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MatricesLinear algebraCollegeabout 0:44

Inverse of [[2,0,0],[0,3,0],[0,0,4]]: animated lesson

A narrated, step-by-step matrices video lesson with captions. Final answer: A⁻¹ = [[1/2, 0, 0], [0, 1/3, 0], [0, 0, 1/4]].

Video with a synthetic English voice and captions. Pick another language for subtitles in that language, or open the interactive version.

Video narrated by a synthetic voice (Piper TTS, en_US-ljspeech-high voice (LJ Speech dataset, public domain)); the steps come from our maths engine and the answer was checked independently. Download MP4 (0.3 MB) · Captions (WebVTT)

Steps on screen

  1. 1. The problem
    Worked example
    [200030004]−1\begin{bmatrix}2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4\end{bmatrix}^{-1}
  2. 2. Steps, part 1
    1. det⁡A=24≠0\det A = 24 \neq 0
      Check that the determinant is non-zero
    2. [200100030010004001]\begin{bmatrix}2 & 0 & 0 & 1 & 0 & 0 \\ 0 & 3 & 0 & 0 & 1 & 0 \\ 0 & 0 & 4 & 0 & 0 & 1\end{bmatrix}
      Write the augmented matrix [A | I]
    3. R1→12R1: [1001200030010004001]R_{1} \to \tfrac{1}{2} R_{1}:\ \begin{bmatrix}1 & 0 & 0 & \frac{1}{2} & 0 & 0 \\ 0 & 3 & 0 & 0 & 1 & 0 \\ 0 & 0 & 4 & 0 & 0 & 1\end{bmatrix}
      Divide R1 by 2
    4. R2→13R2: [10012000100130004001]R_{2} \to \tfrac{1}{3} R_{2}:\ \begin{bmatrix}1 & 0 & 0 & \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 & 0 & \frac{1}{3} & 0 \\ 0 & 0 & 4 & 0 & 0 & 1\end{bmatrix}
      Divide R2 by 3
  3. 3. Steps, part 2
    1. R3→14R3: [100120001001300010014]R_{3} \to \tfrac{1}{4} R_{3}:\ \begin{bmatrix}1 & 0 & 0 & \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 & 0 & \frac{1}{3} & 0 \\ 0 & 0 & 1 & 0 & 0 & \frac{1}{4}\end{bmatrix}
      Divide R3 by 4
    2. A−1=[120001300014]A^{-1} = \begin{bmatrix}\frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{4}\end{bmatrix}
      The right half is now the inverse
    3. AA−1=[100010001]A A^{-1} = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}
      Check: A·A⁻¹ = I
  4. 4. The answer
    A−1=[120001300014]A^{-1} = \begin{bmatrix}\frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{4}\end{bmatrix}

    A⁻¹ = [[1/2, 0, 0], [0, 1/3, 0], [0, 0, 1/4]]

  5. 5. Check
    • Multiplying A by the result gives the identity matrix exactly.

Transcript

The problem. Here is the problem: [[2, 0, 0], [0, 3, 0], [0, 0, 4]]. Find the inverse of a 3 times 3 matrix.

Steps, part 1. Check that the determinant is non-zero. Write the augmented matrix [A | I]. Divide R1 by 2. Divide R2 by 3.

Steps, part 2. Divide R3 by 4. The right half is now the inverse. Check: A times A to the power minus 1 equals I.

The answer. So the answer is A to the power minus 1 equals [[1 over 2, 0, 0], [0, 1 over 3, 0], [0, 0, 1 over 4]]. The determinant is 24, so A is invertible. Every entry of the inverse is exact.

Check. Multiplying A by the result gives the identity matrix exactly.

About matrices

Determinants, inverses, products, transposes and eigenvalues of matrices.

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Inverse of [[2,0,0],[0,3,0],[0,0,4]]: Matrices Lesson