Determinants

Determinant of a Square Matrix

Understand

For a \(2\times2\) matrix,

\[\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc.\]

For a \(3\times3\) matrix we expand along any row or column, using the sign pattern

\[\begin{vmatrix} + & - & + \\ - & + & - \\ + & - & + \end{vmatrix}\]

Along the first row: \(|A| = a_{11}M_{11} - a_{12}M_{12} + a_{13}M_{13}\), where \(M_{ij}\) is the 2 × 2 determinant left after deleting row \(i\) and column \(j\).

Useful facts

For an \(n\times n\) matrix, \(|kA| = k^n|A|\).
\(|A^T| = |A|\) and \(|AB| = |A|\,|B|\).
A matrix is singular if \(|A| = 0\), non-singular otherwise.

Key Concepts

Choose the row/column with the most zeros.
Signs follow a chessboard pattern starting with + at the top left.

Formula Bank

2 × 2 determinant

\[\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc\]

Main diagonal product minus the other diagonal product.

Scalar multiple

\[|kA| = k^{n}|A| \quad (A \text{ of order } n)\]

Each of the \(n\) rows contributes a factor \(k\). Also \(|AB| = |A||B|\) and \(|A^T| = |A|\).

Determinant rules

\[|A^T|=|A|,\quad |AB|=|A|\,|B|,\quad |kA|=k^n|A|,\quad |A^{-1}|=\frac{1}{|A|}\]

Key Points

Invertible ⇔ non-zero determinant

A square matrix is invertible exactly when \(|A| \ne 0\).

Expand along zeros

Choose the row or column with the most zeros – fewer calculations, fewer slips.

Common Mistakes

Writing |kA| = k|A|

For a 3 × 3 matrix with \(|A| = 4\), claiming \(|2A| = 8\).

Correct: \(|kA| = k^n|A|\), so \(|2A| = 2^3 \times 4 = 32\).

Ignoring the sign pattern

Expanding \(a_{11}M_{11} + a_{12}M_{12} + a_{13}M_{13}\) with all plus signs.

Correct: use + − + along row 1 (chessboard pattern), i.e. multiply by cofactors, not minors.

Scalar multiple of a matrix

For a \(3\times3\) matrix, \(|2A|=2^3|A|=8|A|\), not \(2|A|\). A scalar multiplies every row.

Solved Examples

Expanding a 3 × 3 determinant

Evaluate \(\begin{vmatrix} 2 & 3 & 1 \\ 0 & 1 & 4 \\ 1 & 0 & 2 \end{vmatrix}\).

Expand along row 1: \(2(2 - 0) - 3(0 - 4) + 1(0 - 1)\).
\(= 4 + 12 - 1\).

Final Answer: \(15\)

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Topic Summary

2×2: \(ad - bc\). 3×3: expand with + − + signs. \(|kA| = k^n|A|\). Singular ⇔ \(|A| = 0\).

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