Introduction

The determinant of a 3×3 matrix plays a vital role in solving systems of linear equations, checking invertibility, and understanding geometric transformations.

This article explains how to compute the determinant of matrix
\(\begin{bmatrix}1 & 3 & 0\\ 2 & 6 & 4\\ -1 & -1 & 2\end{bmatrix}\)
using the standard 3×3 determinant formula, with each step written for competitive‑exam style questions.

Formula for 3×3 determinant

For a general 3×3 matrix
\[
A = \begin{bmatrix}
a & b & c\\
d & e & f\\
g & h & i
\end{bmatrix},
\]
the determinant is
\[
\det(A) = aei + bfg + cdh – ceg – bdi – afh. \tag{1}
\]

This expression comes from expanding the determinant using the Leibniz formula or Sarrus’s rule for 3×3 matrices.

Step‑by‑step solution for the given matrix

The given matrix is
\[
A = \begin{bmatrix}
1 & 3 & 0\\
2 & 6 & 4\\
-1 & -1 & 2
\end{bmatrix}.
\]

Match it with the general form:

• \(a = 1,\ b = 3,\ c = 0\)
• \(d = 2,\ e = 6,\ f = 4\)
• \(g = -1,\ h = -1,\ i = 2\)

1. Positive diagonal products

• \(aei = 1 \cdot 6 \cdot 2 = 12\)
• \(bfg = 3 \cdot 4 \cdot (-1) = -12\)
• \(cdh = 0 \cdot 2 \cdot (-1) = 0\)

Sum of positive terms \(= 12 + (-12) + 0 = 0\).

2. Negative diagonal products

• \(ceg = 0 \cdot 6 \cdot (-1) = 0\)
• \(bdi = 3 \cdot 2 \cdot 2 = 12\)
• \(afh = 1 \cdot 4 \cdot (-1) = -4\)

Sum of negative terms \(= 0 + 12 + (-4) = 8\).

3. Apply the determinant formula

\[
\det(A) = \text{(sum of positive)} – \text{(sum of negative)} = 0 – 8 = -8.
\]

4. Verification by cofactor expansion

Using cofactor expansion along the first row,

\[
\det(A) =
1\begin{vmatrix}6 & 4\\ -1 & 2\end{vmatrix}
– 3\begin{vmatrix}2 & 4\\ -1 & 2\end{vmatrix}
+ 0\begin{vmatrix}2 & 6\\ -1 & -1\end{vmatrix}.
\]

• \(\begin{vmatrix}6 & 4\\ -1 & 2\end{vmatrix} = 6 \cdot 2 – 4 \cdot (-1) = 12 + 4 = 16\)
• \(\begin{vmatrix}2 & 4\\ -1 & 2\end{vmatrix} = 2 \cdot 2 – 4 \cdot (-1) = 4 + 4 = 8\)

So
\[
\det(A) = 1 \cdot 16 – 3 \cdot 8 + 0 = 16 – 24 = -8.
\]

Thus, the determinant of the matrix
\(\begin{bmatrix}1 & 3 & 0\\ 2 & 6 & 4\\ -1 & -1 & 2\end{bmatrix}\)
equals \(-8\).

Key points for exams

• Use the standard 3×3 determinant formula or cofactor expansion along a row or column with a zero entry to simplify calculations.
• For matrix [1 3 0; 2 6 4; -1 -1 2], careful substitution and arithmetic lead to the determinant value \(-8\), which also confirms the matrix is invertible because the determinant is non‑zero.