Properties of determinants — all 10 properties with examples and applications

medium CBSE JEE-MAIN 4 min read

Question

State all 10 properties of determinants and show how each property helps simplify determinant evaluation. Demonstrate with a 3×33 \times 3 determinant.

(CBSE 12 Board + JEE Main pattern)


Solution — Step by Step

Here is the complete list. We will use a determinant DD of matrix AA of order nn:

#PropertyWhat it means
1Transpose$
2Row swapSwapping two rows changes sign: DDD \to -D
3Identical rowsIf two rows are identical, D=0D = 0
4Scalar multiple of rowFactor out kk from one row: D=kDD' = kD
5Sum propertyIf a row is sum of two sets of elements, DD splits into sum of two determinants
6Row operation (add multiple)RiRi+kRjR_i \to R_i + kR_j does not change DD
7All zeros in a rowD=0D = 0
8Proportional rowsIf Ri=kRjR_i = kR_j, then D=0D = 0
9Scalar multiplication of matrix$
10Product property$

Consider:

D=123456789D = \begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix}

Apply R2R2R1R_2 \to R_2 - R_1 and R3R3R1R_3 \to R_3 - R_1 (Property 6 — value unchanged):

D=123333666D = \begin{vmatrix} 1 & 2 & 3 \\ 3 & 3 & 3 \\ 6 & 6 & 6 \end{vmatrix}

Now R3=2R2R_3 = 2R_2 (proportional rows — Property 8), so D=0D = 0.

Suppose we need 3A|3A| where AA is 3×33 \times 3 and A=5|A| = 5.

By Property 9: 3A=33A=27×5=135|3A| = 3^3 \cdot |A| = 27 \times 5 = \mathbf{135}

Students who write 3A=3A=15|3A| = 3|A| = 15 lose marks every year. The power is nn, not 1.

flowchart TD
    A["Given a Determinant"] --> B{"Any row/col all zeros?"}
    B -- Yes --> Z["D = 0"]
    B -- No --> C{"Two rows identical or proportional?"}
    C -- Yes --> Z
    C -- No --> D{"Can we simplify using R_i → R_i + kR_j?"}
    D -- Yes --> E["Apply row operations (D unchanged)"]
    E --> F{"Upper triangular form?"}
    F -- Yes --> G["D = product of diagonal"]
    F -- No --> D
    D -- No --> H["Expand along row/col with most zeros"]
    H --> I["Compute cofactors"]
    I --> J["Final Answer"]
    G --> J

Why This Works

Determinants encode the “signed volume” of the parallelepiped formed by row (or column) vectors. When two rows are identical, the parallelepiped collapses to a lower dimension — volume becomes zero. When we swap two rows, the orientation flips — hence the sign change.

Row operations of the type RiRi+kRjR_i \to R_i + kR_j correspond to shearing, which preserves volume. That is why this operation does not change the determinant value, and it is our primary weapon for simplification.

The scalar multiplication property kA=knA|kA| = k^n|A| comes from the fact that each of the nn rows gets multiplied by kk, and Property 4 says each such multiplication pulls out one factor of kk.


Alternative Method

For small determinants, Sarrus’ Rule (only for 3×33 \times 3) gives a quick computation without row operations:

D=a11(a22a33a23a32)a12(a21a33a23a31)+a13(a21a32a22a31)D = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31})

But for JEE, the row-reduction approach is faster and less error-prone for larger or parameter-heavy determinants.

In JEE Main, most determinant questions can be solved in under 2 minutes if you apply Property 6 first to create zeros, then expand along the row/column with the most zeros. Practice making the first column all zeros except one entry.


Common Mistake

The number one error: confusing kA|kA| with kAk|A|. For a 3×33 \times 3 matrix, 2A=8A|2A| = 8|A|, not 2A2|A|. The exponent equals the order of the matrix. This has appeared as a direct MCQ in CBSE boards and JEE Main multiple times — and students still get it wrong under exam pressure.

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