Properties of determinants

A $3\times3$ determinant has six terms in its full expansion, and a student who expands every one of them by hand, every time, is doing far more arithmetic than the exam rewards. The properties of determinants are a small set of rules — about swapping rows, spotting repeated rows, and adding multiples of one row to another — that let you simplify a determinant before you expand it, often turning a messy calculation into one with a whole row of zeros. None of the eight properties need to be memorised as a disconnected list: most follow from just two ideas, that swapping rows flips the sign, and that a repeated row forces the value to zero. This article restates those properties tightly, proves the one used most often, and reworks the two Pass-1 examples with the extra step-by-step care a board exam demands.

The core insight

Row and column operations — swap, scale, or add a multiple of one row to another — let you create zeros in a determinant before expanding it, without changing (or with a predictable change to) its value.

The core idea

The following properties hold for determinants of order 2 and 3, the orders in this syllabus, and they are meant to be applied, not proved in full generality.

Transpose property. $|A| = |A^T|$ — interchanging every row with the corresponding column leaves the value unchanged. This is why every property stated for rows applies equally to columns.
Zero row/column. If every entry of one row (or column) is $0$ , the determinant is $0$ .
Repeated row/column. If two rows (or two columns) are identical, the determinant is $0$ .
Row/column interchange. Swapping any two rows (or columns) multiplies the determinant by $-1$ — the magnitude is unchanged, only the sign flips.
Scalar multiple. Multiplying every entry of one row by a scalar $k$ multiplies the whole determinant by $k$ :

\begin{vmatrix} ka_1 & kb_1 & kc_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = k\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}

where: $a_i, b_i, c_i$ are the entries of row $i$ , and $k$ is any real scalar factored out of row 1 only.

Proportional rows. If one row is a scalar multiple of another, the determinant is $0$ — factor the scalar out (scalar-multiple property), and the two rows left behind are identical (repeated-row property), which is $0$ .
Sum splitting. If every entry of one row is written as a sum of two terms, the determinant splits into the sum of two determinants, with every other row unchanged.
Invariance under row addition. Adding a scalar multiple of one row to another row does not change the determinant's value at all. This is the property used most often, to manufacture zeros before expanding.

Figure

Two three by three grids show that swapping two rows flips the sign while adding a multiple of one row to another leaves the value unchanged

Row addition leaves the determinant unchanged

Derivation AAdding a multiple of one row to another does not change the determinant

The goal is to show, using only the sum-splitting property and the repeated-row property already stated above, that replacing row 2 by row 2 plus $k$ times row 1 leaves a $3\times3$ determinant exactly unchanged in value — the property that justifies every row-reduction step used when simplifying a determinant before expanding it.

Step 1. Write the modified determinant with row 2 replaced by $(a_2+ka_1,\ b_2+kb_1,\ c_2+kc_1)$ , rows 1 and 3 left as they are:

\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2+ka_1 & b_2+kb_1 & c_2+kc_1 \\ a_3 & b_3 & c_3 \end{vmatrix}

Step 2. Apply the sum-splitting property to row 2, since every entry of that row is a sum of two terms. The determinant splits into two determinants, each keeping rows 1 and 3 unchanged:

= \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} + \begin{vmatrix} a_1 & b_1 & c_1 \\ ka_1 & kb_1 & kc_1 \\ a_3 & b_3 & c_3 \end{vmatrix}

Step 3. In the second determinant, factor $k$ out of row 2 using the scalar-multiple property. That leaves a determinant whose row 2 is now identical to row 1:

\begin{vmatrix} a_1 & b_1 & c_1 \\ ka_1 & kb_1 & kc_1 \\ a_3 & b_3 & c_3 \end{vmatrix} = k\begin{vmatrix} a_1 & b_1 & c_1 \\ a_1 & b_1 & c_1 \\ a_3 & b_3 & c_3 \end{vmatrix}

Step 4. By the repeated-row property, a determinant with two identical rows is $0$ , so the whole second term vanishes, no matter what $k$ is. What survives from Step 2 is only the first determinant, the original one — proving $\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2+ka_1 & b_2+kb_1 & c_2+kc_1 \\ a_3 & b_3 & c_3 \end{vmatrix} = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}$ , exactly as claimed.

Worked example

Evaluate $\begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 10 \end{vmatrix}$ using row operations to simplify before expanding.

Step 1. Apply $R_2 \to R_2 - 4R_1$ , allowed with no change in value by the property just proved: $R_2 = (4-4(1),\ 5-4(2),\ 6-4(3)) = (0,\ -3,\ -6)$ .

Step 2. Apply $R_3 \to R_3 - 7R_1$ , the same property with $k=7$ : $R_3 = (7-7(1),\ 8-7(2),\ 10-7(3)) = (0,\ -6,\ -11)$ .

Step 3. The determinant is now $\begin{vmatrix} 1 & 2 & 3 \\ 0 & -3 & -6 \\ 0 & -6 & -11 \end{vmatrix}$ , with two zeros in column 1. Expand along column 1, so only the row-1 cofactor survives:

|A| = 1\begin{vmatrix} -3 & -6 \\ -6 & -11 \end{vmatrix} - 0 + 0

Step 4. Evaluate the surviving $2\times2$ determinant: $(-3)(-11) - (-6)(-6) = 33 - 36 = -3$ .

Step 5. So $|A| = 1(-3) = -3$ . The determinant equals $-3$ .

Competitive stretch

For $\begin{vmatrix} 2 & 4 & 6 \\ 1 & 3 & 5 \\ 1 & 3 & 5 \end{vmatrix}$ , notice rows 2 and 3 are identical $(1,3,5)$ before doing any arithmetic — by the repeated-row property this determinant is $0$ instantly, with no expansion at all. Spotting this in under five seconds, rather than expanding all six terms, is exactly the skill these properties reward in a timed paper.

Common traps

Using an already-modified row when applying a second row operation, instead of the original row, can silently corrupt the value — always keep track of which version of a row is current.
Treating a row swap as having "no effect" — it flips the sign, so forgetting the minus sign is a common scoring error.
Multiplying a row by $k$ to simplify arithmetic and forgetting that the whole determinant is then also multiplied by $k$ , so the final answer must be divided back by $k$ to compensate.
Trying to derive all eight properties from scratch under exam pressure — the syllabus expects you to apply them fluently, with only the row-addition property (proved above) needing to be understood, not the rest.

Ask the tutor: Have a determinant that looks messy to expand directly — ask the AI tutor which row operation simplifies it fastest.

Why does adding a multiple of one row to another not change the determinant, when scaling a row does?

Scaling one row by $k$ genuinely changes the geometry the determinant measures, so the value scales by $k$ too. Adding $k$ times row 1 to row 2 is different: as the derivation above shows, that operation splits into the original determinant plus a second determinant that has two proportional (here, identical after factoring) rows, which is always zero. So only the unchanged original term survives — the "extra" piece you added contributes nothing to the value.

Do I have to prove these properties in the exam, or can I just use them?

You can just use them. The CBSE syllabus for this chapter expects you to apply the properties — spotting a repeated row, creating zeros with a row operation, tracking a sign flip from a swap — to evaluate determinants faster. Only the row-addition property is worth being able to justify briefly if asked, since it is the one that does the real work in simplification; the others are typically quoted and used directly.

How do I know which row operation to try first on a messy determinant?

Look for a way to create a zero, or better, two zeros in one row or column, so that expanding along it involves only one nonzero cofactor. A common strategy is to eliminate the entries below the leading entry of column 1 using $R_2 \to R_2 - kR_1$ and $R_3 \to R_3 - mR_1$ , exactly as in the worked example — this is safe because it never changes the value. Before doing any arithmetic at all, always scan for an identical or proportional row first, since that ends the problem immediately with a zero.