Matrices & Determinants
Comprehensive coverage of matrix algebra, determinant properties, inverse matrices, Cramer's rule, and systems of linear equations at JEE Advanced level.
0% complete5 units
01
Matrix Algebra
Types of matrices, matrix operations, transpose, symmetric and skew-symmetric matrices.
Types & Operations
Definition — Matrix
A matrix is a rectangular array of numbers arranged in rows and columns. An $m \times n$ matrix $A = [a_{ij}]$ has $m$ rows and $n$ columns.
- Square matrix: $m = n$. Diagonal matrix: $a_{ij} = 0$ for $i \neq j$.
- Identity matrix: $I_n$ with $a_{ii} = 1$, $a_{ij} = 0$ for $i \neq j$.
- Scalar matrix: $kI_n$. Null/Zero matrix: all entries $0$.
- Triangular: Upper ($a_{ij} = 0$ for $i > j$) or lower ($a_{ij} = 0$ for $i < j$).
Matrix Multiplication
If $A$ is $m \times n$ and $B$ is $n \times p$, then $AB$ is $m \times p$ with $(AB)_{ij} = \sum_{k=1}^{n} a_{ik}b_{kj}$.
Key: $AB \neq BA$ in general. $AB = 0$ does not imply $A = 0$ or $B = 0$.
Key: $AB \neq BA$ in general. $AB = 0$ does not imply $A = 0$ or $B = 0$.
Transpose & Symmetric/Skew-Symmetric
$(A^T)_{ij} = a_{ji}$. Properties: $(AB)^T = B^T A^T$, $(A^T)^T = A$.
Symmetric: $A^T = A$. Skew-symmetric: $A^T = -A$ (diagonal entries must be $0$).
JEE Trick: Every square matrix $A$ can be written as $A = \frac{A + A^T}{2} + \frac{A - A^T}{2}$ (symmetric + skew-symmetric).
Symmetric: $A^T = A$. Skew-symmetric: $A^T = -A$ (diagonal entries must be $0$).
JEE Trick: Every square matrix $A$ can be written as $A = \frac{A + A^T}{2} + \frac{A - A^T}{2}$ (symmetric + skew-symmetric).
📝 Example
If $A$ is a $3 \times 3$ skew-symmetric matrix, find $\det(A)$.
$A^T = -A \Rightarrow \det(A^T) = \det(-A) \Rightarrow \det(A) = (-1)^3 \det(A) = -\det(A)$. So $2\det(A) = 0$, giving $\det(A) = 0$.
General rule: For odd-order skew-symmetric matrices, the determinant is always $0$.
General rule: For odd-order skew-symmetric matrices, the determinant is always $0$.
02
Determinants
Properties of determinants, cofactor expansion, area of triangle, and computational tricks for JEE.
Properties & Expansion
Determinant of a 3x3 Matrix
$\det(A) = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix} = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31})$
Key Properties
P1: $\det(A^T) = \det(A)$
P2: Swapping two rows/columns changes the sign.
P3: If two rows/columns are identical, $\det = 0$.
P4: $\det(kA) = k^n \det(A)$ for $n \times n$ matrix.
P5: $\det(AB) = \det(A) \cdot \det(B)$.
P6: Adding a scalar multiple of one row to another does not change the determinant.
JEE Trick: Factor out common elements from rows/columns before expanding.
P2: Swapping two rows/columns changes the sign.
P3: If two rows/columns are identical, $\det = 0$.
P4: $\det(kA) = k^n \det(A)$ for $n \times n$ matrix.
P5: $\det(AB) = \det(A) \cdot \det(B)$.
P6: Adding a scalar multiple of one row to another does not change the determinant.
JEE Trick: Factor out common elements from rows/columns before expanding.
Area of Triangle
Area of triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$:
$\text{Area} = \frac{1}{2}\left|\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}\right|$
If area $= 0$, the three points are collinear.
$\text{Area} = \frac{1}{2}\left|\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}\right|$
If area $= 0$, the three points are collinear.
📝 Example
Evaluate: $\begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix}$
This is the Vandermonde determinant. $R_2 \to R_2 - R_1$, $R_3 \to R_3 - R_1$:
$= \begin{vmatrix} 1 & a & a^2 \\ 0 & b-a & b^2-a^2 \\ 0 & c-a & c^2-a^2 \end{vmatrix} = (b-a)(c^2-a^2) - (c-a)(b^2-a^2)$
$= (b-a)(c-a)(c+a) - (c-a)(b-a)(b+a) = (b-a)(c-a)[(c+a)-(b+a)]$
$= \boxed{(a-b)(b-c)(c-a)}$.
$= \begin{vmatrix} 1 & a & a^2 \\ 0 & b-a & b^2-a^2 \\ 0 & c-a & c^2-a^2 \end{vmatrix} = (b-a)(c^2-a^2) - (c-a)(b^2-a^2)$
$= (b-a)(c-a)(c+a) - (c-a)(b-a)(b+a) = (b-a)(c-a)[(c+a)-(b+a)]$
$= \boxed{(a-b)(b-c)(c-a)}$.
03
Inverse of a Matrix
Adjoint, inverse using adjoint method, properties of inverse, and the Cayley-Hamilton theorem.
Adjoint & Inverse
Adjoint
$\text{adj}(A) = [\text{cofactor matrix of } A]^T$. The $(i,j)$-entry of $\text{adj}(A)$ is $C_{ji}$ (cofactor of $a_{ji}$).
Inverse Formula
$A^{-1} = \frac{1}{\det(A)} \text{adj}(A)$, provided $\det(A) \neq 0$.
Key identities: $A \cdot \text{adj}(A) = \det(A) \cdot I$, $\det(\text{adj}(A)) = [\det(A)]^{n-1}$.
Key identities: $A \cdot \text{adj}(A) = \det(A) \cdot I$, $\det(\text{adj}(A)) = [\det(A)]^{n-1}$.
Properties of Inverse
$(AB)^{-1} = B^{-1}A^{-1}$
$(A^T)^{-1} = (A^{-1})^T$
$\det(A^{-1}) = \frac{1}{\det(A)}$
$(kA)^{-1} = \frac{1}{k}A^{-1}$
$(A^T)^{-1} = (A^{-1})^T$
$\det(A^{-1}) = \frac{1}{\det(A)}$
$(kA)^{-1} = \frac{1}{k}A^{-1}$
Cayley-Hamilton Theorem
Every square matrix satisfies its own characteristic equation. If $p(\lambda) = \det(A - \lambda I) = 0$ is the characteristic polynomial, then $p(A) = O$ (zero matrix).
JEE Application: Use this to find $A^{-1}$ or higher powers of $A$ by reducing them using the characteristic equation.
JEE Application: Use this to find $A^{-1}$ or higher powers of $A$ by reducing them using the characteristic equation.
📝 Example
If $A = \begin{pmatrix} 2 & 1 \\ 1 & 3 \end{pmatrix}$, find $A^{-1}$ using Cayley-Hamilton.
Characteristic equation: $\lambda^2 - 5\lambda + 5 = 0$ (trace $= 5$, det $= 5$).
By Cayley-Hamilton: $A^2 - 5A + 5I = O \Rightarrow A^2 = 5A - 5I$.
Multiply by $A^{-1}$: $A = 5I - 5A^{-1} \Rightarrow A^{-1} = \frac{1}{5}(5I - A) = I - \frac{A}{5} = \frac{1}{5}\begin{pmatrix} 3 & -1 \\ -1 & 2 \end{pmatrix}$.
By Cayley-Hamilton: $A^2 - 5A + 5I = O \Rightarrow A^2 = 5A - 5I$.
Multiply by $A^{-1}$: $A = 5I - 5A^{-1} \Rightarrow A^{-1} = \frac{1}{5}(5I - A) = I - \frac{A}{5} = \frac{1}{5}\begin{pmatrix} 3 & -1 \\ -1 & 2 \end{pmatrix}$.
04
Systems of Linear Equations
Cramer's rule, consistency of systems, homogeneous systems — tested extensively in JEE Advanced.
Cramer's Rule & Consistency
Cramer's Rule (for $n$ equations in $n$ unknowns)
For $AX = B$ where $A$ is $n \times n$:
If $\det(A) \neq 0$: unique solution $x_i = \frac{D_i}{D}$ where $D = \det(A)$ and $D_i$ is obtained by replacing the $i$-th column of $A$ with $B$.
If $\det(A) \neq 0$: unique solution $x_i = \frac{D_i}{D}$ where $D = \det(A)$ and $D_i$ is obtained by replacing the $i$-th column of $A$ with $B$.
Non-Homogeneous System $AX = B$
Case 1: $D \neq 0$ — unique solution (consistent).
Case 2: $D = 0$ and at least one $D_i \neq 0$ — no solution (inconsistent).
Case 3: $D = 0$ and all $D_i = 0$ — infinitely many solutions or no solution (check further).
Case 2: $D = 0$ and at least one $D_i \neq 0$ — no solution (inconsistent).
Case 3: $D = 0$ and all $D_i = 0$ — infinitely many solutions or no solution (check further).
Homogeneous System $AX = O$
Always has the trivial solution $X = O$.
Non-trivial solutions exist iff $\det(A) = 0$.
JEE Trick: If a homogeneous system has a non-trivial solution, then $\det(A) = 0$. This is frequently used to find unknown parameters.
Non-trivial solutions exist iff $\det(A) = 0$.
JEE Trick: If a homogeneous system has a non-trivial solution, then $\det(A) = 0$. This is frequently used to find unknown parameters.
📝 Example
For what value of $k$ does the system $x + y + z = 0$, $x + ky + 3z = 0$, $x + 3y + kz = 0$ have a non-trivial solution?
$\det(A) = \begin{vmatrix} 1 & 1 & 1 \\ 1 & k & 3 \\ 1 & 3 & k \end{vmatrix} = 1(k^2-9) - 1(k-3) + 1(3-k) = k^2 - 9 - k + 3 + 3 - k = k^2 - 2k - 3 = (k-3)(k+1) = 0$.
So $k = 3$ or $k = -1$.
So $k = 3$ or $k = -1$.
05
Rank & Applications
Rank of a matrix, echelon form, and applications to systems — JEE Advanced level.
Rank of a Matrix
Definition — Rank
The rank of a matrix $A$, denoted $\rho(A)$, is the order of the largest non-zero minor (or equivalently, the number of non-zero rows in the row echelon form).
Rank and System Consistency
For $AX = B$ ($A$ is $m \times n$, augmented matrix $[A|B]$):
Consistent iff $\rho(A) = \rho([A|B])$.
Unique solution iff $\rho(A) = \rho([A|B]) = n$.
Infinite solutions iff $\rho(A) = \rho([A|B]) < n$ (with $n - \rho(A)$ free variables).
Consistent iff $\rho(A) = \rho([A|B])$.
Unique solution iff $\rho(A) = \rho([A|B]) = n$.
Infinite solutions iff $\rho(A) = \rho([A|B]) < n$ (with $n - \rho(A)$ free variables).
JEE Useful Facts
$\rho(A) \leq \min(m, n)$
$\rho(AB) \leq \min(\rho(A), \rho(B))$
$\rho(A + B) \leq \rho(A) + \rho(B)$
For $n \times n$ matrix: $\rho(A) = n \iff \det(A) \neq 0 \iff A$ is invertible.
$\rho(AB) \leq \min(\rho(A), \rho(B))$
$\rho(A + B) \leq \rho(A) + \rho(B)$
For $n \times n$ matrix: $\rho(A) = n \iff \det(A) \neq 0 \iff A$ is invertible.
📝 Example
Find the rank of $A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 3 & 5 \end{pmatrix}$.
$R_2 \to R_2 - 2R_1$: $\begin{pmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 1 & 3 & 5 \end{pmatrix}$. $R_3 \to R_3 - R_1$: $\begin{pmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 0 & 1 & 2 \end{pmatrix}$. Two non-zero rows, so $\rho(A) = \boxed{2}$.
★ Key Takeaways
- $\det(kA) = k^n \det(A)$ — students frequently forget the power of $n$. This is a top JEE trap.
- For odd-order skew-symmetric matrices, the determinant is always $0$ — use this as a shortcut.
- Cayley-Hamilton theorem is powerful for finding $A^{-1}$ and $A^n$ without direct computation.
- Homogeneous systems always have the trivial solution. Non-trivial solutions exist iff the determinant is $0$.
- Always check rank conditions for system consistency: $\rho(A) = \rho([A|B])$ for consistency, and $= n$ for uniqueness.
✎ Practice Problems
Problem 1
If $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$, find $A^2 - 5A + 2I$.
Show Solution ▼
$A^2 = \begin{pmatrix} 7 & 10 \\ 15 & 22 \end{pmatrix}$, $5A = \begin{pmatrix} 5 & 10 \\ 15 & 20 \end{pmatrix}$, $2I = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$. $A^2 - 5A + 2I = \begin{pmatrix} 4 & 0 \\ 0 & 4 \end{pmatrix} = 4I$. (Note: Cayley-Hamilton gives $A^2 - 5A + (-2)I = O$ since $\det(A)=-2$, trace $=5$. So $A^2 - 5A + 2I = 4I$.)
Problem 2
Evaluate $\begin{vmatrix} a+b & b+c & c+a \\ b+c & c+a & a+b \\ c+a & a+b & b+c \end{vmatrix}$.
Show Solution ▼
Add all three columns: each entry becomes $2(a+b+c)$. Factor out $2(a+b+c)$ from $C_1$. Then $C_2 \to C_2 - C_1$, $C_3 \to C_3 - C_1$. After simplification, the determinant $= 2(a^3 + b^3 + c^3 - 3abc)$.
Problem 3
If $A$ is a $3 \times 3$ matrix with $\det(A) = 5$, find $\det(2A^{-1})$.
Show Solution ▼
$\det(2A^{-1}) = 2^3 \cdot \det(A^{-1}) = 8 \cdot \frac{1}{\det(A)} = \frac{8}{5}$.
Problem 4
If $A^2 = A$ (idempotent matrix), what are the possible values of $\det(A)$?
Show Solution ▼
$A^2 = A \Rightarrow \det(A^2) = \det(A) \Rightarrow [\det(A)]^2 = \det(A) \Rightarrow \det(A)[\det(A) - 1] = 0$. So $\det(A) = 0$ or $\det(A) = 1$.
Problem 5
Find the value of $\lambda$ for which the system $x + y + z = 6$, $x + 2y + 3z = 10$, $x + 2y + \lambda z = 12$ is inconsistent.
Show Solution ▼
$D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & \lambda \end{vmatrix} = 1(2\lambda - 6) - 1(\lambda - 3) + 1(2-2) = 2\lambda - 6 - \lambda + 3 = \lambda - 3$. For inconsistency, $D = 0$, so $\lambda = 3$. Check: with $\lambda = 3$, equations 2 and 3 become $x+2y+3z = 10$ and $x+2y+3z = 12$ (contradiction). So $\lambda = 3$ gives an inconsistent system.
🎯 Interactive MCQs
1. If $A$ is an invertible $3 \times 3$ matrix, then $\det(\text{adj}(A))$ equals:
A $\det(A)$
B $[\det(A)]^2$
C $[\det(A)]^3$
D $\frac{1}{\det(A)}$
2. If $A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 4 \end{pmatrix}$, then $A^{-1}$ is:
A $\frac{1}{6}\begin{pmatrix} 6 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 2 & 1 \end{pmatrix}$
B $\frac{1}{6}\begin{pmatrix} 6 & 0 & 0 \\ 0 & 4 & 2 \\ 0 & -1 & 1 \end{pmatrix}$
C $\frac{1}{4}\begin{pmatrix} 4 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 2 & 1 \end{pmatrix}$
D Does not exist
3. The system $x + 2y - z = 1$, $3x + y + 2z = 4$, $x - 3y + 4z = 2$ has:
A Infinitely many solutions
B A unique solution
C No solution
D Only the trivial solution
4. If $A$ is a $3 \times 3$ matrix such that $A^2 = 2A - I$, then $A^3$ equals:
A $3A - 2I$
B $4A - 3I$
C $8A - 7I$
D $A$
5. The value of $\begin{vmatrix} x & y & x+y \\ y & x+y & x \\ x+y & x & y \end{vmatrix}$ is:
A $0$
B $-2(x^3 + y^3)$
C $2(x^3 + y^3)$
D $3xy(x+y)$