Solving systems with the inverse and Cramer's rule

What you will be able to do

• write a system of linear equations as $AX = B$;
• solve a system with a non-zero determinant by the inverse;
• decide whether a system has one solution, none or infinitely many;
• use Cramer's rule, as taught in A-Level and IB.

A system as one matrix equation

A shop sells pens and notebooks. Two pens and one notebook cost $11$ rupees. Five pens and three notebooks cost $28$ rupees. With $p$ for the price of a pen and $n$ for a notebook:

Matrix multiplication packs both equations into one:

$A$ is the coefficient matrix, $X$ holds the unknowns and $B$ the constants.

Solving with the inverse

If $|A| \ne 0$, multiply both sides on the left by $A^{-1}$:

Here $|A| = 6 - 5 = 1$ and $A^{-1} =

$. So

A pen costs $5$ rupees and a notebook costs $1$ rupee. Always multiply by $A^{-1}$ on the left of $B$, because the order of matrix products matters.

The same method works for three unknowns. A museum sells child tickets ($x$ euros), student tickets ($y$ euros) and adult tickets ($z$ euros). Three groups pay $110$, $80$ and $60$ euros:

The determinant is $1$ and $A^{-1} =

$, so$X = A^{-1}B = (110 - 160 + 60,\ 80 - 60,\ -110 + 80 + 60) = (10, 20, 30)$. A child ticket costs$10$euros, a student ticket$20$and an adult ticket$30$.

The three cases

For two unknowns, each equation is a straight line. Two lines can cross at one point, be parallel, or be the same line. These are the three cases.

• $|A| \ne 0$: one solution. The lines cross, and $X = A^{-1}B$ is the crossing point. The system is consistent, and the solution is unique.
• $|A| = 0$ and the equations contradict each other: no solution. The system is inconsistent. Example: $x + y = 2$ and $2x + 2y = 5$.
• $|A| = 0$ and the equations agree: infinitely many solutions. Example: $x + y = 2$ and $2x + 2y = 4$. Every point of the line is a solution.

When $|A| = 0$, compute $(\text{adj}\,A)B$. If it is not the zero column, there is no solution. If it is the zero column, there may be no solution or infinitely many, and you must look at the equations directly.

For three unknowns, each equation is a plane. Take $x + y + z = 6$ and $x + 2y + 3z = 14$ with a third equation $2x + 3y + 4z = c$. The coefficients of the third equation are the sum of the first two, so $|A| = 0$. If $c = 20$ (the sum $6 + 14$), the third plane contains the line where the first two meet: infinitely many solutions $(t - 2,\ 8 - 2t,\ t)$ for any number $t$. If $c = 21$, the three planes have no common point and there is no solution. A-Level Further Maths and IB call the two situations a sheaf (or a common line) and a prism (no common point).

Homogeneous systems

A system with all constants $0$ is homogeneous: $AX = O$. It always has the solution $X = O$. It has a non-zero solution exactly when $|A| = 0$. For example, $x + 2y = 0$ and $3x + ky = 0$ has non-zero solutions only when $k - 6 = 0$, that is, $k = 6$.

Cramer's rule

A-Level Further Maths and IB write the solution with determinants. If $|A| \ne 0$, then

where $\Delta = |A|$, and $\Delta_i$ is $|A|$ with column $i$ replaced by the constants $B$. The rule follows from $X = \dfrac{1}{|A|}(\text{adj}\,A)B$. For the pens and notebooks,

Cramer's rule is neat for two unknowns. For more it needs many determinants, so the inverse or row operations are quicker. It does not work at all when $\Delta = 0$.

A worked example

For which values of $k$ does $x + 2y = 3$, $2x + ky = 6$ have a unique solution? For $k = 4$, what happens?

1. The determinant. $|A| = k - 4$.
2. A unique solution needs $|A| \ne 0$, so $k \ne 4$.
3. For $k = 4$. The second equation is $2x + 4y = 6$, which is twice the first. The lines are the same line, so there are infinitely many solutions.

Practise with new numbers

Try it in Python

The program solves $AX = B$ when $|A| \ne 0$. Otherwise it compares the rank of $A$ with the rank of the augmented matrix, to tell no solution from infinitely many. It runs the museum system and the plane system of the lesson.

Common mistakes

• Multiplying on the wrong side. The solution is $A^{-1}B$, not $BA^{-1}$.
• Using the inverse when $|A| = 0$. There is no inverse. Decide between no solution and infinitely many.
• Saying that $|A| = 0$ always means no solution. It may mean infinitely many.
• Mixing the order of the unknowns. Keep $x, y, z$ in the same order in every equation.
• Dividing by $\Delta = 0$ in Cramer's rule. The rule only works when $\Delta \ne 0$.

Remember these

• $AX = B$ has the unique solution $X = A^{-1}B$ when $|A| \ne 0$.
• When $|A| = 0$, the system has no solution or infinitely many.
• $AX = O$ has a non-zero solution exactly when $|A| = 0$.
• Cramer's rule: $x_i = \Delta_i / \Delta$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Write $2x + y = 5$ and $3x + 2y = 8$ as $AX = B$ and solve it with the inverse.

$.

=

$, so$x = 2$and$y = 1$.

2. Does $x + y = 2$, $2x + 2y = 5$ have a solution?

3. For which $k$ does $x + 2y = 3$, $2x + ky = 6$ have infinitely many solutions?

4. For which $k$ does $x + 2y = 0$, $3x + ky = 0$ have a non-zero solution?

5. Use Cramer's rule to solve $x + y = 5$, $x - y = 1$.

= -2$.

}{-2} = \dfrac{-6}{-2} = 3$and$y = \dfrac{

}{-2} = \dfrac{-4}{-2} = 2$.

References

• NCERT, Mathematics Textbook for Class XII, Part I, Chapter 4 (Determinants), the section on applications to systems of linear equations.
• OpenStax, Precalculus 2e, Section 9.7 (Solving Systems with Inverses) and Section 9.8 (Solving Systems with Cramer's Rule).