Matrix math solver and matrix calculator with steps

What a matrix calculator with steps does differently

An ordinary matrix calculator prints a grid of numbers and stops. A matrix calculator with steps writes out each move with a one-line reason, the way a worked example in a textbook does, so you can find the line where your own work went wrong. The four examples below are real output for a 2×2 inverse, a 3×3 determinant, a matrix product and a linear system. Other operations, such as a transpose or eigenvalues, are not shown here; you can type them, and if the solver cannot do a problem it says so and the attempt does not use a free solve.

How to use it

1. Type the matrix with one pair of square brackets per row inside an outer pair, for example [[1, 2], [3, 4]].
2. Say what you want: find the inverse of, find the determinant of, multiply, or solve the system using a matrix.
3. For a linear system, type the equations separated by commas, such as 2x + y = 7, x - 3y = -7.
4. Read the steps, and fix any misread entry in the box.

Matrix math solver: worked examples with steps

Each example is the solver's own output, and each final answer was checked separately with a computer algebra system. Use the link under an example to run it on the board above.

1. Find the inverse of the matrix [[4, 7], [2, 6]]

1. Identify the matrix elementsA = [[a, b], [c, d]] = [[4, 7], [2, 6]]Assign variable labels to each element of the 2x2 matrix.
2. Calculate the determinantdet(A) = ad - bc = (4)(6) - (7)(2) = 24 - 14 = 10The determinant of a 2x2 matrix is computed using ad - bc. Since the determinant is non-zero, the matrix is invertible.
3. Set up the adjugate matrixadj(A) = [[d, -b], [-c, a]] = [[6, -7], [-2, 4]]Swap the diagonal elements and negate the off-diagonal elements to form the adjugate matrix.
4. Multiply by the reciprocal of the determinantA^(-1) = (1/10) * [[6, -7], [-2, 4]] = [[3/5, -7/10], [-1/5, 2/5]]Multiply each entry in the adjugate matrix by 1 over the determinant and simplify fractions.

Answer: [[3/5, -7/10], [-1/5, 2/5]]

Check: Multiply A by A^{-1}: [[4, 7], [2, 6]] * [[0.6, -0.7], [-0.2, 0.4]] = [[4(0.6)+7(-0.2), 4(-0.7)+7(0.4)], [2(0.6)+6(-0.2), 2(-0.7)+6(0.4)]] = [[2.4-1.4, -2.8+2.8], [1.2-1.2, -1.4+2.4]] = [[1, 0], [0, 1]], which yields the identity matrix.

Open this example on the board above →

2. Find the determinant of the 3x3 matrix [[2, 0, 1], [1, 3, 2], [1, 1, 1]]

1. Set up cofactor expansion along the first rowdet(A) = 2 * det([[3, 2], [1, 1]]) - 0 * det([[1, 2], [1, 1]]) + 1 * det([[1, 3], [1, 1]])Expanding along the first row takes advantage of the zero entry to simplify calculations.
2. Evaluate the first 2x2 minordet([[3, 2], [1, 1]]) = 3*1 - 2*1 = 1Use the formula ad - bc for the determinant of a 2x2 matrix.
3. Evaluate the second non-zero 2x2 minordet([[1, 3], [1, 1]]) = 1*1 - 3*1 = -2Apply the 2x2 determinant formula to the minor corresponding to the entry in row 1, column 3.
4. Combine the termsdet(A) = 2*(1) - 0 + 1*(-2) = 0Multiply each minor by its corresponding cofactor and sum the results.

Answer: 0

Check: Compute using Sarrus' rule: (2*3*1 + 0*2*1 + 1*1*1) - (1*3*1 + 2*2*1 + 0*1*1) = (6 + 0 + 1) - (3 + 4 + 0) = 7 - 7 = 0. Both methods agree.

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4. Solve the system using a matrix: 2x + y = 7, x - 3y = -7

1. Write the augmented matrix[ [2, 1 | 7], [1, -3 | -7] ]Represent the coefficients and constants of the linear system in an augmented matrix form.
2. Swap row 1 and row 2Swap R1 and R2: [ [1, -3 | -7], [2, 1 | 7] ]Placing a 1 in the top-left pivot position simplifies subsequent row operations.
3. Eliminate the x-term in row 2R2 -> R2 - 2*R1: [ [1, -3 | -7], [0, 7 | 21] ]Subtract 2 times the first row from the second row to create a zero in the first column.
4. Scale row 2 to make the pivot 1R2 -> (1/7)*R2: [ [1, -3 | -7], [0, 1 | 3] ]Divide row 2 by 7 to solve directly for y.
5. Eliminate the y-term in row 1R1 -> R1 + 3*R2: [ [1, 0 | 2], [0, 1 | 3] ]Add 3 times the second row to the first row to obtain reduced row echelon form (RREF).

Answer: x = 2, y = 3

Check: Substitute x = 2 and y = 3 into both equations: 2(2) + 3 = 7 (holds), and 2 - 3(3) = 2 - 9 = -7 (holds). Both equations are satisfied.

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Inverse of a 2×2 matrix in one paragraph

For a matrix with rows (a, b) and (c, d), compute the determinant ad − bc first. If it is zero the matrix has no inverse. Otherwise swap a and d, change the signs of b and c, and divide every entry by the determinant. Multiplying the original matrix by your answer should give the identity matrix, which is the check the solver writes in its last line.

When the determinant is zero

The second example has determinant 0, so its rows are linearly dependent and the matrix has no inverse. Expanding along a row that contains a zero, as the solver does there, saves work.

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