Matrix Power Calculator

Published on: October 19 2025
Final Answer: Free • Full Steps: Pro

This Matrix Power Calculator raises a square matrix to a whole-number power and shows the working line by line. A matrix power is repeated multiplication, so \(A^{3}\) means \(A\) multiplied by itself twice over, not each entry cubed. Only square matrices can be raised to a power, since the shapes have to keep matching at every step. Following the same pattern every time makes a matrix power easy to check and repeat.

Step-by-step method

  1. Set up the problem.
  2. Write the formula.
  3. Write the power out as repeated multiplication.
  4. Multiply one factor at a time.

Formula:

\(A^{n} = \underbrace{A \times A \times \cdots \times A}_{n \text{ factors}}\)
See Example 1 Hide Example 1

Problem

\(A^{2} = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}^{2}\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Write the formula.
  3. Write the power out as repeated multiplication.
  4. Multiply one factor at a time.

Step 1

Step 1 - Set up the problem.

In this problem: Matrix A is \(2 \times 2\), so it is square and can be multiplied by itself. We are raising it to the power \(2\).

\(A^{2} = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}^{2}\)

Step 2

Step 2 - Write the formula.

In this problem: A matrix power is repeated multiplication, so the matrix is multiplied by itself once for every unit of the exponent.

\(A^{n} = \underbrace{A \times A \times \cdots \times A}_{n \text{ factors}}\)

Step 3

Step 3 - Write the power out as repeated multiplication.

In this problem: Writing the power out in full puts \(A\) down \(2\) times.

\(A^{2} = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \times \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\)

Step 4

Step 4 - Multiply one factor at a time.

In this problem: Multiplying row by column: \(1 \cdot 1 + 2 \cdot 3 = 7\), \(1 \cdot 2 + 2 \cdot 4 = 10\), \(3 \cdot 1 + 4 \cdot 3 = 15\), \(3 \cdot 2 + 4 \cdot 4 = 22\).

\(A^{2} = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \times \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} 7 & 10 \\ 15 & 22 \end{pmatrix}\)

Final Answer

\(A^{2} = \begin{pmatrix} 7 & 10 \\ 15 & 22 \end{pmatrix}\)
See Example 2 Hide Example 2

Problem

\(A^{3} = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}^{3}\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Write the formula.
  3. Write the power out as repeated multiplication.
  4. Multiply one factor at a time.

Step 1

Step 1 - Set up the problem.

In this problem: Matrix A is \(2 \times 2\), so it is square and can be multiplied by itself. We are raising it to the power \(3\).

\(A^{3} = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}^{3}\)

Step 2

Step 2 - Write the formula.

In this problem: A matrix power is repeated multiplication, so the matrix is multiplied by itself once for every unit of the exponent.

\(A^{n} = \underbrace{A \times A \times \cdots \times A}_{n \text{ factors}}\)

Step 3

Step 3 - Write the power out as repeated multiplication.

In this problem: Writing the power out in full puts \(A\) down \(3\) times.

\(A^{3} = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix} \times \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix} \times \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}\)

Step 4

Step 4 - Multiply one factor at a time.

In this problem: \(2\) multiplications are still to be done, so each one is taken in turn.

\(A^{3} = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix} \times \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix} \times \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}\)

Step 4a

Step 4a - Multiply one factor at a time.

In this problem: Multiplying row by column: \(2 \cdot 2 + 1 \cdot 0 = 4\), \(2 \cdot 1 + 1 \cdot 3 = 5\), \(0 \cdot 2 + 3 \cdot 0 = 0\), \(0 \cdot 1 + 3 \cdot 3 = 9\).

\(A^{2} = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix} \times \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix} 4 & 5 \\ 0 & 9 \end{pmatrix}\)

Step 4b

Step 4b - Multiply one factor at a time.

In this problem: Multiplying row by column: \(4 \cdot 2 + 5 \cdot 0 = 8\), \(4 \cdot 1 + 5 \cdot 3 = 19\), \(0 \cdot 2 + 9 \cdot 0 = 0\), \(0 \cdot 1 + 9 \cdot 3 = 27\).

\(A^{3} = \begin{pmatrix} 4 & 5 \\ 0 & 9 \end{pmatrix} \times \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix} 8 & 19 \\ 0 & 27 \end{pmatrix}\)

Final Answer

\(A^{3} = \begin{pmatrix} 8 & 19 \\ 0 & 27 \end{pmatrix}\)
Matrix A

Example Problems

Click one to load it into the box and solve it.