Determinants

Determinant via Row Reduction

Compute the determinant of a 3x3 matrix by reducing it to upper triangular form using row elimination. The determinant equals the product of the diagonal entries of the triangular result. Each elimination step shows the pivot, multiplier, and updated row. The answer is reconciled with the cofactor-expansion result for the same matrix.

row reduction and det Elementary row operations (add a multiple of one row to another) do not change the determinant. Reducing A to upper triangular form U gives det(A) = u_11 * u_22 * u_33.

Example

Reduce to triangular form and multiply the diagonal entries.

highlighted = computed this step

Step 1 — Set up

Start with the given matrix data.

A=[21−1032104]A=\begin{bmatrix}2&1&-1\\0&3&2\\1&0&4\end{bmatrix}

Step 2 — Update row 3

Update row 3 using row 1: R3 <- R3 - 1/2*R1.

A=[21−10320-1/29/2]A=\begin{bmatrix}2&1&-1\\0&3&2\\\hl{0}&\hl{-1/2}&\hl{9/2}\end{bmatrix}

Step 3 — Update row 3

Update row 3 using row 2: R3 <- R3 + 1/6*R2.

A=[21−10320029/6]A=\begin{bmatrix}2&1&-1\\0&3&2\\\hl{0}&\hl{0}&\hl{29/6}\end{bmatrix}

Step 4 — Diagonal product

Multiply the diagonal entries: 2 * 3 * 29/6 = 29.

A=[21−103200296],2⋅3⋅296=29A=\begin{bmatrix}2&1&-1\\0&3&2\\0&0&\frac{29}{6}\end{bmatrix},\quad 2\cdot 3\cdot \frac{29}{6}=\hl{29}

Step 5 — Result

Read the determinant: det = 29.

A=[21−103200296],det=29,det⁡(A)=29A=\begin{bmatrix}2&1&-1\\0&3&2\\0&0&\frac{29}{6}\end{bmatrix},\quad det=\hl{29},\quad \det(A)=29