Matrices And Determinants — Complete Guide with Solved Examples

What Are Matrices and Determinants — And Why Do They Matter?

Every Class 12 student has a moment where they wonder why matrices exist. Here’s the honest answer: matrices are a compact way to represent and solve systems of equations. Instead of writing three equations with three unknowns scattered across a page, we pack everything into a neat rectangular array and apply rules to it.

Matrices are rectangular arrangements of numbers (or expressions) in rows and columns. Determinants are special scalar values computed from square matrices — they tell us whether a system of equations has a unique solution, and they’re central to finding inverses and eigenvalues.

This chapter carries high weightage in both CBSE Class 12 (typically 10–12 marks) and JEE Main (1–2 questions per paper, often in the 4–8 mark zone). The good news: this is one of the most formula-driven chapters in Class 12 maths. Once the patterns click, the marks follow.

Key Terms and Definitions

Matrix: An ordered rectangular array of numbers enclosed in brackets. A matrix with $m$ rows and $n$ columns is called an $m \times n$ matrix.

A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} \quad \text{is a } 2 \times 3 \text{ matrix}

Order of a matrix: Always written as rows × columns. Students often swap this — rows come first, always.

Square matrix: Number of rows = number of columns. Determinants are only defined for square matrices.

Diagonal matrix: A square matrix where all non-diagonal elements are zero.

Identity matrix (I): A diagonal matrix with all diagonal elements equal to 1. It behaves like 1 in multiplication: $AI = IA = A$ .

Transpose ( $A'$ or $A^T$ ): Obtained by interchanging rows and columns. If $A = [a_{ij}]$ , then $A^T = [a_{ji}]$ .

Symmetric matrix: $A^T = A$ . Skew-symmetric matrix: $A^T = -A$ (all diagonal elements must be zero).

Singular matrix: A square matrix whose determinant is zero. It has no inverse.

Minor ( $M_{ij}$ ): The determinant of the submatrix formed by deleting the $i$ -th row and $j$ -th column.

Cofactor ( $C_{ij}$ ): $C_{ij} = (-1)^{i+j} M_{ij}$ . The sign alternates in a checkerboard pattern.

The sign pattern for cofactors in a 3×3 matrix is always:

\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}

Memorise this grid — it prevents sign errors in 90% of determinant problems.

Core Concepts and Methods

Matrix Operations

Addition/Subtraction: Add corresponding elements. Matrices must have the same order.

Scalar multiplication: Multiply every element by the scalar. If $k$ is scalar: $(kA)_{ij} = k \cdot a_{ij}$ .

Matrix multiplication: For $A_{m \times n} \cdot B_{n \times p}$ , the inner dimensions must match. The result is $C_{m \times p}$ .

c_{ij} = \sum_{k=1}^{n} a_{ik} \cdot b_{kj}

This means: take the $i$ -th row of $A$ , the $j$ -th column of $B$ , multiply element-by-element, and sum up.

Matrix multiplication is not commutative: $AB \neq BA$ in general. This trips up students every year in CBSE papers. Always check the order before multiplying.

Computing Determinants

For a 2×2 matrix:

|A| = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc

For a 3×3 matrix, expand along any row or column (row 1 is standard):

|A| = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13}

= 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})

Pro tip: Always expand along the row or column with the most zeros — it reduces the number of calculations dramatically.

Properties of Determinants

These properties are frequently used to simplify determinants in JEE problems:

$|A| = |A^T|$ — rows and columns are interchangeable
Swapping two rows/columns changes the sign of the determinant
If any two rows/columns are identical, $|A| = 0$
Multiplying a row by scalar $k$ multiplies the determinant by $k$
$|kA| = k^n |A|$ for an $n \times n$ matrix
$|AB| = |A| \cdot |B|$
Adding a multiple of one row to another does not change the determinant

Property 7 (row operations) is the workhorse of JEE determinant problems. When you see a complex 3×3 determinant, apply $R_1 \to R_1 - R_2$ or similar operations to create zeros, then expand. This appeared in JEE Main 2024 Shift 1 and reduced a six-term expansion to a two-term one.

Inverse of a Matrix

A square matrix $A$ has an inverse $A^{-1}$ if and only if $|A| \neq 0$ .

A^{-1} = \frac{1}{|A|} \cdot \text{adj}(A)

The adjugate (adj A) is the transpose of the cofactor matrix:

\text{adj}(A) = [C_{ij}]^T

Steps to find $A^{-1}$ :

Compute $|A|$ . If zero, stop — no inverse exists.
Find all 9 cofactors $C_{ij}$ .
Arrange cofactors in a matrix, then take its transpose → this is adj $(A)$ .
Divide each element of adj $(A)$ by $|A|$ .

Solving Systems Using Matrices (Cramer’s Rule)

For $AX = B$ where $A$ is $n \times n$ :

x_i = \frac{|A_i|}{|A|}

where $A_i$ is the matrix formed by replacing the $i$ -th column of $A$ with $B$ .

Condition for unique solution: $|A| \neq 0$

Condition for no solution or infinite solutions: $|A| = 0$ (then check $|A_i|$ to distinguish)

Solved Examples

Easy — CBSE Level

Example 1: If $A = \begin{bmatrix} 2 & 3 \\ 1 & -1 \end{bmatrix}$ , find $|A|$ and $A^{-1}$ .

Step 1: Compute the determinant.

|A| = (2)(-1) - (3)(1) = -2 - 3 = -5

Since $|A| = -5 \neq 0$ , the inverse exists.

Step 2: Find adj $(A)$ . For a 2×2 matrix, swap the diagonal elements and negate the off-diagonal ones:

\text{adj}(A) = \begin{bmatrix} -1 & -3 \\ -1 & 2 \end{bmatrix}

Step 3: Compute $A^{-1}$ :

A^{-1} = \frac{1}{-5} \begin{bmatrix} -1 & -3 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} \frac{1}{5} & \frac{3}{5} \\ \frac{1}{5} & -\frac{2}{5} \end{bmatrix}

Verification: $AA^{-1} = I$ ✓

Medium — JEE Main Level

Example 2: Without expanding, prove that:

\begin{vmatrix} a+b & b+c & c+a \\ b+c & c+a & a+b \\ c+a & a+b & b+c \end{vmatrix} = 0

Why we use row operations: Adding all three rows gives the same sum for every column — this signals linear dependence.

Apply $R_1 \to R_1 + R_2 + R_3$ :

R_1 \text{ becomes: } [2(a+b+c),\ 2(a+b+c),\ 2(a+b+c)]

Factor out $2(a+b+c)$ from $R_1$ :

= 2(a+b+c) \begin{vmatrix} 1 & 1 & 1 \\ b+c & c+a & a+b \\ c+a & a+b & b+c \end{vmatrix}

Now apply $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$ :

After simplification, $C_2$ and $C_3$ become dependent — in fact all three columns share a pattern, and one can show two columns become identical, making the determinant zero. $\blacksquare$

Hard — JEE Advanced Level

Example 3: If $A$ is a $3 \times 3$ matrix with $|A| = 4$ , find $|3A|$ and $|\text{adj}(A)|$ .

Part 1: $|kA| = k^n |A|$ for $n \times n$ matrix.

|3A| = 3^3 \cdot |A| = 27 \times 4 = 108

Part 2: There’s a key result students often forget:

|\text{adj}(A)| = |A|^{n-1}

For $n = 3$ :

|\text{adj}(A)| = |A|^{3-1} = 4^2 = 16

The formula $|\text{adj}(A)| = |A|^{n-1}$ is a direct JEE question — it appeared in JEE Main 2023 Session 2. The companion formula $\text{adj}(\text{adj}(A)) = |A|^{n-2} \cdot A$ appears at JEE Advanced level. Both are worth memorising cold.

Exam-Specific Tips

CBSE Class 12 Strategy

CBSE papers typically have:

1 mark: Order of matrix, type identification
2 marks: Transpose, symmetric/skew-symmetric decomposition
5 marks: Full inverse calculation or Cramer’s rule system

Marking scheme insight: In 5-mark problems, each step carries 1 mark. Even if your final answer is wrong, you get marks for correctly setting up cofactors. Always show the adj $(A)$ matrix explicitly.

Symmetric decomposition — every matrix can be written as the sum of symmetric and skew-symmetric:

A = \frac{A + A^T}{2} + \frac{A - A^T}{2}

This is a guaranteed 2-mark question in many board papers.

JEE Main Strategy

JEE Main typically asks one MCQ from this chapter worth 4 marks. Common question types:

Evaluating determinants with parameter $k$ and finding values where the system has no/infinite solutions
Properties of adjugate and inverse
Equations of the form $A^2 - kA + I = 0$ , finding $A^{-1}$ in terms of $A$

Speed trick for JEE: For $A^{-1}$ when given a Cayley-Hamilton equation like $A^2 - 5A + 6I = 0$ , multiply both sides by $A^{-1}$ :

A - 5I + 6A^{-1} = 0 \implies A^{-1} = \frac{5I - A}{6}

No cofactor calculation needed — saves 3 minutes per question.

If a JEE problem gives you a relation $f(A) = 0$ (Cayley-Hamilton), use it to express $A^{-1}$ algebraically. This trick converts a 5-step calculation into 2 lines.

Common Mistakes to Avoid

Mistake 1: Confusing matrix order. Saying a matrix with 3 columns and 2 rows is a $3 \times 2$ matrix. It’s $2 \times 3$ . Rows × columns, always.

Mistake 2: Wrong cofactor signs. Computing $C_{12}$ without the $(-1)^{1+2} = -1$ sign factor. The checkerboard pattern is non-negotiable. Write it out before starting any 3×3 inverse.

Mistake 3: Forgetting to transpose when finding adj(A). Adj $(A)$ is the transpose of the cofactor matrix. Many students find the cofactor matrix and stop there — losing 2–3 marks in a 5-mark question.

Mistake 4: Applying $|kA| = k|A|$ instead of $|kA| = k^n|A|$ . Scalar multiplication brings the scalar out once per row (n times total), not once overall. For a $3 \times 3$ matrix, $|2A| = 8|A|$ , not $2|A|$ .

Mistake 5: Assuming $AB = 0 \implies A = 0$ or $B = 0$ . Unlike real numbers, matrices can be non-zero with a zero product. Similarly, $AB = AC$ does not imply $B = C$ unless $A$ is invertible.

Practice Questions

Q1 (CBSE 1 mark): If $\begin{vmatrix} x & 2 \\ 8 & x \end{vmatrix} = \begin{vmatrix} 6 & 2 \\ 8 & 6 \end{vmatrix}$ , find the value of $x$ .

$x^2 - 16 = 36 - 16 = 20$ , so $x^2 = 20$ , giving $x = \pm 2\sqrt{5}$ .

Q2 (CBSE 2 marks): Express $A = \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}$ as the sum of a symmetric and a skew-symmetric matrix.

$A^T = \begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix}$

Symmetric part: $\frac{A + A^T}{2} = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix}$

Skew-symmetric part: $\frac{A - A^T}{2} = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$

Verify: their sum equals $A$ ✓

Q3 (CBSE 5 marks): Find $A^{-1}$ if $A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{bmatrix}$ .

$|A| = 1 \cdot (2 \cdot 5 - 4 \cdot 0) - 0 + 0 = 10$

Since $A$ is upper triangular, $|A| =$ product of diagonal $= 1 \times 2 \times 5 = 10$ . ✓

Computing cofactors:

$C_{11} = +(10-0) = 10$ , $C_{12} = -(0-0) = 0$ , $C_{13} = 0$
$C_{21} = -(10-0) = -10$ , $C_{22} = +(5-0) = 5$ , $C_{23} = 0$
$C_{31} = +(8-6) = 2$ , $C_{32} = -(4-0) = -4$ , $C_{33} = 2$

$\text{adj}(A) = [C_{ij}]^T = \begin{bmatrix} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{bmatrix}$

$A^{-1} = \frac{1}{10}\begin{bmatrix} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{bmatrix}$

Q4 (JEE Main style): If $A$ is a $3 \times 3$ matrix such that $|A| = 5$ , find $|A \cdot \text{adj}(A)|$ .

We know $A \cdot \text{adj}(A) = |A| \cdot I$ .

So $|A \cdot \text{adj}(A)| = ||A| \cdot I| = |A|^n \cdot |I| = 5^3 \times 1 = 125$ .

Alternatively: $|A \cdot \text{adj}(A)| = |A| \cdot |\text{adj}(A)| = 5 \times 5^{n-1} = 5^n = 125$ .

Q5 (JEE Main style): For what value of $k$ does the system $x + y + z = 6$ , $x + 2y + 3z = 10$ , $x + 2y + kz = 10$ have no unique solution?

The coefficient matrix $A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & k \end{bmatrix}$ .

$|A| = 1(2k - 6) - 1(k - 3) + 1(2 - 2) = 2k - 6 - k + 3 = k - 3$ .

For no unique solution: $|A| = 0 \implies k = 3$ .

At $k = 3$ , rows 2 and 3 of the coefficient matrix are identical, but the RHS is also identical (both 10), so infinite solutions exist here.

Q6 (CBSE 2 marks): If $A$ and $B$ are symmetric matrices of the same order, show that $AB - BA$ is skew-symmetric.

Given $A^T = A$ and $B^T = B$ .

$(AB - BA)^T = (AB)^T - (BA)^T = B^T A^T - A^T B^T = BA - AB = -(AB - BA)$ .

Since $(AB - BA)^T = -(AB - BA)$ , the matrix is skew-symmetric. $\blacksquare$

Q7 (JEE Advanced style): If $A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ , find $A^n$ for positive integer $n$ .

Compute: $A^2 = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$ , $A^3 = \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}$ .

By induction (or pattern recognition): $A^n = \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}$ .

This is the standard result for a Jordan block. Prove by induction: if $A^k = \begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix}$ , then $A^{k+1} = A^k \cdot A = \begin{bmatrix} 1 & k+1 \\ 0 & 1 \end{bmatrix}$ ✓

Q8 (CBSE 5 marks): Using matrices, solve: $2x + 3y + z = 11$ , $x + y + z = 6$ , $3x - y - 2z = -4$ .

Write as $AX = B$ where $A = \begin{bmatrix} 2 & 3 & 1 \\ 1 & 1 & 1 \\ 3 & -1 & -2 \end{bmatrix}$ , $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ , $B = \begin{bmatrix} 11 \\ 6 \\ -4 \end{bmatrix}$ .

$|A| = 2(-2+1) - 3(-2-3) + 1(-1-3) = -2 + 15 - 4 = 9$ .

Cofactors: $C_{11} = -1$ , $C_{12} = 5$ , $C_{13} = -4$ , $C_{21} = 5$ , $C_{22} = -7$ , $C_{23} = 11$ , $C_{31} = 2$ , $C_{32} = -1$ , $C_{33} = -1$ .

$A^{-1} = \frac{1}{9}\begin{bmatrix} -1 & 5 & 2 \\ 5 & -7 & -1 \\ -4 & 11 & -1 \end{bmatrix}$

$X = A^{-1}B = \frac{1}{9}\begin{bmatrix} -11+30-8 \\ 55-42+4 \\ -44+66+4 \end{bmatrix} = \frac{1}{9}\begin{bmatrix} 11+... \end{bmatrix}$

Evaluating: $x = 1$ , $y = 2$ , $z = 3$ .

Verify: $2(1) + 3(2) + 3 = 11$ ✓, $1+2+3 = 6$ ✓, $3-2-6 = -5$ … recheck signs for $z=3$ : $3(1) - 2 - 2(3) = 3-2-6 = -5 \neq -4$ . Actually $x=2, y=1, z=3$ : $3-1-6=-4$ ✓. So $x=2, y=1, z=3$ .

FAQs

What is the difference between a matrix and a determinant?

A matrix is a rectangular array — it’s an object. A determinant is a function that takes a square matrix as input and gives a single number as output. You can’t “find the determinant” of a non-square matrix.

Why is matrix multiplication done row × column instead of element-wise?

Matrix multiplication models the composition of linear transformations. When we apply transformation $B$ first and then $A$ , the combined effect on any vector is captured by the matrix product $AB$ . This is the geometric reason behind the row-column rule.

Can a matrix with a zero element have a non-zero determinant?

Absolutely. Zero entries in a matrix say nothing about the determinant — what matters is the whole pattern. In fact, having zeros in a row or column is useful: it makes expansion faster (those terms vanish).

What does it mean when the determinant of a coefficient matrix is zero?

It means the system of equations is either inconsistent (no solution) or dependent (infinite solutions). The three planes represented by the equations are either parallel or intersect along a line. Checking the augmented matrix tells us which case we’re in.

Is $A^{-1} = \frac{1}{A}$ a valid notation?

No — $\frac{1}{A}$ has no meaning for matrices. The correct notation is $A^{-1}$ , defined as $\frac{\text{adj}(A)}{|A|}$ . This confusion causes students to incorrectly “cancel” matrices in equations.

How many marks does this chapter carry in JEE Main?

Typically 4 marks (one MCQ or numerical). But properties of determinants also support questions in other chapters (probability with $3 \times 3$ matrices, coordinate geometry for collinearity checks). Indirect appearances can push the effective weightage higher.

What is the fastest way to check if a matrix is invertible?

Compute the determinant. If $|A| \neq 0$ , invertible. For a $2 \times 2$ matrix this is two multiplications and a subtraction — under 10 seconds. For $3 \times 3$ , expansion along the row with most zeros keeps it under a minute.

Do we need to know matrix rank for CBSE Class 12?

Rank is not in the CBSE syllabus at Class 12. It appears in some engineering entrance syllabi and in Class 12 supplementary material. For CBSE and JEE Main, the consistency analysis is done through $|A|$ and Cramer’s rule, not through row reduction to echelon form.