How to Solve: Matrices (Inverse, Elementary Operations, System of Equations, Matrix Multiplication) – IIT JEE Guide

How to Solve: Matrices (Inverse, Elementary Operations, System of Equations, Matrix Multiplication) – IIT JEE Guide

Introduction

Mastering matrices unlocks 10-15 marks in IIT JEE (Main + Advanced)—enough to push you into the top 1%. Whether it’s solving systems of equations in seconds, cracking cryptography problems, or acing linear algebra in engineering, matrices are the secret weapon of high scorers.

What You Need To Know First

1. Basic algebra (solving linear equations, determinants).
2. Row operations (swapping, scaling, adding rows).
3. Determinants (how to compute 2×2 and 3×3).

Key Vocabulary

Formulas To Know

1. Matrix Multiplication (A × B)

• Formula: ( (AB){ij} = \sum )}^{n} A_{ik} B_{kj
• Variables:
• ( A ) = ( m \times n ) matrix
• ( B ) = ( n \times p ) matrix
• ( AB ) = ( m \times p ) matrix
• MEMORISE THIS: Order matters! ( AB \neq BA ) in most cases.

2. Inverse of a 2×2 Matrix

• Formula: ( A^{-1} = \frac{1}{\det A} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} ) where ( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} )
• MEMORISE THIS: Only works if ( \det A \neq 0 ).

3. Determinant of a 3×3 Matrix (Rule of Sarrus)

• Formula: ( \det A = a(ei - fh) - b(di - fg) + c(dh - eg) ) for ( A = \begin{bmatrix} a & b & c \ d & e & f \ g & h & i \end{bmatrix} )
• MEMORISE THIS: Expand along the first row.

4. Solving ( AX = B ) Using Inverse

• Formula: ( X = A^{-1}B )
• MEMORISE THIS: Only works if ( A ) is non-singular (det ≠ 0).

5. Elementary Row Operations (ERO)

• Types:
• Swap two rows (( R_i \leftrightarrow R_j )).
• Multiply a row by a non-zero scalar (( kR_i )).
• Add a multiple of one row to another (( R_i + kR_j )).
• MEMORISE THIS: These preserve the solution set of ( AX = B ).

Step-by-Step Method

How to Find the Inverse of a Matrix (Using Elementary Operations)

Goal: Find ( A^{-1} ) such that ( A \cdot A^{-1} = I ).

1. Write the augmented matrix ( [A | I] ).
2. Apply EROs to transform ( A ) into ( I ).
3. The right side becomes ( A^{-1} ).
4. Check: Multiply ( A \cdot A^{-1} ). Should get ( I ).

Example: Find ( A^{-1} ) for ( A = \begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix} ).

How to Solve a System of Equations Using Matrices

Goal: Solve ( AX = B ) where ( A ) is a coefficient matrix.

1. Write the augmented matrix ( [A | B] ).
2. Apply EROs to get ( [I | X] ).
3. Read the solution from the right side.

Example: Solve ( x + 2y = 5 ), ( 3x + 4y = 11 ).

How to Multiply Two Matrices

Goal: Compute ( AB ).

1. Check dimensions: ( A ) is ( m \times n ), ( B ) is ( n \times p ). Result is ( m \times p ).
2. For each element ( (AB)_{ij} ), multiply row ( i ) of ( A ) by column ( j ) of ( B ) and sum.
3. Write the result in the ( i )-th row, ( j )-th column.

Example: Multiply ( A = \begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix} ) and ( B = \begin{bmatrix} 5 & 6 \ 7 & 8 \end{bmatrix} ).

Worked Examples

Example 1 – Basic: Find the Inverse of a 2×2 Matrix

Problem: Find ( A^{-1} ) for ( A = \begin{bmatrix} 2 & 3 \ 1 & 4 \end{bmatrix} ).

Solution: 1. Compute det(A):
( \det A = (2)(4) - (3)(1) = 8 - 3 = 5 \neq 0 ) → Inverse exists. 2. Apply formula:
( A^{-1} = \frac{1}{5} \begin{bmatrix} 4 & -3 \ -1 & 2 \end{bmatrix} ). 3. Final answer:
( A^{-1} = \begin{bmatrix} \frac{4}{5} & -\frac{3}{5} \ -\frac{1}{5} & \frac{2}{5} \end{bmatrix} ).

What we did and why: - Used the 2×2 inverse formula because it’s the fastest method. - Checked det ≠ 0 to ensure the inverse exists.

Example 2 – Medium: Solve a System Using Inverse

Problem: Solve ( 2x + 3y = 5 ), ( x + 4y = 6 ).

Solution: 1. Write in matrix form:
( A = \begin{bmatrix} 2 & 3 \ 1 & 4 \end{bmatrix} ), ( X = \begin{bmatrix} x \ y \end{bmatrix} ), ( B = \begin{bmatrix} 5 \ 6 \end{bmatrix} ). 2. Find ( A^{-1} ):
( \det A = (2)(4) - (3)(1) = 5 ).
( A^{-1} = \frac{1}{5} \begin{bmatrix} 4 & -3 \ -1 & 2 \end{bmatrix} ). 3. Compute ( X = A^{-1}B ):
( X = \frac{1}{5} \begin{bmatrix} 4 & -3 \ -1 & 2 \end{bmatrix} \begin{bmatrix} 5 \ 6 \end{bmatrix} = \frac{1}{5} \begin{bmatrix} 20 - 18 \ -5 + 12 \end{bmatrix} = \frac{1}{5} \begin{bmatrix} 2 \ 7 \end{bmatrix} ). 4. Final answer:
( x = \frac{2}{5} ), ( y = \frac{7}{5} ).

What we did and why: - Used matrix inversion because it’s efficient for small systems. - Verified det ≠ 0 to ensure a unique solution.

Example 3 – Exam-Style: Matrix Multiplication with Variables

Problem: If ( A = \begin{bmatrix} 1 & x \ 0 & 1 \end{bmatrix} ) and ( B = \begin{bmatrix} 2 & 3 \ 1 & 4 \end{bmatrix} ), find ( x ) such that ( AB = BA ).

Solution: 1. Compute ( AB ):
( AB = \begin{bmatrix} 1 & x \ 0 & 1 \end{bmatrix} \begin{bmatrix} 2 & 3 \ 1 & 4 \end{bmatrix} = \begin{bmatrix} 2 + x & 3 + 4x \ 1 & 4 \end{bmatrix} ). 2. Compute ( BA ):
( BA = \begin{bmatrix} 2 & 3 \ 1 & 4 \end{bmatrix} \begin{bmatrix} 1 & x \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 2x + 3 \ 1 & x + 4 \end{bmatrix} ). 3. Set ( AB = BA ):
( \begin{bmatrix} 2 + x & 3 + 4x \ 1 & 4 \end{bmatrix} = \begin{bmatrix} 2 & 2x + 3 \ 1 & x + 4 \end{bmatrix} ). 4. Equate elements:
- ( 3 + 4x = 2x + 3 ) → ( 2x = 0 ) → ( x = 0 ).
- ( 4 = x + 4 ) → ( x = 0 ). 5. Final answer:
( x = 0 ).

What we did and why: - Multiplied matrices carefully (order matters!). - Set corresponding elements equal to solve for ( x ).

Common Mistakes

Exam Traps

1-Minute Recap (Night Before Exam)

"Listen up—this is your 60-second matrix survival guide for JEE.

1. Inverse? Check det ≠ 0 first. For 2×2, use the formula. For bigger matrices, row reduce ( [A | I] ).
2. System of equations? Write ( AX = B ), then row reduce ( [A | B] ) or use ( X = A^{-1}B ).
3. Matrix multiplication? Dimensions must match (columns of first = rows of second). Multiply row by column.
4. Elementary operations? Swap, scale, or add rows—never multiply a row by zero.
5. Exam traps? Watch for det = 0, wrong dimensions, and hidden variables.

You’ve got this. Now go crush those matrices!