Determinant evaluation

The $2\times2$ case is the building block. For a $2\times2$ matrix, the determinant is the product of the leading diagonal minus the product of the other diagonal:

where: $a,b,c,d$ are the matrix entries in row-then-column order. Every $3\times3$ expansion reduces to several of these.

The $3\times3$ case expands along a row or column using cofactors. The cofactor of entry $a_{ij}$ is $C_{ij}=(-1)^{i+j}M_{ij}$, where $M_{ij}$ is the $2\times2$ minor left after deleting row $i$ and column $j$. Expanding along the first row,

where: each bracketed $2\times2$ array is the minor obtained by striking out the row and column of the entry it multiplies, and the alternating $+,-,+$ signs are exactly the cofactor signs $(-1)^{i+j}$ for the first row ($i=1$, $j=1,2,3$). The same formula works along any row or any column — a smart choice of row or column (one with a zero already in it) cuts the arithmetic down immediately.

Five properties do most of the work in exam problems.

• Interchanging two rows (or columns) changes the sign of the determinant.
• If two rows (or columns) are identical or proportional, the determinant is zero.
• Scaling every entry of one row (or column) by $k$ scales the determinant by $k$; scaling the whole $n\times n$ matrix by $k$ scales the determinant by $k^n$, so $|kA|=k^n|A|$.
• Adding a scalar multiple of one row (or column) to another leaves the determinant unchanged — written $R_i \to R_i + kR_j$. This is the main simplification tool.
• $|A^T|=|A|$, and $|AB|=|A|\,|B|$ for square matrices of the same order.

The sign grid below is worth memorising cold, since a plus/minus slip on the middle cofactor is the single most common arithmetic error in this topic.

The claim $R_i \to R_i + kR_j$ leaves a determinant unchanged is used constantly, but it is worth proving rather than trusting blindly, since seeing why it holds is what stops students from misapplying it to operations that also scale a row, which is a genuinely different and value-changing move.

Step 1. Take a general $3\times3$ determinant and expand it along the row that will be modified, say row 2, using the cofactor formula: $|A| = a_2 C_{21} + b_2 C_{22} + c_2 C_{23}$, where $C_{21},C_{22},C_{23}$ are the cofactors built from rows 1 and 3 only — they do not involve row 2 at all.

Step 2. Apply $R_2 \to R_2 + kR_1$, so the new row 2 entries become $(a_2+ka_1,\ b_2+kb_1,\ c_2+kc_1)$. Expand the new determinant along this same row: $|A'| = (a_2+ka_1)C_{21} + (b_2+kb_1)C_{22} + (c_2+kc_1)C_{23}$, using the same cofactors $C_{21},C_{22},C_{23}$, because those cofactors are built only from rows 1 and 3, which have not changed.

Step 3. Expand the brackets and regroup: $|A'| = \left(a_2C_{21}+b_2C_{22}+c_2C_{23}\right) + k\left(a_1C_{21}+b_1C_{22}+c_1C_{23}\right)$. The first bracket is exactly $|A|$ from Step 1. The second bracket, $a_1C_{21}+b_1C_{22}+c_1C_{23}$, is row 1's entries multiplied by row 2's cofactors — this is the expansion of a determinant with row 1 repeated in both row 1 and row 2, which is zero by the identical-rows property.

Step 4. So $|A'| = |A| + k\cdot 0 = |A|$. The operation $R_i \to R_i + kR_j$ genuinely leaves the determinant unchanged, and the proof shows exactly why: it only works because the row being added is a different, untouched row, and the row being changed is not simultaneously being scaled. If instead you write $R_2 \to 2R_2 - R_1$, the row-2 part is scaled by $2$ first, and the determinant scales by that factor before the zero-sum cancellation applies — which is precisely the trap flagged below.

Evaluate $\begin{vmatrix}2 & 3 & 1\\ 1 & -1 & 2\\ 3 & 2 & 0\end{vmatrix}$ first by direct cofactor expansion along the first row, then re-derive the same value using row operations to confirm both methods agree.

Step 1 — direct expansion, find the three $2\times2$ minors.

Step 2 — apply cofactor signs and combine with row-1 entries.

So direct expansion gives $\mathbf{15}$.

Step 3 — now simplify with row operations instead. Apply $R_2 \to R_2 - \tfrac{1}{2}R_1$ and $R_3 \to R_3 - \tfrac{3}{2}R_1$ to create zeros in the first column (these are legal by the proof above, since row 1 is untouched and unscaled):

Step 4 — expand the new determinant along the now-mostly-zero first column.

Both routes agree: the determinant is $15$. Direct expansion needed six multiplications across three minors; the row-operation route needed two operations to create zeros but then only a single $2\times2$ minor to finish — the trade-off worth recognising in an exam.

• Forgetting the alternating sign pattern $(-1)^{i+j}$ when expanding — a plus/minus slip on the middle term is the single most common arithmetic error.
• Writing a row operation like $R_2\to 2R_2 - R_1$ and forgetting this scales the determinant by $2$ as well as combining rows — only pure operations $R_i\to R_i+kR_j$, with no scaling of $R_i$ itself, leave the determinant unchanged.
• Mixing row and column operations mid-calculation — pick one style and stay consistent within a single simplification.
• Assuming $|A+B|=|A|+|B|$ — this is false in general; determinants are not additive over matrix addition.

Ask the tutor: ask the AI tutor for a fresh $3\times3$ determinant and try spotting the fastest row or column operation before you expand.