Use cramers rule to solve the system of equations calculator

This calculator solves Systems of Linear Equations using Gaussian Elimination Method, Inverse Matrix Method, or Cramer's rule. Also you can compute a number of solutions in a system of linear equations (analyse the compatibility) using Rouché–Capelli theorem.

Enter coefficients of your system into the input fields. Leave cells empty for variables, which do not participate in your equations. To input fractions use /: 1/3.

  • 2x-2y+z=-3 x+3y-2z=1 3x-y-z=2
  • Leave extra cells empty to enter non-square matrices.
  • You can use decimal (finite and periodic) fractions: 1/3, 3.14, -1.3(56), or 1.2e-4; or arithmetic expressions: 2/3+3*(10-4), (1+x)/y^2, 2^0.5 (=2), 2^(1/3), 2^n, sin(phi), or cos(3.142rad).
  • Use ↵ Enter, Space, ↑↓, ⌫, and Delete to navigate between cells, Ctrl⌘ Cmd+C/Ctrl⌘ Cmd+V to copy/paste matrices.
  • Drag-and-drop matrices from the results, or even from/to a text editor.
  • To learn more about matrices use Wikipedia.

Recall the traveling-based example from section two and the system of equations that we got from it after simplifying each line:

x + y + z = 26;

y - 2z = 6; and

2x - z = 12.

Let's also remind ourselves of the augmented coefficient matrix:

1 1 1 26
| 0 1 -2 6
2 0 -1 12

Before we move on to construct the four matrices used in Cramer's rule for 3x3 systems, let's take some time to describe how we can input the data into the Cramer's rule calculator.

We have three equations to deal with, so let's tell the calculator that by choosing the right option in the "Number of equations" field. This will turn our tool into a 3-variable system of equations solver and show us a picture of what such a system looks like, with a few mysterious symbols, like a₁ or c₂. These denote the coefficients of our system, i.e., the numbers that stand to the left of the variables in each line and the numbers to the right of the = sign.

This notation is listed in the calculator, where you can input the values from the problem that you want to solve. The a's correspond to the numbers next to x's, the b's to y's, the c's to z's, the d's are the numbers on the right, and the indices tell us the number of the row. Note that our Cramer's rule calculator accepts only linear equations. This means that we cannot, for example, have a quadratic one, or an expression with a variable under a square root.

Observe that those coefficients are also in our augmented coefficient matrix. In fact, it is enough to copy them into the Cramer's rule calculator. For example, the first row of the matrix has numbers 1, 1, 1, and 26. It corresponds to the first equation, which has coefficients denoted with the subscript . Therefore, we have:

a₁ = 1, b₁ = 1, c₁ = 1, d₁ = 26.

Similarly, the other two rows give us:

a₂ = 0, b₂ = 1, c₂ = -2, d₂ = 6; and

a₃ = 2, b₃ = 0, c₃ = -1, d₃ = 12.

Once you get all that data into the Cramer's rule calculator, it should spit out the values of the four determinants, followed by the solution to the system. Let's look how it figured that out.

As we've mentioned in the section above, Cramer's rule for 2x2 and 3x3 systems means that we have to calculate the determinants of a few matrices. The first one, the so-called main one, is simply the coefficient matrix that we've defined in section two:

1 1 1
W = | 0 1 -2
2 0 -1

To construct the other three, we'll need to exchange one column of this matrix with the fourth extra column of the augmented coefficient matrix, which in our case has numbers 26, 6, and 12. To get Wₓ, the x matrix, we put those numbers instead of the column corresponding to the x variable, namely the first one:

26 1 1
Wₓ = | 6 1 -2
12 0 -1

Similarly, we get

1 26 1
Wᵧ = | 0 6 -2
2 12 -1
1 1 26
Wz = | 0 1 6
2 0 12

All we need to do is use the determinant formula from the section above for all four of the matrices:

|W| = 1*1*(-1) + 1*(-2)*2 + 1*0*0 - 1*1*2 - 1*0*(-1) - 1*(-2)*0 = -7,

|Wₓ| = 26*1*(-1) + 1*(-2)*12 + 1*6*0 - 1*1*26 - 1*6*(-1) - 26*(-2)*0 = -56,

|Wᵧ| = 1*26*(-1) + 26*(-2)*2 + 1*0*12 - 1*6*2 - 26*0*(-1) - 1*(-2)*12 = -98,

|Wz| = 1112 + 162 + 2600 - 2612 - 1012 - 160 = -28.

Lastly, we use Cramer's rule for 3x3 systems and obtain the solution:

x = |Wₓ| / |W| = -56 / (-7) = 8,

y = |Wᵧ| / |W| = -98 / (-7) = 14,

z = |Wz| / |W| = -28 / (-7) = 4.

Going back to the problem we started with, this means that the bike is equal to 8, the car is equal to 14, and the plane is equal to 4.

It makes you want to do some traveling, doesn't it?