Algebra Of Determinants

MathsDeterminantsFor JEE aspirants

A determinant is a scalar value computed from a square array of numbers, used in JEE Mathematics to solve linear systems, find areas, and test whether vectors are linearly independent. For a $2 \times 2$ determinant, the value is $a_{1} b_{2} - a_{2} b_{1}$ ; for a $3 \times 3$ determinant, we use cofactor expansion along any row or column. This concept covers order, minors, cofactors, expansion, the product of two determinants, and the power-cofactor formula - the algebraic foundations you will apply throughout the Determinants and Matrices unit.

Key Formulas - Quick Reference

Second-order determinant: $a_{1} a_{2} b_{1} b_{2} = a_{1} b_{2} - a_{2} b_{1}$
Third-order determinant (expansion along $R_{1}$ ): $Δ = a_{1} (b_{2} c_{3} - c_{2} b_{3}) - b_{1} (a_{2} c_{3} - c_{2} a_{3}) + c_{1} (a_{2} b_{3} - b_{2} a_{3})$
Cofactor of element $a_{ij}$ : $A_{ij} = (- 1)^{i + j} \cdot M_{ij}$ , where $M_{ij}$ is the corresponding minor
Row-expansion identity: $Δ = a_{i} A_{i} + b_{i} B_{i} + c_{i} C_{i}$ (products of elements with their own cofactors)
Cross-row cofactor identity: $a_{i} A_{j} + b_{i} B_{j} + c_{i} C_{j} = 0$ when $i \neq = j$
Product of two determinants (row by row): $∣ a_{ij} ∣ \cdot ∣ α_{ij} ∣ = ∣ \sum_{k} a_{ik} α_{j k} ∣$
Power-cofactor formula: if $Δ = ∣ a_{ij} ∣$ is of order $n$ , then $∣ A_{ij} ∣ = Δ^{n - 1}$

1. What is a Determinant?

Consider the pair of homogeneous linear equations $a_{1} x + b_{1} y = 0$ and $a_{2} x + b_{2} y = 0$ . Solving them gives $\frac{- a _{1}}{b _{1}} = \frac{y}{x} = \frac{- a _{2}}{b _{2}}$ , which leads to the elimination condition $a_{1} b_{2} - a_{2} b_{1} = 0$ . We express this compactly as:

$a_{1} a_{2} b_{1} b_{2} = 0$

The expression

a_{1} a_{2} b_{1} b_{2}

is called a determinant of order two, and equals

a_{1} b_{2} - a_{2} b_{1}

Third-order Determinants

A determinant of order three consisting of 3 rows and 3 columns is written as:

$Δ = a_{1} a_{2} a_{3} b_{1} b_{2} b_{3} c_{1} c_{2} c_{3}$

and is evaluated as:

$Δ = a_{1} b_{2} b_{3} c_{2} c_{3} - b_{1} a_{2} a_{3} c_{2} c_{3} + c_{1} a_{2} a_{3} b_{2} b_{3}$

$= a_{1} (b_{2} c_{3} - c_{2} b_{3}) - b_{1} (a_{2} c_{3} - c_{2} a_{3}) + c_{1} (a_{2} b_{3} - b_{2} a_{3})$

The numbers $a_{i}$ , $b_{i}$ , $c_{i}$ (for $i = 1, 2, 3$ ) are called the elements of the determinant.

2. Minors and Cofactors

Minor: The determinant obtained by deleting the $i^{th}$ row and $j^{th}$ column is called the minor of the element at position $(i, j)$ , denoted $M_{ij}$ .

Cofactor: The cofactor of the element $a_{ij}$ is $A_{ij} = (- 1)^{i + j} M_{ij}$ .

Sign Chessboard

The factor $(- 1)^{i + j}$ produces an alternating sign pattern:

Figure 1: Cofactor sign pattern for a

3 \times 3

determinant. Each position gets

(- 1)^{i + j}

Expansion by Cofactors

Note that we can write:

$Δ = a_{1} A_{1} + b_{1} B_{1} + c_{1} C_{1}$

where $A_{1}$ , $B_{1}$ , $C_{1}$ are the cofactors of $a_{1}$ , $b_{1}$ , $c_{1}$ respectively. That is, the sum of products of the elements of any row (or column) with the corresponding cofactors equals the value of the determinant.

We can expand the determinant through any row or column:

$Δ = a_{2} A_{2} + b_{2} B_{2} + c_{2} C_{2} = a_{1} A_{1} + a_{2} A_{2} + a_{3} A_{3}, etc.$

Also, the sum of products of elements of any row (or column) with the cofactors of a different row (or column) is zero:

$a_{1} A_{2} + b_{1} B_{2} + c_{1} C_{2} = 0$

In compact form:

$a_{i} A_{j} + b_{i} B_{j} + c_{i} C_{j} = {Δ 0 if i = j if i \neq = j$

Note: These results are true for determinants of any order.

Solved Example 1

Problem: Find the cofactors of all the elements of

147 2 - 5 0 36 - 1

Solution:

(i) Minor of $1 = - 5 0 6 - 1 = 5$ , so cofactor of $1 = (- 1)^{1 + 1} (5) = 5$ .

(ii) Minor of $2 = 47 6 - 1 = - 4 - 42 = - 46$ , so cofactor of $2 = (- 1)^{1 + 2} (- 46) = 46$ .

(iii) Minor of $3 = 47 - 5 0 = 35$ , so cofactor of $3 = (- 1)^{1 + 3} (35) = 35$ .

(iv) Minor of $4 = 20 3 - 1 = - 2$ , so cofactor of $4 = (- 1)^{2 + 1} (- 2) = 2$ .

(v) Minor of $- 5 = 17 3 - 1 = - 22$ , so cofactor of $- 5 = (- 1)^{2 + 2} (- 22) = - 22$ .

The remaining cofactors are found similarly. Each sign follows the chessboard $(- 1)^{i + j}$ .

3. Product of Two Determinants

Two determinants of the same order can be multiplied. The standard row-by-row multiplication rule is:

$a_{1} a_{2} a_{3} b_{1} b_{2} b_{3} c_{1} c_{2} c_{3} \cdot α_{1} α_{2} α_{3} β_{1} β_{2} β_{3} γ_{1} γ_{2} γ_{3} = a_{1} α_{1} + b_{1} β_{1} + c_{1} γ_{1} a_{2} α_{1} + b_{2} β_{1} + c_{2} γ_{1} a_{3} α_{1} + b_{3} β_{1} + c_{3} γ_{1} a_{1} α_{2} + b_{1} β_{2} + c_{1} γ_{2} a_{2} α_{2} + b_{2} β_{2} + c_{2} γ_{2} a_{3} α_{2} + b_{3} β_{2} + c_{3} γ_{2} a_{1} α_{3} + b_{1} β_{3} + c_{1} γ_{3} a_{2} α_{3} + b_{2} β_{3} + c_{2} γ_{3} a_{3} α_{3} + b_{3} β_{3} + c_{3} γ_{3}$

Here we have multiplied rows by rows. We can also multiply rows by columns, columns by rows, or columns by columns - each gives the same product.

Power Cofactor Formula

Δ = ∣ a_{ij} ∣

is a determinant of order

n

, then the value of the determinant

∣ A_{ij} ∣

(whose entries are the cofactors of

a_{ij}

) is

Δ^{n - 1}

. This is known as the power cofactor formula.

For a $3 \times 3$ determinant $Δ$ , the cofactor determinant equals $Δ^{2}$ . For a $2 \times 2$ , it equals $Δ$ itself.

Solved Example 2

Problem: Prove by multiplication of determinants and the power-cofactor formula that:

0 c b c 0 a b a 0^{2} = b^{2} + c^{2} ab a c ab c^{2} + a^{2} b c a c b c a^{2} + b^{2} = - a^{2} ab a c ab - b^{2} b c a c b c - c^{2} = 4 a^{2} b^{2} c^{2}

Solution:

The first determinant equals $2 ab c$ , and its square is $(2 ab c)^{2} = 4 a^{2} b^{2} c^{2}$ .
The second determinant arises from direct row-to-row multiplication of the first determinant with itself: entry $(i, j)$ is the dot product of row $i$ and row $j$ of the original.
The third determinant is the cofactor determinant of the first. By the power-cofactor formula (with $n = 3$ ), it equals $Δ^{n - 1} = Δ^{2}$ , i.e. the square of the first.

Therefore all three expressions are equal to $4 a^{2} b^{2} c^{2}$ .

Solved Example 3

Problem: Express

Δ = (a - x)^{2} (b - x)^{2} (c - x)^{2} (a - y)^{2} (b - y)^{2} (c - y)^{2} (a - z)^{2} (b - z)^{2} (c - z)^{2}

as the product of two determinants and evaluate it.

Solution:

Expanding each entry: $(a - x)^{2} = a^{2} - 2 a x + x^{2}$ , and similarly for the others. So:

$Δ = a^{2} - 2 a x + x^{2} b^{2} - 2 b x + x^{2} c^{2} - 2 c x + x^{2} a^{2} - 2 a y + y^{2} b^{2} - 2 b y + y^{2} c^{2} - 2 cy + y^{2} a^{2} - 2 a z + z^{2} b^{2} - 2 b z + z^{2} c^{2} - 2 cz + z^{2}$

Each entry $(i, j)$ is the row-column product of the vectors $(a_{i}^{2}, - 2 a_{i}, 1)$ and $(1, x_{j}, x_{j}^{2})$ . So the determinant factorises as:

$Δ = a^{2} b^{2} c^{2} - 2 a - 2 b - 2 c 111 \times 111 x y z x^{2} y^{2} z^{2}$

Both factors are Vandermonde-type. The first equals $- 2 (a - b) (b - c) (c - a)$ (after factoring out $- 2$ from column 2) and the second equals $(y - x) (z - y) (z - x)$ . Combining and simplifying signs:

$Δ = 2 (a - b) (b - c) (c - a) (x - y) (y - z) (z - x)$

Common Mistakes to Avoid

Watch out

Forgetting the sign in the cofactor. The minor $M_{ij}$ is the sub-determinant; the cofactor is $A_{ij} = (- 1)^{i + j} M_{ij}$ . Students often report the minor as the cofactor when the position sign is negative.
Mixing up row-expansion with cross-row. $a_{1} A_{1} + b_{1} B_{1} + c_{1} C_{1} = Δ$ , but $a_{1} A_{2} + b_{1} B_{2} + c_{1} C_{2} = 0$ . Always check that you are multiplying an element with the cofactor of the same row (or column) it belongs to.
Assuming multiplication is only row-by-row. You can also multiply row-by-column, column-by-row, or column-by-column. All give the same product $Δ_{1} Δ_{2}$ .
Applying the power-cofactor formula with the wrong exponent. For order $n$ , the cofactor determinant is $Δ^{n - 1}$ , not $Δ^{n}$ . So a $3 \times 3$ cofactor determinant equals $Δ^{2}$ , and a $2 \times 2$ cofactor determinant equals $Δ$ .
Expanding along a random row when a zero-heavy row is available. Always expand along the row or column with the most zeros - it dramatically cuts work.

Frequently Asked Questions

What is the difference between a matrix and a determinant?

A matrix is an array of numbers arranged in rows and columns; it is denoted with square brackets $[]$ and has no single numerical value. A determinant is a scalar value computed only from a square array of numbers, denoted with vertical bars $∣ ∣$ . Every square matrix has a determinant, but only square matrices do.

How do I decide which row or column to expand along?

Choose the row or column with the most zeros. Each zero eliminates a $2 \times 2$ minor calculation, so a row with two zeros in a $3 \times 3$ determinant reduces the work to a single $2 \times 2$ evaluation. If no row or column has zeros, expand along $R_{1}$ or $C_{1}$ by default.

Why is the cofactor sign given by $(- 1)^{i + j}$ ?

The sign comes from the recursive definition of determinants using permutations. It ensures that swapping any two rows (or columns) flips the sign of the determinant, which is one of the axioms of the determinant function.

What is the power-cofactor formula used for in JEE problems?

It lets you compute the determinant of a cofactor matrix without doing the full expansion. If a problem gives you the cofactor determinant $∣ A_{ij} ∣$ of an order- $n$ determinant and asks for the original, you can recover $Δ$ via $∣ A_{ij} ∣ = Δ^{n - 1}$ . This is a common trick in JEE Advanced problems.

Can two determinants of different orders be multiplied?

No. The product rule for determinants requires both determinants to be of the same order (both $2 \times 2$ , both $3 \times 3$ , etc.), just like matrix multiplication requires compatible dimensions. To multiply determinants of different orders, you would first need to extend the smaller one with padding rows and columns of an identity block.

Is the value of a determinant unique regardless of how I expand it?

Yes. A determinant has a single value; expanding along any row or column, or using any multiplication order for the product rule, gives the same scalar. This is why row-column expansion is a well-defined operation.

How is the determinant of a $2 \times 2$ related to area?

For a $2 \times 2$ determinant with rows representing two vectors in the plane, the absolute value equals the area of the parallelogram spanned by those vectors. The sign indicates orientation. This gives determinants a direct geometric meaning that generalises to volume in $3 \times 3$ and beyond.

Previous year questions on Algebra Of Determinants

16 questions from past papers, each with a step-by-step solution.

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1. What is a Determinant?

Third-order Determinants

2. Minors and Cofactors

Sign Chessboard

Expansion by Cofactors

3. Product of Two Determinants

Power Cofactor Formula

Common Mistakes to Avoid

Frequently Asked Questions

What is the difference between a matrix and a determinant?

How do I decide which row or column to expand along?

Why is the cofactor sign given by (−1)i+j?

What is the power-cofactor formula used for in JEE problems?

Can two determinants of different orders be multiplied?

Is the value of a determinant unique regardless of how I expand it?

How is the determinant of a 2×2 related to area?

Previous year questions on Algebra Of Determinants

Ready to master Determinants?

Why is the cofactor sign given by $(- 1)^{i + j}$ ?

How is the determinant of a $2 \times 2$ related to area?