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MatricesLinear algebraCollegeabout 0:49

Inverse of [[1,2,3],[0,1,4],[5,6,0]]: animated lesson

A narrated, step-by-step matrices video lesson with captions. Final answer: A⁻¹ = [[-24, 18, 5], [20, -15, -4], [-5, 4, 1]].

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
    [123014560]−1\begin{bmatrix}1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0\end{bmatrix}^{-1}
  2. 2. Steps, part 1
    1. det⁡A=1≠0\det A = 1 \neq 0
      Check that the determinant is non-zero
    2. [123100014010560001]\begin{bmatrix}1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & 4 & 0 & 1 & 0 \\ 5 & 6 & 0 & 0 & 0 & 1\end{bmatrix}
      Write the augmented matrix [A | I]
    3. R3→R3−5R1: [1231000140100−4−15−501]R_{3} \to R_{3} - 5R_{1}:\ \begin{bmatrix}1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & 4 & 0 & 1 & 0 \\ 0 & -4 & -15 & -5 & 0 & 1\end{bmatrix}
      Clear column 1 using the pivot in R1
    4. R1→R1−2R2, R3→R3+4R2: [10−51−20014010001−541]R_{1} \to R_{1} - 2R_{2},\ R_{3} \to R_{3} + 4R_{2}:\ \begin{bmatrix}1 & 0 & -5 & 1 & -2 & 0 \\ 0 & 1 & 4 & 0 & 1 & 0 \\ 0 & 0 & 1 & -5 & 4 & 1\end{bmatrix}
      Clear column 2 using the pivot in R2
  3. 3. Steps, part 2
    1. R1→R1+5R3, R2→R2−4R3: [100−2418501020−15−4001−541]R_{1} \to R_{1} + 5R_{3},\ R_{2} \to R_{2} - 4R_{3}:\ \begin{bmatrix}1 & 0 & 0 & -24 & 18 & 5 \\ 0 & 1 & 0 & 20 & -15 & -4 \\ 0 & 0 & 1 & -5 & 4 & 1\end{bmatrix}
      Clear column 3 using the pivot in R3
    2. A−1=[−2418520−15−4−541]A^{-1} = \begin{bmatrix}-24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1\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=[−2418520−15−4−541]A^{-1} = \begin{bmatrix}-24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1\end{bmatrix}

    A⁻¹ = [[-24, 18, 5], [20, -15, -4], [-5, 4, 1]]

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

Transcript

The problem. Here is the problem: [[1, 2, 3], [0, 1, 4], [5, 6, 0]]. 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]. Clear column 1 using the pivot in R1. Clear column 2 using the pivot in R2.

Steps, part 2. Clear column 3 using the pivot in R3. 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 [[-24, 18, 5], [20, negative 15, negative 4], [-5, 4, 1]]. The determinant is 1, 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 [[1,2,3],[0,1,4],[5,6,0]]: Matrices Lesson