Properties of Determinants

Introduction to Properties of Determinants

Evaluating determinants of higher orders such as $3 \times 3$ or $4 \times 4$ by direct expansion is often lengthy and error-prone. To make calculations shorter and smarter, we use standard properties of determinants. These properties help us transform a determinant into a simpler form—usually by creating zeros, identifying proportional rows/columns, or factoring expressions—before expansion.

These properties are extremely important not only for board-level understanding, but also for competitive exams such as JEE Main and JEE Advanced, where speed, pattern recognition, and algebraic simplification matter a lot.

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The Six Fundamental Properties

1. Reflection Property

The value of a determinant remains unchanged when rows are written as columns and columns are written as rows.

• Mathematically, $$|A| = |A^T|.$$ This means the determinant of a matrix is equal to the determinant of its transpose.

2. Switching Property

If any two rows or any two columns of a determinant are interchanged, then the sign of the determinant changes.

• For example, if $R_1 \leftrightarrow R_2$, then the new determinant becomes $-\Delta$.
• Similarly, if $C_1 \leftrightarrow C_3$, the sign also changes.

3. Repetition Property

If any two rows or any two columns are identical, then the determinant is zero.

• More generally, if two rows or two columns are proportional, the determinant is also zero.
• This happens because one row/column becomes linearly dependent on another.

4. Scalar Multiple Property

If each element of one row or one column is multiplied by a constant $k$, then the determinant also gets multiplied by $k$.

• Therefore, a common factor can be taken out from any one row or any one column.
• If all rows of an $n \times n$ determinant are multiplied by $k$, then the determinant gets multiplied by $k^n$.

5. Sum Property

If the elements of any one row or one column are sums of two terms, then the determinant can be split into the sum of two determinants.

• For example, $$\begin{vmatrix} a_1+x & b_1 & c_1 \\ a_2+y & b_2 & c_2 \\ a_3+z & b_3 & c_3 \end{vmatrix} = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} + \begin{vmatrix} x & b_1 & c_1 \\ y & b_2 & c_2 \\ z & b_3 & c_3 \end{vmatrix}.$$ This property is linearity in a row or a column.

6. Invariance Property

If, to each element of any one row or one column, we add the corresponding equimultiples of another row or column, then the value of the determinant remains unchanged.

• Mathematically, operations of the form $$R_i \to R_i + kR_j$$ or $$C_i \to C_i + kC_j$$ do not change the determinant.
• This is one of the most useful properties for creating zeros.
• Caution: If the target row itself is multiplied, such as in $R_i \to mR_i + kR_j$, then the determinant gets multiplied by $m$. So only the addition part keeps the determinant unchanged.

Example 1: Verifying Property 3 (Proportional Rows)

Without expanding, evaluate the determinant: $$\Delta = \begin{vmatrix} 102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \end{vmatrix}$$

Solution: Observe the first row carefully: $$102 = 6 \times 17, \quad 18 = 6 \times 3, \quad 36 = 6 \times 6.$$ So the first row is $6$ times the third row. That is, $$R_1 = 6R_3.$$ Hence two rows are proportional. By Property 3, if any two rows or columns are proportional, the determinant is zero. Therefore, $$\Delta = 0.$$

Answer: $0$

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Example 2: Using Property 6 to simplify

Prove that: $$\begin{vmatrix} a & a+b & a+b+c \\ 2a & 3a+2b & 4a+3b+2c \\ 3a & 6a+3b & 10a+6b+3c \end{vmatrix} = a^3$$

Solution: Let the determinant be $\Delta$. Apply row operations to create zeros in the first column: $$R_2 \to R_2 - 2R_1, \qquad R_3 \to R_3 - 3R_1.$$ Then, $$\Delta = \begin{vmatrix} a & a+b & a+b+c \\ 0 & (3a+2b)-2(a+b) & (4a+3b+2c)-2(a+b+c) \\ 0 & (6a+3b)-3(a+b) & (10a+6b+3c)-3(a+b+c) \end{vmatrix}.$$ Simplifying, $$\Delta = \begin{vmatrix} a & a+b & a+b+c \\ 0 & a & 2a+b \\ 0 & 3a & 7a+3b \end{vmatrix}.$$ Now eliminate the entry $3a$ in the third row, second column by using $$R_3 \to R_3 - 3R_2.$$ So, $$\Delta = \begin{vmatrix} a & a+b & a+b+c \\ 0 & a & 2a+b \\ 0 & 0 & (7a+3b)-3(2a+b) \end{vmatrix} = \begin{vmatrix} a & a+b & a+b+c \\ 0 & a & 2a+b \\ 0 & 0 & a \end{vmatrix}.$$ This is now an upper triangular determinant, so its value is the product of diagonal entries: $$\Delta = a \cdot a \cdot a = a^3.$$

Answer: Proved.

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Example 3: Using the Sum Property (Property 5)

Show that: $$\begin{vmatrix} x & y & z \\ a & b & c \\ x+a & y+b & z+c \end{vmatrix} = 0$$

Solution: Let the determinant be $\Delta$.

Method 1: Using Property 5 The third row is the sum of the first two rows component-wise: $$R_3 = R_1 + R_2.$$ Using the sum property, $$\Delta = \begin{vmatrix} x & y & z \\ a & b & c \\ x & y & z \end{vmatrix} + \begin{vmatrix} x & y & z \\ a & b & c \\ a & b & c \end{vmatrix}.$$ In the first determinant, $R_1$ and $R_3$ are identical, so it is zero. In the second determinant, $R_2$ and $R_3$ are identical, so it is also zero. Hence, $$\Delta = 0 + 0 = 0.$$

Method 2: Using Property 6 Apply $$R_3 \to R_3 - R_1 - R_2.$$ Then, $$\Delta = \begin{vmatrix} x & y & z \\ a & b & c \\ 0 & 0 & 0 \end{vmatrix} = 0.$$ Since one row is entirely zero, the determinant is zero.

Answer: Proved.

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Example 4: Classic Cyclic Determinant

Evaluate: $$\Delta = \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix}$$

Solution: Apply row operations: $$R_2 \to R_2 - R_1, \qquad R_3 \to R_3 - R_1.$$ Then, $$\Delta = \begin{vmatrix} 1 & x & x^2 \\ 0 & y-x & y^2-x^2 \\ 0 & z-x & z^2-x^2 \end{vmatrix}.$$ Factor the difference of squares: $$y^2-x^2 = (y-x)(y+x), \qquad z^2-x^2 = (z-x)(z+x).$$ So, $$\Delta = \begin{vmatrix} 1 & x & x^2 \\ 0 & y-x & (y-x)(y+x) \\ 0 & z-x & (z-x)(z+x) \end{vmatrix}.$$ Take $(y-x)$ common from $R_2$ and $(z-x)$ common from $R_3$: $$\Delta = (y-x)(z-x)\begin{vmatrix} 1 & x & x^2 \\ 0 & 1 & y+x \\ 0 & 1 & z+x \end{vmatrix}.$$ Now expand along the first column: $$\Delta = (y-x)(z-x)\begin{vmatrix} 1 & y+x \\ 1 & z+x \end{vmatrix}.$$ So, $$\Delta = (y-x)(z-x)\big[(z+x)-(y+x)\big] = (y-x)(z-x)(z-y).$$ Now rewrite in the standard cyclic form: $$(y-x) = -(x-y), \qquad (z-y)=-(y-z).$$ Thus, $$\Delta = (x-y)(y-z)(z-x).$$

Answer: $(x-y)(y-z)(z-x)$

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Example 5: Taking $abc$ Common Technique

Evaluate: $$\Delta = \begin{vmatrix} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{vmatrix}$$

Solution: A very effective trick is to factor $a$, $b$, and $c$ from rows after rewriting the entries. Assume temporarily that $a, b, c \neq 0$ while simplifying; the final expression obtained is polynomial and remains valid generally.

Take $a$ common from $R_1$, $b$ from $R_2$, and $c$ from $R_3$: $$\Delta = abc \begin{vmatrix} \frac{1+a}{a} & \frac{1}{a} & \frac{1}{a} \\ \frac{1}{b} & \frac{1+b}{b} & \frac{1}{b} \\ \frac{1}{c} & \frac{1}{c} & \frac{1+c}{c} \end{vmatrix} = abc \begin{vmatrix} 1+\frac{1}{a} & \frac{1}{a} & \frac{1}{a} \\ \frac{1}{b} & 1+\frac{1}{b} & \frac{1}{b} \\ \frac{1}{c} & \frac{1}{c} & 1+\frac{1}{c} \end{vmatrix}.$$ Now apply $$R_1 \to R_1 + R_2 + R_3.$$ Then every entry of the first row becomes $$1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}.$$ So, $$\Delta = abc \begin{vmatrix} S & S & S \\ \frac{1}{b} & 1+\frac{1}{b} & \frac{1}{b} \\ \frac{1}{c} & \frac{1}{c} & 1+\frac{1}{c} \end{vmatrix},$$ where $$S = 1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}.$$ Take $S$ common from $R_1$: $$\Delta = abc \cdot S \begin{vmatrix} 1 & 1 & 1 \\ \frac{1}{b} & 1+\frac{1}{b} & \frac{1}{b} \\ \frac{1}{c} & \frac{1}{c} & 1+\frac{1}{c} \end{vmatrix}.$$ Now apply column operations: $$C_2 \to C_2 - C_1, \qquad C_3 \to C_3 - C_1.$$ Then, $$\Delta = abc \cdot S \begin{vmatrix} 1 & 0 & 0 \\ \frac{1}{b} & 1 & 0 \\ \frac{1}{c} & 0 & 1 \end{vmatrix}.$$ This determinant is upper triangular, so its value is 1. Therefore, $$\Delta = abc \left(1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right).$$ Multiplying out, $$\Delta = abc + bc + ac + ab.$$

Answer: $abc + ab + bc + ca$

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Example 6: Determinant with Complex Roots of Unity

If $\omega$ is a complex cube root of unity, find the value of $$\begin{vmatrix} 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \\ \omega^2 & 1 & \omega \end{vmatrix}.$$

Solution: Recall the standard identities for cube roots of unity: $$\omega^3 = 1, \qquad 1 + \omega + \omega^2 = 0.$$ Apply the column operation $$C_1 \to C_1 + C_2 + C_3.$$ Then the first column becomes $$\begin{bmatrix} 1+\omega+\omega^2 \\ \omega+\omega^2+1 \\ \omega^2+1+\omega \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}.$$ So the determinant becomes a determinant with an entire zero column, hence $$\Delta = 0.$$

Answer: $0$

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Example 7: Proving using Skew-Symmetric Property

Evaluate without expansion: $$\begin{vmatrix} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{vmatrix}$$

Solution: Let the determinant be $\Delta$. The given matrix is skew-symmetric because its transpose is the negative of the matrix itself: $$A^T = -A.$$ Now by Property 1, $$|A| = |A^T|.$$ But since $A^T = -A$, $$|A^T| = |-A|.$$ For a $3 \times 3$ determinant, $$|-A| = (-1)^3|A| = -|A|.$$ Therefore, $$|A| = -|A|.$$ So, $$2|A| = 0 \Rightarrow |A| = 0.$$ Hence, $$\Delta = 0.$$ This proves the standard result that the determinant of a skew-symmetric matrix of odd order is always zero.

Answer: $0$

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Example 8: Circular Operation Trick

Show that: $$\begin{vmatrix} a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{vmatrix} = 0$$

Solution: Let the determinant be $\Delta$. Apply the row operation $$R_1 \to R_1 + R_2 + R_3.$$ Now compute each entry of the new first row: $$(a-b) + (b-c) + (c-a) = 0,$$ $$(b-c) + (c-a) + (a-b) = 0,$$ $$(c-a) + (a-b) + (b-c) = 0.$$ Therefore, $$\Delta = \begin{vmatrix} 0 & 0 & 0 \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{vmatrix}.$$ Since the first row is entirely zero, the determinant is zero. Hence, $$\Delta = 0.$$

Answer: Proved.