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Determinant of [[1,0,2],[2,1,0],[0,3,1]]: animated lesson

A narrated, step-by-step matrices video lesson with captions. Final answer: det(A) = 13.

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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 (0.2 MB) · Captions (WebVTT)

Steps on screen

1. The problem
Worked example
$\det \begin{bmatrix}1 & 0 & 2 \\ 2 & 1 & 0 \\ 0 & 3 & 1\end{bmatrix}$
2. Step by step
1. $A = \begin{bmatrix}1 & 0 & 2 \\ 2 & 1 & 0 \\ 0 & 3 & 1\end{bmatrix}$
  Read the 3×3 matrix
2. $\det A = (1)\begin{vmatrix}1 & 0 \\ 3 & 1\end{vmatrix} - (0)\begin{vmatrix}2 & 0 \\ 0 & 1\end{vmatrix} + (2)\begin{vmatrix}2 & 1 \\ 0 & 3\end{vmatrix}$
  Expand along the first row (cofactor expansion)
3. $M_{11} = 1,\quad M_{12} = 2,\quad M_{13} = 6$
  Evaluate each 2×2 minor
4. $\det A = (1)(1) - (0)(2) + (2)(6) = 13$
  Combine with the alternating signs + − +
3. The answer
$\det A = 13$
det = 13
4. Check
- Computed exactly with fractions; row reduction gives the same value 13.

Transcript

The problem. Here is the problem: [[1, 0, 2], [2, 1, 0], [0, 3, 1]]. Find the determinant of a 3 times 3 matrix.

Step by step. Read the 3 times 3 matrix. Expand along the first row (cofactor expansion). Evaluate each 2 times 2 minor. Combine with the alternating signs plus minus +.

The answer. So the answer is det(A) equals 13. The determinant is 13, which is non-zero, so the matrix is invertible and its rows are linearly independent.

Check. Computed exactly with fractions; row reduction gives the same value 13.

About matrices

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

Determinant of [[1,0,2],[2,1,0],[0,3,1]]: Matrices Lesson