Chapter 3: Matrices

Dear Class 12 Student! Matrices are the foundation of linear algebra. Initially devised to solve systems of linear equations rapidly, they are now the backbone of computer graphics, quantum mechanics, and cryptography. For the Boards, mastering Matrix Multiplication and Symmetric/Skew-Symmetric properties is critical. For JEE, the Trace properties and Special Matrices (Nilpotent, Orthogonal) frequently appear in MCQs. Let's arrange our thoughts!

1. Introduction to Matrices

Definition: A matrix is an ordered rectangular array of numbers or functions. The numbers or functions are called the elements or the entries of the matrix.

Order and Representation Order of a Matrix: A matrix having $m$ rows (horizontal) and $n$ columns (vertical) is said to be of order $\mathbf{m \times n}$ (read as "$m$ by $n$").
General Representation: $$ A = \begin{bmatrix} a_{11} & a_{12} & \dots & a_{1n} \\ a_{21} & a_{22} & \dots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \dots & a_{mn} \end{bmatrix} $$ In compact form: $A = [a_{ij}]_{m \times n}$, where $a_{ij}$ represents the element in the $i^{th}$ row and $j^{th}$ column.

Practice Problem 1: Construction of a Matrix Question: Construct a $2 \times 2$ matrix, $A = [a_{ij}]$, whose elements are given by $a_{ij} = \frac{|i - 2j|}{2}$.

Solution:
For a $2 \times 2$ matrix, the elements are $a_{11}, a_{12}, a_{21}, a_{22}$.
1. $a_{11} = \frac{|1 - 2(1)|}{2} = \frac{|-1|}{2} = \frac{1}{2}$
2. $a_{12} = \frac{|1 - 2(2)|}{2} = \frac{|-3|}{2} = \frac{3}{2}$
3. $a_{21} = \frac{|2 - 2(1)|}{2} = \frac{0}{2} = 0$
4. $a_{22} = \frac{|2 - 2(2)|}{2} = \frac{|-2|}{2} = \frac{2}{2} = 1$
Therefore, the required matrix is: $$ A = \begin{bmatrix} 1/2 & 3/2 \\ 0 & 1 \end{bmatrix} $$

2. Types of Matrices (NCERT Focus)

Column Matrix: A matrix having only one column. Order: $m \times 1$.
Row Matrix: A matrix having only one row. Order: $1 \times n$.
Square Matrix: A matrix in which the number of rows equals the number of columns ($m = n$). A matrix of order $n \times n$ is simply called a square matrix of order $n$.

Diagonal, Scalar, and Identity Matrices (Note: These definitions apply ONLY to Square Matrices).

1. Diagonal Matrix: All its non-diagonal elements are zero. ($a_{ij} = 0$ when $i \neq j$).
Can be denoted as $\text{diag}(d_1, d_2, d_3, \dots)$.

2. Scalar Matrix: A diagonal matrix in which all the principal diagonal elements are exactly equal to a scalar constant $k$. ($a_{ij} = 0$ for $i \neq j$, and $a_{ii} = k$).

3. Identity (Unit) Matrix ($I$): A scalar matrix in which all principal diagonal elements are exactly $1$. $$ I_2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \quad I_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $$

Zero (Null) Matrix ($O$): A matrix in which all elements are zero. It can be of any order (not necessarily square).

Practice Problem 2: Identifying Matrices Question: Classify the following matrix: $A = \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix}$. Can it be called a diagonal matrix?

Solution:
The given matrix is a square matrix ($3 \times 3$). All non-diagonal elements are zero, and all diagonal elements are equal (to 5).
Therefore, it is a Scalar Matrix.
Yes, every scalar matrix is a special case of a Diagonal Matrix. (It is also a square matrix).

3. Special Matrices and Concepts (R.D. Sharma / JEE Focus)

Trace of a Matrix ($Tr(A)$) The Trace of a square matrix $A$ is defined as the sum of its principal diagonal elements.
$$ Tr(A) = \sum_{i=1}^{n} a_{ii} = a_{11} + a_{22} + \dots + a_{nn} $$ Crucial JEE Properties:
1. $Tr(A \pm B) = Tr(A) \pm Tr(B)$
2. $Tr(kA) = k \cdot Tr(A)$
3. $\mathbf{Tr(AB) = Tr(BA)}$ (Even though $AB \neq BA$!)

Advanced Square Matrices

Idempotent Matrix: A matrix $A$ is idempotent if $A^2 = A$. (e.g., Identity matrix $I$).
Nilpotent Matrix: A matrix $A$ is nilpotent if $A^k = O$ for some positive integer $k$. The smallest such $k$ is called the index of nilpotence.
Involutory Matrix: A matrix $A$ is involutory if $A^2 = I$. (e.g., Identity matrix $I$ is involutory).
Orthogonal Matrix: A real square matrix $A$ is orthogonal if $A A' = A' A = I$ (where $A'$ is the transpose).

Practice Problem 3: Trace Properties Question: If $A$ and $B$ are square matrices of order 3, such that $Tr(A) = 4$ and $Tr(B) = -2$, find the trace of the matrix $(2A - 3B)$.

Solution:
Using the linear properties of Trace:
$Tr(2A - 3B) = Tr(2A) - Tr(3B)$
$= 2 \cdot Tr(A) - 3 \cdot Tr(B)$
$= 2(4) - 3(-2)$
$= 8 + 6 = \mathbf{14}$.

Practice Problem 4: Idempotent Matrix Question: Show that the matrix $A = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix}$ is an idempotent matrix.

Solution:
A matrix is idempotent if $A^2 = A$. We must calculate $A \times A$.
$$ A^2 = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix} \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix} $$ Row 1 $\times$ Col 1: $(2)(2) + (-2)(-1) + (-4)(1) = 4 + 2 - 4 = 2$
Row 1 $\times$ Col 2: $(2)(-2) + (-2)(3) + (-4)(-2) = -4 - 6 + 8 = -2$
Row 1 $\times$ Col 3: $(2)(-4) + (-2)(4) + (-4)(-3) = -8 - 8 + 12 = -4$
Continuing this for all rows, we get:
$$ A^2 = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix} = A $$ Since $A^2 = A$, the matrix is idempotent.

4. Equality of Matrices

Two matrices $A = [a_{ij}]$ and $B = [b_{ij}]$ are said to be equal ($A = B$) if and only if:

They are of the exact same order.
Each corresponding element is identically equal ($a_{ij} = b_{ij}$ for all possible values of $i$ and $j$).

Practice Problem 5: Solving Equations using Equality Question: Find the values of $x, y, z,$ and $w$ if: $$ \begin{bmatrix} x - y & 2x + z \\ 2x - y & 3z + w \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix} $$

Solution:
Equating the corresponding elements of the two matrices:
1) $x - y = -1$
2) $2x - y = 0 \implies y = 2x$
Substitute (2) into (1): $x - 2x = -1 \implies -x = -1 \implies \mathbf{x = 1}$.
From (2), $y = 2(1) \implies \mathbf{y = 2}$.
3) $2x + z = 5 \implies 2(1) + z = 5 \implies \mathbf{z = 3}$.
4) $3z + w = 13 \implies 3(3) + w = 13 \implies 9 + w = 13 \implies \mathbf{w = 4}$.

5. Operations on Matrices

A. Addition and Scalar Multiplication

Addition: Valid ONLY if the matrices are of the same order. You simply add the corresponding elements.
Properties: Commutative ($A+B = B+A$), Associative ($(A+B)+C = A+(B+C)$), Additive Identity is the Null Matrix ($A+O = A$), Additive Inverse is $-A$.

Scalar Multiplication: If a matrix $A$ is multiplied by a scalar $k$, then every single element of $A$ is multiplied by $k$. ($[k a_{ij}]$).

B. Multiplication of Matrices (Crucial Skill)

Condition for Matrix Multiplication Two matrices $A$ and $B$ can be multiplied to find the product $AB$ ONLY IF the number of columns in $A$ equals the number of rows in $B$.
If $A$ is of order $(m \times \mathbf{n})$ and $B$ is of order $(\mathbf{n} \times p)$, then $AB$ is defined and its order will be $\mathbf{m \times p}$.

Row-by-Column Rule: To find the element in the $i^{th}$ row and $j^{th}$ column of the product matrix $AB$, you take the dot product of the $i^{th}$ row of matrix $A$ and the $j^{th}$ column of matrix $B$.

Practice Problem 6: Matrix Multiplication Question: Let $A = \begin{bmatrix} 1 & -2 & 3 \\ -4 & 2 & 5 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 3 \\ 4 & 5 \\ 2 & 1 \end{bmatrix}$. Find $AB$.

Solution:
Order of $A$ is $2 \times 3$. Order of $B$ is $3 \times 2$.
Columns of $A$ (3) = Rows of $B$ (3). Matrix multiplication is possible. The resulting matrix $AB$ will be of order $2 \times 2$.
Let $AB = C = \begin{bmatrix} c_{11} & c_{12} \\ c_{21} & c_{22} \end{bmatrix}$.

$c_{11} = (\text{Row 1 of A}) \cdot (\text{Col 1 of B}) = (1)(2) + (-2)(4) + (3)(2) = 2 - 8 + 6 = 0$
$c_{12} = (\text{Row 1 of A}) \cdot (\text{Col 2 of B}) = (1)(3) + (-2)(5) + (3)(1) = 3 - 10 + 3 = -4$
$c_{21} = (\text{Row 2 of A}) \cdot (\text{Col 1 of B}) = (-4)(2) + (2)(4) + (5)(2) = -8 + 8 + 10 = 10$
$c_{22} = (\text{Row 2 of A}) \cdot (\text{Col 2 of B}) = (-4)(3) + (2)(5) + (5)(1) = -12 + 10 + 5 = 3$

$$ AB = \begin{bmatrix} 0 & -4 \\ 10 & 3 \end{bmatrix} $$

Properties of Matrix Multiplication 1. Non-Commutative: In general, $AB \neq BA$. Even if $AB$ is defined, $BA$ might not be defined. Even if both are defined, they are rarely equal.
2. Associative: $(AB)C = A(BC)$, provided dimensions allow multiplication.
3. Distributive: $A(B + C) = AB + AC$.
4. Multiplicative Identity: For any square matrix $A$ of order $n$, $AI = IA = A$.

Practice Problem 7: Zero Divisors Anomaly Question: In real numbers, if $ab = 0$, then either $a = 0$ or $b = 0$. Show by an example that this is NOT true for matrices.

Solution:
Let $A = \begin{bmatrix} 0 & -1 \\ 0 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 3 & 5 \\ 0 & 0 \end{bmatrix}$.
Clearly, $A \neq O$ and $B \neq O$.
Now, compute $AB$:
$$ AB = \begin{bmatrix} 0(3) + (-1)(0) & 0(5) + (-1)(0) \\ 0(3) + 2(0) & 0(5) + 2(0) \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O $$
Thus, the product of two non-zero matrices can be a zero matrix.

6. Transpose of a Matrix

Definition: The matrix obtained by interchanging the rows and columns of matrix $A$ is called the transpose of $A$. It is denoted by $A'$ or $A^T$.
If $A = [a_{ij}]_{m \times n}$, then $A' = [a_{ji}]_{n \times m}$.

Properties of Transpose (Must Memorize) 1. $(A')' = A$
2. $(kA)' = kA'$ (where $k$ is a scalar)
3. $(A + B)' = A' + B'$
4. Reversal Law for Transpose: $\mathbf{(AB)' = B' A'}$ (Extremely crucial for algebraic proofs).

Practice Problem 8: Verifying Reversal Law Question: If $A = \begin{bmatrix} -2 \\ 4 \\ 5 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 3 & -6 \end{bmatrix}$, verify that $(AB)' = B'A'$.

Solution:
1. Order of $A$ is $3 \times 1$. Order of $B$ is $1 \times 3$. $AB$ will be $3 \times 3$.
$$ AB = \begin{bmatrix} -2(1) & -2(3) & -2(-6) \\ 4(1) & 4(3) & 4(-6) \\ 5(1) & 5(3) & 5(-6) \end{bmatrix} = \begin{bmatrix} -2 & -6 & 12 \\ 4 & 12 & -24 \\ 5 & 15 & -30 \end{bmatrix} $$ 2. Take transpose: $(AB)' = \begin{bmatrix} -2 & 4 & 5 \\ -6 & 12 & 15 \\ 12 & -24 & -30 \end{bmatrix}$

3. Now, $B' = \begin{bmatrix} 1 \\ 3 \\ -6 \end{bmatrix}$ ($3 \times 1$) and $A' = \begin{bmatrix} -2 & 4 & 5 \end{bmatrix}$ ($1 \times 3$).
4. Compute $B'A'$:
$$ B'A' = \begin{bmatrix} 1(-2) & 1(4) & 1(5) \\ 3(-2) & 3(4) & 3(5) \\ -6(-2) & -6(4) & -6(5) \end{bmatrix} = \begin{bmatrix} -2 & 4 & 5 \\ -6 & 12 & 15 \\ 12 & -24 & -30 \end{bmatrix} $$
Hence, $(AB)' = B'A'$ is verified.

7. Symmetric and Skew-Symmetric Matrices

1. Symmetric Matrix: A square matrix $A = [a_{ij}]$ is said to be symmetric if $A' = A$.
This means $a_{ij} = a_{ji}$ for all possible values of $i$ and $j$. The matrix is symmetric across its main diagonal.

2. Skew-Symmetric Matrix: A square matrix $A = [a_{ij}]$ is said to be skew-symmetric if $A' = -A$.
This means $a_{ij} = -a_{ji}$ for all possible values of $i$ and $j$.

Crucial Property of Skew-Symmetric Matrices In a skew-symmetric matrix, we know $a_{ij} = -a_{ji}$.
For the principal diagonal elements, $i = j$.
Therefore, $a_{ii} = -a_{ii} \implies 2a_{ii} = 0 \implies \mathbf{a_{ii} = 0}$.
Conclusion: All principal diagonal elements of a skew-symmetric matrix are ALWAYS strictly zero.

Important Theorems (Standard Board Proofs)

Theorem 1: For any square matrix $A$ with real entries, $(A + A')$ is a symmetric matrix, and $(A - A')$ is a skew-symmetric matrix.
Proof Sketch: Let $P = A + A'$. Then $P' = (A + A')' = A' + (A')' = A' + A = P$. Hence symmetric.

Theorem 2: Any square matrix can be expressed as the sum of a symmetric and a skew-symmetric matrix.
Formula: $$ A = \frac{1}{2}(A + A') + \frac{1}{2}(A - A') $$

Practice Problem 9: Expressing as a Sum Question: Express the matrix $A = \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}$ as the sum of a symmetric and a skew-symmetric matrix.

Solution:
1. Find $A'$: $A' = \begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix}$.
2. Let $P = \frac{1}{2}(A + A')$ (Symmetric part):
$P = \frac{1}{2} \left( \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix} + \begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix} \right) = \frac{1}{2} \begin{bmatrix} 6 & 6 \\ 6 & -2 \end{bmatrix} = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix}$.
(Notice $P' = P$, so it is symmetric).

3. Let $Q = \frac{1}{2}(A - A')$ (Skew-Symmetric part):
$Q = \frac{1}{2} \left( \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix} - \begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix} \right) = \frac{1}{2} \begin{bmatrix} 0 & 4 \\ -4 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$.
(Notice $Q' = -Q$ and diagonals are zero, so it is skew-symmetric).

4. Final Expression: $A = P + Q = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix} + \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$.

8. Elementary Operations (Transformations) of a Matrix

This is one of the most important and most confusing areas for students, so learn it as a rule-system rather than by memory. There are six elementary operations: 3 row operations and 3 column operations.

All 6 Elementary Operations (Full Notation) Row Operations:
1. Interchange rows: $R_i \leftrightarrow R_j$
2. Multiply a row by non-zero number: $R_i \rightarrow kR_i,\ (k\neq0)$
3. Add multiple of one row to another row: $R_i \rightarrow R_i + kR_j$

Column Operations:
4. Interchange columns: $C_i \leftrightarrow C_j$
5. Multiply a column by non-zero number: $C_i \rightarrow kC_i,\ (k\neq0)$
6. Add multiple of one column to another column: $C_i \rightarrow C_i + kC_j$

Meaning of Symbols (No Confusion Zone) $R_i$ means "$i^{th}$ row", $C_j$ means "$j^{th}$ column".
Arrow "$\rightarrow$" means "replace by".
Example: $R_2 \rightarrow R_2 - 3R_1$ means row 2 changes, row 1 stays fixed.
"$\leftrightarrow$" means swap both entries (interchange).

Golden Rule: In one step, change only the target row/column written on the left side of the arrow.

Why Students Get Confused (and How to Avoid It)

Mistake 1: Changing both rows in $R_2 \rightarrow R_2 + 2R_1$. Only $R_2$ changes.
Mistake 2: Using $k=0$ in $R_i \rightarrow kR_i$. Not allowed (operation must be reversible).
Mistake 3: Mixing row operation with column computation in same step.
Mistake 4: Writing wrong sign in $R_i \rightarrow R_i - kR_j$ while eliminating entries.

Exam Strategy for Elimination To make an entry zero (say 6 under pivot 2), choose operation like:
$R_2 \rightarrow R_2 - 3R_1$ because $6-3(2)=0$.
For columns, same logic: if entry is in a column, use $C_i \rightarrow C_i + kC_j$ appropriately.

Practice Problem 8A: Row Interchange Question: Apply $R_1 \leftrightarrow R_3$ on $$ A=\begin{bmatrix}1&2&3\\4&5&6\\7&8&9\end{bmatrix}. $$

Solution:
Swap row 1 and row 3, row 2 stays unchanged: $$ \begin{bmatrix}7&8&9\\4&5&6\\1&2&3\end{bmatrix}. $$

Practice Problem 8B: Row Scaling + Replacement Question: For $$ A=\begin{bmatrix}2&-1\\5&3\end{bmatrix}, $$ perform in order: (i) $R_1\rightarrow \frac12R_1$, (ii) $R_2\rightarrow R_2-5R_1$.

Solution:
After (i): $$ \begin{bmatrix}1&-1/2\\5&3\end{bmatrix}. $$ After (ii): $$ R_2=[5,3]-5[1,-1/2]=[0,11/2]. $$ Final matrix: $$ \begin{bmatrix}1&-1/2\\0&11/2\end{bmatrix}. $$

Practice Problem 8C: Column Interchange Question: Apply $C_1 \leftrightarrow C_3$ on $$ A=\begin{bmatrix}2&0&5\\1&4&-1\\3&2&7\end{bmatrix}. $$

Solution:
Swap first and third columns: $$ \begin{bmatrix}5&0&2\\-1&4&1\\7&2&3\end{bmatrix}. $$

Practice Problem 8D: Column Replacement Question: If $$ A=\begin{bmatrix}1&2&3\\0&1&4\\2&-1&5\end{bmatrix}, $$ find matrix after $C_3 \rightarrow C_3 - 2C_1$.

Solution:
$C_3=[3,4,5]^T,\ C_1=[1,0,2]^T$.
New $C_3=[3,4,5]^T-2[1,0,2]^T=[1,4,1]^T$.
So result is $$ \begin{bmatrix}1&2&1\\0&1&4\\2&-1&1\end{bmatrix}. $$

Practice Problem 8E: Identify Incorrect Step Question: A student does $R_2 \rightarrow R_2-2R_1$ on $$ \begin{bmatrix}1&3\\2&7\end{bmatrix} $$ and writes new matrix as $\begin{bmatrix}1&3\\0&1\end{bmatrix}$. Is this correct?

Solution:
Compute correctly: $$ R_2=[2,7]-2[1,3]=[0,1]. $$ Row 1 remains unchanged. So new matrix is $$ \begin{bmatrix}1&3\\0&1\end{bmatrix}. $$ Yes, this particular step is correct. It is a valid row replacement operation.

9. Invertible Matrices and Inverse via Elementary Operations

Definition: Let $A$ be a square matrix of order $n$. If there exists another square matrix $B$ of the same order such that $AB = BA = I$, then $B$ is called the inverse of $A$, denoted by $A^{-1}$.
Key Facts: - Rectangular matrices do NOT possess an inverse. - Inverse of a square matrix, if it exists, is absolutely unique. - Reversal Law for Inverse: $(AB)^{-1} = B^{-1} A^{-1}$.

Algorithm: Inverse via Elementary Row Operations To find $A^{-1}$ using row operations:
1. Write the equation $A = IA$.
2. Perform a sequence of elementary row operations on the matrix $A$ on the LHS, and simultaneously on the identity matrix $I$ on the RHS. Leave the right-most $A$ untouched.
3. Continue these operations until the LHS matrix $A$ transforms into the Identity matrix $I$.
4. The equation will transform into $I = BA$. The resulting matrix $B$ on the RHS is $A^{-1}$.

Rule of Thumb for Non-Existence: If, during the row operations, all elements of any one or more rows on the LHS become strictly zero, it signifies that $A^{-1}$ does NOT exist.

Practice Problem 10: Finding Inverse Question: Find the inverse of the matrix $A = \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix}$ using elementary row operations.

Solution:
Write $A = IA \implies \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} A$

Step 1: We need a 0 where the '2' is in the second row. Apply $R_2 \rightarrow R_2 - 2R_1$.
LHS: $\begin{bmatrix} 1 & 2 \\ (2 - 2) & (-1 - 4) \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & -5 \end{bmatrix}$
RHS: $\begin{bmatrix} 1 & 0 \\ (0 - 2) & (1 - 0) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix}$
Equation: $\begin{bmatrix} 1 & 2 \\ 0 & -5 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix} A$

Step 2: We need a 1 where the '-5' is in the second row. Apply $R_2 \rightarrow \frac{1}{-5} R_2$.
Equation: $\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 2/5 & -1/5 \end{bmatrix} A$

Step 3: We need a 0 where the '2' is in the first row. Apply $R_1 \rightarrow R_1 - 2R_2$.
LHS: $\begin{bmatrix} (1 - 0) & (2 - 2) \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$
RHS: $\begin{bmatrix} (1 - 4/5) & (0 - (-2/5)) \\ 2/5 & -1/5 \end{bmatrix} = \begin{bmatrix} 1/5 & 2/5 \\ 2/5 & -1/5 \end{bmatrix}$

We have reached $I = BA$. Therefore, the inverse is:
$$ A^{-1} = \begin{bmatrix} 1/5 & 2/5 \\ 2/5 & -1/5 \end{bmatrix} = \frac{1}{5} \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} $$

10. High-Scoring Matrix Identities and Result Patterns

Board + Competitive Algebra Patterns 1. Non-Commutativity Caution: In general, $AB \neq BA$. So usual algebraic expansions must be written carefully.

2. Correct Expansions:
$(A + B)^2 = A^2 + AB + BA + B^2$
$(A - B)^2 = A^2 - AB - BA + B^2$
$(A + B)(A - B) = A^2 - AB + BA - B^2$

3. When $AB = BA$: Matrix algebra behaves like ordinary algebra for expansions:
$(A + B)^2 = A^2 + 2AB + B^2$, and $(A + B)(A - B) = A^2 - B^2$.

4. Inverse Product Rule: $(AB)^{-1} = B^{-1}A^{-1}$ (order reverses).
5. Transpose Product Rule: $(AB)' = B'A'$ (order reverses).
6. Inverse of Transpose: $(A')^{-1} = (A^{-1})'$ (for invertible square matrices).

Exam Trap Alert Never write $(A+B)^2 = A^2 + 2AB + B^2$ unless it is explicitly given or proven that $AB = BA$. This one line is one of the most common board and entrance mistakes.

Practice Problem 11: Expansion Check Question: If $A=\begin{bmatrix}1&1\\0&1\end{bmatrix}$ and $B=\begin{bmatrix}1&0\\1&1\end{bmatrix}$, compute $(A+B)^2$ and verify that it is not equal to $A^2+2AB+B^2$.

Solution:
First, $A+B=\begin{bmatrix}2&1\\1&2\end{bmatrix}$. So, $$ (A+B)^2 = \begin{bmatrix}2&1\\1&2\end{bmatrix}\begin{bmatrix}2&1\\1&2\end{bmatrix} = \begin{bmatrix}5&4\\4&5\end{bmatrix} $$ Now, compute individual parts:
$$ A^2=\begin{bmatrix}1&2\\0&1\end{bmatrix}, \quad B^2=\begin{bmatrix}1&0\\2&1\end{bmatrix}, \quad AB=\begin{bmatrix}2&1\\1&1\end{bmatrix} $$ Hence, $$ A^2+2AB+B^2= \begin{bmatrix}1&2\\0&1\end{bmatrix} +2\begin{bmatrix}2&1\\1&1\end{bmatrix} +\begin{bmatrix}1&0\\2&1\end{bmatrix} = \begin{bmatrix}6&4\\4&4\end{bmatrix} $$ Therefore, $\begin{bmatrix}5&4\\4&5\end{bmatrix}\neq\begin{bmatrix}6&4\\4&4\end{bmatrix}$. Verified.

11. Solving Matrix Equations (Very Important)

Questions of the form $AX=B$ and $XA=B$ are standard in NCERT, R.S. Aggarwal and R.D. Sharma. The side of multiplication matters.

Method Summary If $AX=B$ and $A$ is invertible, multiply by $A^{-1}$ from the left:
$$ X = A^{-1}B $$ If $XA=B$ and $A$ is invertible, multiply by $A^{-1}$ from the right:
$$ X = BA^{-1} $$ Key point: $A^{-1}B$ and $BA^{-1}$ are generally different.

Practice Problem 12: Solve AX = B Question: Solve for $X$ if $$ A=\begin{bmatrix}1&2\\3&4\end{bmatrix}, \quad B=\begin{bmatrix}5&1\\11&3\end{bmatrix}, \quad AX=B.$$

Solution:
$X=A^{-1}B$. For $A=\begin{bmatrix}1&2\\3&4\end{bmatrix}$, $$ A^{-1}=\frac{1}{(1)(4)-(2)(3)}\begin{bmatrix}4&-2\\-3&1\end{bmatrix} =\frac{1}{-2}\begin{bmatrix}4&-2\\-3&1\end{bmatrix} =\begin{bmatrix}-2&1\\3/2&-1/2\end{bmatrix}. $$ Then $$ X=A^{-1}B= \begin{bmatrix}-2&1\\3/2&-1/2\end{bmatrix} \begin{bmatrix}5&1\\11&3\end{bmatrix} = \begin{bmatrix} 1&1\\ 2&0 \end{bmatrix}. $$ Therefore, $X=\begin{bmatrix}1&1\\2&0\end{bmatrix}$.

Practice Problem 13: Solve XA = B Question: Solve for $X$ if $$ A=\begin{bmatrix}2&1\\1&1\end{bmatrix}, \quad B=\begin{bmatrix}7&4\\5&3\end{bmatrix}, \quad XA=B.$$

Solution:
Here $X=BA^{-1}$. Since $\det(A)=2\cdot1-1\cdot1=1$, $$ A^{-1}=\begin{bmatrix}1&-1\\-1&2\end{bmatrix}. $$ So, $$ X=BA^{-1} =\begin{bmatrix}7&4\\5&3\end{bmatrix} \begin{bmatrix}1&-1\\-1&2\end{bmatrix} =\begin{bmatrix}3&1\\2&1\end{bmatrix}. $$ Therefore, $X=\begin{bmatrix}3&1\\2&1\end{bmatrix}$.

12. Elementary Matrices (R.D. Sharma Enrichment)

An elementary matrix is obtained by performing exactly one elementary row operation on an identity matrix of suitable order.

Core Idea If an elementary matrix $E$ is premultiplied with $A$, i.e., $EA$, then the same row operation is performed on $A$.
This is the theoretical foundation behind inverse through row operations.

Practice Problem 14: Using Elementary Matrix Question: Let $E=\begin{bmatrix}1&0\\-3&1\end{bmatrix}$ and $A=\begin{bmatrix}2&1\\5&4\end{bmatrix}$. Find $EA$ and identify the row operation.

Solution:
$$ EA= \begin{bmatrix}1&0\\-3&1\end{bmatrix} \begin{bmatrix}2&1\\5&4\end{bmatrix} = \begin{bmatrix} 2&1\\ -1&1 \end{bmatrix}. $$ This corresponds to row operation $R_2 \rightarrow R_2 - 3R_1$ on matrix $A$.

13. Matrix Form of Linear Equations

A system can be written as $AX=B$, where $A$ is coefficient matrix, $X$ variable matrix and $B$ constant matrix.

Practice Problem 15: Equation to Matrix Form + Solve Question: Write and solve using matrix method: $$ x+y=5, \quad 2x-y=4. $$

Solution:
Matrix form: $$ \begin{bmatrix}1&1\\2&-1\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}5\\4\end{bmatrix}. $$ Let $A=\begin{bmatrix}1&1\\2&-1\end{bmatrix}$, $X=\begin{bmatrix}x\\y\end{bmatrix}$, $B=\begin{bmatrix}5\\4\end{bmatrix}$. Then $X=A^{-1}B$. Since $\det(A)=-3$, $$ A^{-1}=\frac{1}{-3}\begin{bmatrix}-1&-1\\-2&1\end{bmatrix} =\begin{bmatrix}1/3&1/3\\2/3&-1/3\end{bmatrix}. $$ Therefore, $$ X=\begin{bmatrix}1/3&1/3\\2/3&-1/3\end{bmatrix}\begin{bmatrix}5\\4\end{bmatrix} =\begin{bmatrix}3\\2\end{bmatrix}. $$ Hence, $x=3,\ y=2$.

14. Additional Mixed Practice (NCERT + R.S. + R.D. Level)

Practice Problem 16: Symmetric Condition Question: Find $a,b$ if matrix $A=\begin{bmatrix}2&a\\b&5\end{bmatrix}$ is symmetric.

Solution:
For symmetric matrix, $A'=A$. So corresponding off-diagonal entries must be equal: $a=b$.
Hence required condition is $a=b$.

Practice Problem 17: Skew-Symmetric Condition Question: Find $p,q$ so that $$ A=\begin{bmatrix}0&2&-1\\p&0&4\\1&q&0\end{bmatrix} $$ is skew-symmetric.

Solution:
For skew-symmetric matrix, $a_{ij}=-a_{ji}$.
From $(1,2)$ and $(2,1)$ entries: $p=-2$.
From $(2,3)$ and $(3,2)$ entries: $q=-4$.
Diagonal entries are already zero, so condition satisfied.
Therefore, $p=-2,\ q=-4$.

Practice Problem 18: Conceptual Proof Question: If $A$ is symmetric and $B$ is skew-symmetric (same order), prove that $(AB+BA)$ is skew-symmetric.

Solution:
Given $A'=A$ and $B'=-B$.
$$ (AB+BA)'=(AB)'+(BA)'=B'A'+A'B' $$ $$ =(-B)A+A(-B)=-(BA+AB)=-(AB+BA). $$ Hence $(AB+BA)'=-(AB+BA)$, so $AB+BA$ is skew-symmetric.

Practice Problem 19: Identity Application Question: Let $A$ be involutory ($A^2=I$). Show that $(I+A)(I-A)=O$.

Solution:
$$ (I+A)(I-A)=I-A+A-A^2=I-A^2=I-I=O. $$ Hence proved.

Practice Problem 20: Trace and Product Question: If $$ A=\begin{bmatrix}1&2\\0&3\end{bmatrix}, \quad B=\begin{bmatrix}4&0\\5&1\end{bmatrix}, $$ verify that $Tr(AB)=Tr(BA)$.

Solution:
$$ AB=\begin{bmatrix}14&2\\15&3\end{bmatrix}\Rightarrow Tr(AB)=14+3=17. $$ $$ BA=\begin{bmatrix}4&8\\5&13\end{bmatrix}\Rightarrow Tr(BA)=4+13=17. $$ Therefore, $Tr(AB)=Tr(BA)=17$.

Final Revision Checklist 1. Matrix order, entry notation, and classification (row/column/square/diagonal/scalar/identity/zero).
2. Matrix equality and solving parameter-based equality questions.
3. Matrix operations with strict dimension check for multiplication.
4. Transpose rules and reversal law $(AB)'=B'A'$.
5. Symmetric and skew-symmetric definitions, diagonal condition for skew-symmetric.
6. Decomposition formula: $A=\frac{1}{2}(A+A')+\frac{1}{2}(A-A')$.
7. Elementary row operations and inverse through transformation.
8. Solving $AX=B$ and $XA=B$ correctly with multiplication side awareness.
9. Trace properties and high-frequency special matrix concepts (idempotent, involutory, nilpotent, orthogonal).
10. Non-commutativity-aware algebraic expansions to avoid board mistakes.