Explainer: Solving a System of Two Equations Using a Matrix Inverse

In this explainer, we will learn how to solve a system of two linear equations using the inverse of the matrix of coefficients.

The translation between systems and matrix equations is straightforward. Consider the system 2𝑥+3𝑦2𝑧=5,𝑥2𝑦+4𝑧=2,𝑥+2𝑦3𝑧=1.

We can check that the values 𝑥=12, 𝑦=7, 𝑧=1 solve this system by verifying that it produces identities 2(12)+3(7)2(1)=2421+2=5 and (12)2(7)+4(1)=12+144=2 and 12+2(7)3(1)=1214+3=1.

We can also write the system as an equation between two 3×1 matrices: 2𝑥+3𝑦2𝑧𝑥2𝑦+4𝑧𝑥+2𝑦3𝑧=521.

Then seeing the left matrix as a product of a 3×3 matrix and a 3×1 matrix separates the coefficients from the unknowns: 2𝑥+3𝑦2𝑧𝑥2𝑦+4𝑧𝑥+2𝑦3𝑧=232124123𝑥𝑦𝑧.

In other words, we have expressed our system as the matrix equation 232124123𝑥𝑦𝑧=521 with the form 𝑀𝑋=𝑅, where 𝑀 is a 3×3 matrix, while 𝑋 and 𝑅 are both 3×1 matrices. Here, 𝑋 is the (matrix of) unknowns and 𝑅 the right-hand side of the equation, and this is exactly like the equation 2𝑥=6, which has form 𝑚𝑥=𝑟 with numbers instead of matrices.

This is the simplest kind of linear equation and we know that provided the coefficient 𝑚 (here 2) is nonzero, we can solve this by multiplying the equation through by the multiplicative inverse of that coefficient: 2𝑥=6 is the same as 21(2𝑥)=21(6)𝑥=126=3.

We can use the same method here too. In the place of requiring that 𝑚0 so that 𝑚1 exists, we demand that 𝑀 have an inverse matrix 𝑀1 so that 𝑀1𝑀=𝐼, the 3×3 identity matrix.

Now that we have 𝑀=232124123,𝑀1=258146011, we continue to solve for 𝑋: 232124123𝑥𝑦𝑧=521 is the same as 𝑀1232124123𝑥𝑦𝑧=𝑀1521258146011232124123theidentitymatrix𝑥𝑦𝑧=258146011521𝑥𝑦𝑧=(2)(5)+(5)(2)+(8)(1)(1)(5)+(4)(2)+(6)(1)(0)(5)+(1)(2)+(1)(1)=1271, which gives the same solution, this time assembled into a 3×1 matrix.

Example 1: Expressing a Set of Simultaneous Equations as a Matrix Equation

Express the given set of simultaneous equations as a matrix equation. 7𝑥3𝑦+6𝑧=5,5𝑥2𝑦+2𝑧=11,2𝑥3𝑦+8𝑧=10.

Answer

The important thing is to be sure that the system of equations has the unknowns in the same order for each equation. Then the coefficient matrix 𝑀 is easy to read off the matrix equation: 736522238𝑥𝑦𝑧=51110.

Translating in the other direction is a similar exercise.

Example 2: Identifying a Set of Simultaneous Equations from a Matrix Equation

Write down the set of simultaneous equations that could be solved using the matrix equation 224111256𝑝𝑞𝑟=41410.

Answer

Writing the three components of the matrix equation, we get the following system on unknowns 𝑝,𝑞,𝑟: 2𝑝+2𝑞+4𝑟=4,𝑝𝑞𝑟=14,2𝑝5𝑞+6𝑟=10, which, of course, is the system of equations. As a list, it becomes 2𝑝+1𝑞+4𝑟=4,𝑝𝑞𝑟=14,2𝑝5𝑞+6𝑟=10.

Generally, the bulk of the work is in inverting the matrix.

Example 3: Solving a Matrix Equation by Finding the Inverse

Given that 166024247𝑥𝑦𝑧=829, find the values of 𝑥,𝑦, and 𝑧.

Answer

Of course, the solution exists provided that the 3×3 coefficient matrix above is invertible. This turns out to be the case, with 1660242471=111911611411522211211411111 such that 𝑥𝑦𝑧=111911611411522211211411111829=453.

Therefore, the system is solved by 𝑥=4,𝑦=5,𝑧=3.

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