Systems of equations

What you will be able to do

After this lesson you can

• write a system of linear equations as a matrix equation,
• solve it with np.linalg.solve and check the answer,
• explain when a system has one solution, none, or many, and
• solve problems with three unknowns, such as the currents of a circuit.

Two equations, two lines

You know how to solve $2x + y = 7$ and $x - y = -1$ by hand: add them, to get $3x = 6$, so $x = 2$ and $y = 3$. Each equation is a straight line, and the solution is the point where the lines cross. With three or ten unknowns, solving by hand takes too long. NumPy does it in one line.

First, we write the system with a matrix. The coefficients make a table $A$, the unknowns make a column $x$, and the right sides make a column $b$:

Multiplying out the left side gives the two equations. We write this short as $Ax = b$.

Matrices in NumPy

Predict first. Is A * B the same as A @ B? Then run it.

• A matrix is a two-dimensional array, as in chapter 1.
• A * B multiplies place by place, as for every array: [[0 2] [3 0]].
• A @ B is the matrix product: each place is the sum of the products of a row of A and a column of B. Here it gives [[2 1] [4 3]]. These are different! The matrix product is the one that is needed for equations.
• A @ v multiplies a matrix by a vector: $1 \times 1 + 2 \times 1 = 3$ and $3 \times 1 + 4 \times 1 = 7$.
• A.T is the transpose (rows and columns exchanged). np.eye(2) is the identity matrix, which changes nothing in a product.
• np.linalg.det(A) is the determinant: $1 \times 4 - 2 \times 3 = -2$. It is a single number that says whether the system can be solved.

Solving

Predict first. The solution is $x = 2$, $y = 3$. What will A @ solution be? Then run it.

• np.linalg.solve(A, b) finds the vector $x$ for which $Ax = b$. The answer is [2. 3.], as we found by hand.
• Always check. A @ solution must give back b. The test np.allclose(A @ solution, b) says True. (It uses allclose because of the tiny rounding errors.)

Use solve and not the inverse: you may know that $x = A^{-1}b$, and np.linalg.inv(A) @ b gives the same answer, but solve is faster and more accurate.

Word problems

Three unknowns: a circuit

The solution is $I_1 = 3$ A, $I_2 = 2$ A and $I_3 = 1$ A. The program checks it another way: the 3 ohm and 6 ohm resistors in parallel have the resistance $1 / (1/3 + 1/6) = 2$ ohm, so the total resistance is $2 + 2 = 4$ ohm and the current is $12 / 4 = 3$ A. The two methods agree.

Set a to 0.5: the two lines have the same slope and never meet. That is a singular matrix.

No solution, or many solutions

The picture shows three things that can happen with two lines. In the middle and on the right, the lines are parallel: their slopes are equal, and the determinant is zero.

Common mistakes

• *Using ` for a matrix product.** Use @`.
• Putting the coefficients in a different order in different rows. Each column must belong to the same unknown.
• Forgetting to check the answer. np.allclose(A @ x, b).
• Using integers and comparing with ==. Use np.allclose.
• Ignoring a singular matrix. It tells you something about the problem.

Practice

Start with two quick questions. Each New numbers gives you a fresh one, so you can practise as long as you like.

Then the labs and the problems:

1. Warm-up. Open the lab Solve the system. You write a safe solver and a checker.
2. Build. Open the problem A diet of three foods and the problem No solution or many solutions.

Remember these

• $Ax = b$: a matrix of the coefficients, a column of the unknowns, a column of the right sides.
• A @ B is the matrix product. A * B is place by place.
• np.linalg.solve(A, b) solves, and np.allclose(A @ x, b) checks.
• Determinant not zero: one solution. Determinant zero: none or many, and NumPy raises a LinAlgError.

Check yourself

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

1. How do you write $x + 2y = 5$ and $3x - y = 1$ as $Ax = b$?

$,$x =

$,$b =

$.

2. What is the determinant of $

$, and what does it tell you?

3. Which is the matrix product, A * B or A @ B?

4. Why use np.allclose to check a solution?