\Delta}$$

For two unknowns, use $\Delta_1$ and $\Delta_2$ only.

Exam­ple 3#

Solve $2x + 3y = 8$, $x - y = -1$ by Cramer's rule.

$$\Delta = \begin{vmatrix} 2 & 3 \\ 1 & -1 \end{vmatrix} = -5, \quad \Delta_1 = \begin{vmatrix} 8 & 3 \\ -1 & -1 \end{vmatrix} = -8 + 3 = -5, \quad \Delta_2 = \begin{vmatrix} 2 & 8 \\ 1 & -1 \end{vmatrix} = -2 - 8 = -10$$

So $\displaystyle x = \frac{-5}{-5} = 1$ and $\displaystyle y = \frac{-10}{-5} = 2$, as before.

Exam­ple 4#

Solve the sys­tem of Exam­ple 2 by Cramer's rule.

$\Delta = 9$ from Exam­ple 2. Replac­ing each col­umn in turn by $\begin{bmatrix} 8 \\ 3 \\ 0 \end{bmatrix}$:

$$\Delta_1 = \begin{vmatrix} 8 & 2 & 1 \\ 3 & -1 & 1 \\ 0 & 1 & -1 \end{vmatrix} = 8(1 - 1) - 2(-3 - 0) + 1(3 - 0) = 0 + 6 + 3 = 9$$

$$\Delta_2 = \begin{vmatrix} 1 & 8 & 1 \\ 2 & 3 & 1 \\ 1 & 0 & -1 \end{vmatrix} = 1(-3 - 0) - 8(-2 - 1) + 1(0 - 3) = -3 + 24 - 3 = 18$$

$$\Delta_3 = \begin{vmatrix} 1 & 2 & 8 \\ 2 & -1 & 3 \\ 1 & 1 & 0 \end{vmatrix} = 1(0 - 3) - 2(0 - 3) + 8(2 + 1) = -3 + 6 + 24 = 27$$

So $\displaystyle x = \frac{9}{9} = 1$, $\displaystyle y = \frac{18}{9} = 2$, $\displaystyle z = \frac{27}{9} = 3$.

Method 3: Gauss-Jor­dan (row oper­a­tions)#

Write the aug­mented matrix $[A \mid B]$: the coef­fi­cient matrix with the con­stants as an extra col­umn. Then use row oper­a­tions, which do not change the solu­tions:

• inter­change two rows;
• mul­ti­ply a row by a non-zero num­ber;
• add a mul­ti­ple of one row to another.

Aim to turn the left part into $I$; the last col­umn then shows the solu­tion.

Exam­ple 5#

Solve the sys­tem of Exam­ple 2 by the Gauss-Jor­dan method.

$$\left[\begin{array}{ccc|c} 1 & 2 & 1 & 8 \\ 2 & -1 & 1 & 3 \\ 1 & 1 & -1 & 0 \end{array}\right]$$

$R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - R_1$:

$$\left[\begin{array}{ccc|c} 1 & 2 & 1 & 8 \\ 0 & -5 & -1 & -13 \\ 0 & -1 & -2 & -8 \end{array}\right]$$

Inter­change $R_2$ and $R_3$, then mul­ti­ply the new $R_2$ by $-1$:

$$\left[\begin{array}{ccc|c} 1 & 2 & 1 & 8 \\ 0 & 1 & 2 & 8 \\ 0 & -5 & -1 & -13 \end{array}\right]$$

$R_1 \to R_1 - 2R_2$ and $R_3 \to R_3 + 5R_2$:

$$\left[\begin{array}{ccc|c} 1 & 0 & -3 & -8 \\ 0 & 1 & 2 & 8 \\ 0 & 0 & 9 & 27 \end{array}\right]$$

$\displaystyle R_3 \to \frac{1}{9}R_3$, then $R_1 \to R_1 + 3R_3$ and $R_2 \to R_2 - 2R_3$:

$$\left[\begin{array}{ccc|c} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 3 \end{array}\right]$$

So $x = 1$, $y = 2$, $z = 3$.

When is there exactly one solu­tion?#

• If $|A| \neq 0$, the sys­tem has exactly one solu­tion, and all three meth­ods find it.
• If $|A| = 0$, the inverse and Cramer's rule can­not be used. The sys­tem then has either no solu­tion or infi­nitely many solu­tions. For exam­ple, $x + 2y = 3$ and $2x + 4y = 7$ have $|A| = 4 - 4 = 0$; dou­bling the first gives $2x + 4y = 6$, which con­tra­dicts the sec­ond, so there is no solu­tion. But $x + 2y = 3$ and $2x + 4y = 6$ are the same equa­tion, with infi­nitely many solu­tions.

Exam­ple 6#

Sita buys $2$ pens and $3$ note­books for ₹110, and Ravi buys $3$ pens and $1$ note­book for ₹60. Find the price of a pen and of a note­book.

Let a pen cost ₹$p$ and a note­book ₹$n$: $2p + 3n = 110$ and $3p + n = 60$.

$$\Delta = \begin{vmatrix} 2 & 3 \\ 3 & 1 \end{vmatrix} = 2 - 9 = -7, \quad \Delta_1 = \begin{vmatrix} 110 & 3 \\ 60 & 1 \end{vmatrix} = 110 - 180 = -70, \quad \Delta_2 = \begin{vmatrix} 2 & 110 \\ 3 & 60 \end{vmatrix} = 120 - 330 = -210$$

So $\displaystyle p = \frac{-70}{-7} = 10$ and $\displaystyle n = \frac{-210}{-7} = 30$: a pen costs ₹10 and a note­book ₹30.

Prac­tice#

1. Solve $3x - 2y = 4$, $x + 4y = 6$ by the inverse matrix.
2. Solve $5x + 2y = 1$, $3x + y = 1$ by Cramer's rule.
3. Solve $x + y + z = 9$, $2x - y + z = 5$, $x + 2y - z = 4$ by Cramer's rule.
4. Solve $2x + y - z = 3$, $x - y + z = 6$, $x + 2y + 3z = 7$ by any method.
5. Does the sys­tem $x + 2y = 3$, $2x + 4y = 5$ have a unique solu­tion? Explain.
6. Two kg of apples and one kg of oranges cost ₹130; one kg of apples and three kg of oranges cost ₹140. Find the price per kg of each.
7. The ages of Ravi, Sita and Gopi add up to $66$. Ravi is $4$ years older than Sita, and Sita is $7$ years older than Gopi. Find their ages using matri­ces.
8. Solve the sys­tem of ques­tion 3 by the Gauss-Jor­dan method.

Answers#