Solve 3x + y = 5, x + y = 3 using the inverse matrix Step by Step

Step by step

The question

Question: solve 3x + y = 5 and x + y = 3 using the inverse matrix.

3x + y = 5

We want the x and y that make both true.

x + y = 3

1 Write it as AX = B

The numbers in front of x and y make the matrix A.

The letters make X.

The right sides make B.

Row times column gives back the equations.

[3, 1; 1, 1] × [x; y] = [5; 3]

To get X alone, we multiply by the inverse of A: X = A⁻¹B.

AX = B → X = A⁻¹B

Careful!

The inverse goes in front of B.

With matrices, the order matters.

Careful: X = A⁻¹B, not BA⁻¹

2 Find the inverse

First D, the determinant of A.

D = |3, 1; 1, 1|

So D = 3 × 1 − 1 × 1.

D = 3 × 1 − 1 × 1

3 times 1 is 3.

1 times 1 is 1.

D = 3 − 1

3 minus 1 is 2.

So D = 2.

D = 2

Swap the top left and bottom right.

Change the signs of the other two.

Then times 1 over D.

A⁻¹ = 12 × [1, −1; −1, 3]

So A⁻¹ = rows 1/2, −1/2 and −1/2, 3/2.

A⁻¹ = [1 over 2, −1 over 2; −1 over 2, 3 over 2]

3 Multiply

X is A⁻¹ times B.

[x; y] = [1 over 2, −1 over 2; −1 over 2, 3 over 2] × [5; 3]

= [1 over 2 × 5 + (−1 over 2) × 3; −1 over 2 × 5 + 3 over 2 × 3]

Work them out: x = 1 and y = 2.

[x; y] = [1; 2]

So the answer is (1, 2).

(x, y) = (1, 2)

4 Check

Let's check in every equation.

We put x = 1, y = 2 into the first one.

Equation 1 3(1) + 2 = 3 + 2 = 5

We get 5, the same as the right side.

Now equation 2.

Equation 2 1 + 2 = 3

We get 3, the same as the right side.

Both work.

So the answer is right!

Let's look back.

First: write it as AX = B.

Then: find the inverse.

Then: multiply.

Last: check.

Answer