Matrices Ex 3.4 Class 12 NCERT: Finding Inverse Using Elementary Operations

Understanding Elementary Operations on Matrices

Elementary operations, also known as elementary transformations, are a set of three fundamental operations that can be performed on the rows or columns of a matrix. These operations are crucial because they allow us to transform a matrix into a simpler form without changing its fundamental properties, especially when finding its inverse. It's vital to remember that once you choose to work with row operations, you must stick to only row operations throughout the process for finding the inverse; similarly for column operations. Mixing them is a common mistake and will lead to incorrect results.

There are three types of elementary operations:

1. Interchange of any two rows (or columns): If R_i and R_j are the i-th and j-th rows, this operation is denoted by R_i ↔ R_j. For columns, it's C_i ↔ C_j. For example, swapping Row 1 and Row 2.
2. Multiplication of the elements of any row (or column) by a non-zero scalar: If R_i is the i-th row and 'k' is a non-zero scalar, this operation is denoted by R_i → kR_i. For columns, C_i → kC_i. This means every element in that row or column is multiplied by 'k'. This operation is often used to make a leading element '1'.
3. Addition to the elements of any row (or column) the corresponding elements of any other row (or column) multiplied by a non-zero scalar: This is the most frequently used operation. If R_i and R_j are the i-th and j-th rows, and 'k' is a non-zero scalar, this operation is denoted by R_i → R_i + kR_j. For columns, C_i → C_i + kC_j. This operation is instrumental in making elements zero in the desired positions, a key step in transforming a matrix into the identity matrix.

Key Definitions for Inverse Matrices

Invertible Matrix

A square matrix A is said to be invertible if there exists another square matrix B of the same order such that AB = BA = I, where I is the identity matrix of the same order. Matrix B is called the inverse of A, denoted as A⁻¹.

Identity Matrix (I)

A square matrix in which all the principal diagonal elements are 1 and all other elements are 0. For example, for order 2, I = [[1, 0], [0, 1]].

Singular Matrix

A square matrix is called a singular matrix if its determinant is zero (det(A) = 0). A singular matrix does NOT have an inverse.

Non-Singular Matrix

A square matrix is called a non-singular matrix if its determinant is non-zero (det(A) ≠ 0). Only non-singular matrices have inverses.

Step-by-Step Method for Finding Inverse (Elementary Row Operations)

1. Step 1: Set up the Augmented Matrix — Start by writing the given square matrix A as an augmented matrix [A | I], where I is the identity matrix of the same order as A. Our goal is to transform A into I using elementary row operations, and simultaneously, I will transform into A⁻¹.
2. Step 2: Transform the First Column — Focus on the first column of matrix A. Use elementary row operations to make the element at position (1,1) (top-left) equal to 1. Then, use this '1' to make all other elements in the first column equal to 0. For example, R_i → R_i - (a_i1)R_1.
3. Step 3: Transform the Second Column — Move to the second column. First, make the element at position (2,2) equal to 1 using R_2 → kR_2. Then, use this '1' to make all other elements in the second column (i.e., (1,2) and (3,2) for a 3x3 matrix) equal to 0.
4. Step 4: Transform Subsequent Columns (if applicable) — Repeat the process for subsequent columns. For a 3x3 matrix, you would then transform the third column: make (3,3) equal to 1, then use it to make (1,3) and (2,3) equal to 0.
5. Step 5: Identify the Inverse Matrix — Once the left part of the augmented matrix becomes the identity matrix (I), the right part will be the inverse of A, i.e., [I | A⁻¹]. If at any point, a row (or column) of the left part consists entirely of zeros, then the inverse does not exist (the matrix is singular).
6. Step 6: Verify (Optional but Recommended) — To confirm your answer, multiply the original matrix A by your calculated inverse A⁻¹. If your calculations are correct, the product AA⁻¹ should be equal to the identity matrix I.

Worked Examples: Finding Matrix Inverse

• Example 1: Find the inverse of A = [[1, 2], [2, 3]] using elementary row operations. Step 1: Set up the augmented matrix [A | I]. [[1, 2 | 1, 0], [2, 3 | 0, 1]] Step 2: Make a_11 = 1 (already is). Make a_21 = 0. Operation: R_2 → R_2 - 2R_1 [[1, 2 | 1, 0], [2 - 2(1), 3 - 2(2) | 0 - 2(1), 1 - 2(0)]] [[1, 2 | 1, 0], [0, -1 | -2, 1]] Step 3: Make a_22 = 1. Operation: R_2 → (-1)R_2 [[1, 2 | 1, 0], [0, (-1)(-1) | (-2)(-1), (1)(-1)]] [[1, 2 | 1, 0], [0, 1 | 2, -1]] Step 4: Make a_12 = 0. Operation: R_1 → R_1 - 2R_2 [[1 - 2(0), 2 - 2(1) | 1 - 2(2), 0 - 2(-1)], [0, 1 | 2, -1]] [[1, 0 | 1 - 4, 0 + 2], [0, 1 | 2, -1]] [[1, 0 | -3, 2], [0, 1 | 2, -1]] Step 5: The left side is I, so the right side is A⁻¹. Final answer: A⁻¹ = [[-3, 2], [2, -1]]
• Example 2: Find the inverse of B = [[0, 1, 2], [1, 2, 3], [3, 1, 1]] using elementary row operations. Step 1: Set up the augmented matrix [B | I]. [[0, 1, 2 | 1, 0, 0], [1, 2, 3 | 0, 1, 0], [3, 1, 1 | 0, 0, 1]] Step 2: Make b_11 = 1. (Swap R_1 and R_2) Operation: R_1 ↔ R_2 [[1, 2, 3 | 0, 1, 0], [0, 1, 2 | 1, 0, 0], [3, 1, 1 | 0, 0, 1]] Step 3: Make b_31 = 0. Operation: R_3 → R_3 - 3R_1 [[1, 2, 3 | 0, 1, 0], [0, 1, 2 | 1, 0, 0], [3 - 3(1), 1 - 3(2), 1 - 3(3) | 0 - 3(0), 0 - 3(1), 1 - 3(0)]] [[1, 2, 3 | 0, 1, 0], [0, 1, 2 | 1, 0, 0], [0, -5, -8 | 0, -3, 1]] Step 4: Make b_22 = 1 (already is). Make b_12 = 0 and b_32 = 0. Operation: R_1 → R_1 - 2R_2 [[1 - 2(0), 2 - 2(1), 3 - 2(2) | 0 - 2(1), 1 - 2(0), 0 - 2(0)], [0, 1, 2 | 1, 0, 0], [0, -5, -8 | 0, -3, 1]] [[1, 0, -1 | -2, 1, 0], [0, 1, 2 | 1, 0, 0], [0, -5, -8 | 0, -3, 1]] Operation: R_3 → R_3 + 5R_2 [[1, 0, -1 | -2, 1, 0], [0, 1, 2 | 1, 0, 0], [0 + 5(0), -5 + 5(1), -8 + 5(2) | 0 + 5(1), -3 + 5(0), 1 + 5(0)]] [[1, 0, -1 | -2, 1, 0], [0, 1, 2 | 1, 0, 0], [0, 0, 2 | 5, -3, 1]] Step 5: Make b_33 = 1. Operation: R_3 → (1/2)R_3 [[1, 0, -1 | -2, 1, 0], [0, 1, 2 | 1, 0, 0], [0, 0, 1 | 5/2, -3/2, 1/2]] Step 6: Make b_13 = 0 and b_23 = 0. Operation: R_1 → R_1 + R_3 [[1 + 0, 0 + 0, -1 + 1 | -2 + 5/2, 1 - 3/2, 0 + 1/2], [0, 1, 2 | 1, 0, 0], [0, 0, 1 | 5/2, -3/2, 1/2]] [[1, 0, 0 | 1/2, -1/2, 1/2], [0, 1, 2 | 1, 0, 0], [0, 0, 1 | 5/2, -3/2, 1/2]] Operation: R_2 → R_2 - 2R_3 [[1, 0, 0 | 1/2, -1/2, 1/2], [0 - 0, 1 - 0, 2 - 2(1) | 1 - 2(5/2), 0 - 2(-3/2), 0 - 2(1/2)], [0, 0, 1 | 5/2, -3/2, 1/2]] [[1, 0, 0 | 1/2, -1/2, 1/2], [0, 1, 0 | 1 - 5, 0 + 3, 0 - 1], [0, 0, 1 | 5/2, -3/2, 1/2]] [[1, 0, 0 | 1/2, -1/2, 1/2], [0, 1, 0 | -4, 3, -1], [0, 0, 1 | 5/2, -3/2, 1/2]] Step 7: The left side is I, so the right side is B⁻¹. Final answer: B⁻¹ = [[1/2, -1/2, 1/2], [-4, 3, -1], [5/2, -3/2, 1/2]]

Exam Tips and Common Pitfalls

Finding the inverse using elementary operations is a methodical process, but it's prone to small errors if not done carefully. Here are some essential tips for your CBSE exams:

• Consistency is Key: Always use either elementary row operations throughout the entire problem or elementary column operations throughout. Never mix them within a single problem; this is a common mistake and will lead to an incorrect answer.
• Check for Singularity Early: Before beginning the lengthy process, quickly calculate the determinant of the given matrix. If det(A) = 0, the matrix is singular, and its inverse does not exist. You can state this and save time, rather than performing operations endlessly.
• Systematic Approach: For a 3x3 matrix, follow a systematic path. A good strategy is to first transform the first column (make (1,1) a '1', then (2,1) and (3,1) '0's). Then move to the second column (make (2,2) a '1', then (1,2) and (3,2) '0's), and finally the third column. This avoids undoing previous operations.
• Arithmetic Precision: Double-check every addition, subtraction, and multiplication step. A single arithmetic error can propagate and ruin the entire solution. Use fractions rather than decimals to maintain precision.
• Practice, Practice, Practice: The only way to master this topic and increase speed is through consistent practice. Work through various examples from your NCERT textbook and reference books.
• Verify Your Answer: If time permits in the exam, always verify your inverse by multiplying A with A⁻¹. The result should be the identity matrix I. This provides an excellent check for your solution.

Practice Questions with Solutions

• Q: Find the inverse of A = [[2, 1], [1, 1]] using elementary row operations. A: Step 1: Write A = IA. [[2, 1 | 1, 0], [1, 1 | 0, 1]] Step 2: R_1 ↔ R_2 [[1, 1 | 0, 1], [2, 1 | 1, 0]] Step 3: R_2 → R_2 - 2R_1 [[1, 1 | 0, 1], [0, -1 | 1, -2]] Step 4: R_2 → (-1)R_2 [[1, 1 | 0, 1], [0, 1 | -1, 2]] Step 5: R_1 → R_1 - R_2 [[1, 0 | 1, -1], [0, 1 | -1, 2]] Final answer: A⁻¹ = [[1, -1], [-1, 2]]
• Q: Find the inverse of B = [[1, 3], [2, 7]] using elementary row operations. A: Step 1: Write B = IB. [[1, 3 | 1, 0], [2, 7 | 0, 1]] Step 2: R_2 → R_2 - 2R_1 [[1, 3 | 1, 0], [0, 1 | -2, 1]] Step 3: R_1 → R_1 - 3R_2 [[1, 0 | 1 - 3(-2), 0 - 3(1)], [0, 1 | -2, 1]] [[1, 0 | 7, -3], [0, 1 | -2, 1]] Final answer: B⁻¹ = [[7, -3], [-2, 1]]
• Q: Find the inverse of C = [[1, 0, 0], [0, 2, 0], [0, 0, 3]] using elementary row operations. A: Step 1: Write C = IC. [[1, 0, 0 | 1, 0, 0], [0, 2, 0 | 0, 1, 0], [0, 0, 3 | 0, 0, 1]] Step 2: R_2 → (1/2)R_2 [[1, 0, 0 | 1, 0, 0], [0, 1, 0 | 0, 1/2, 0], [0, 0, 3 | 0, 0, 1]] Step 3: R_3 → (1/3)R_3 [[1, 0, 0 | 1, 0, 0], [0, 1, 0 | 0, 1/2, 0], [0, 0, 1 | 0, 0, 1/3]] Final answer: C⁻¹ = [[1, 0, 0], [0, 1/2, 0], [0, 0, 1/3]]
• Q: Find the inverse of D = [[2, 3], [5, 7]] using elementary row operations. A: Step 1: Write D = ID. [[2, 3 | 1, 0], [5, 7 | 0, 1]] Step 2: R_1 → (1/2)R_1 [[1, 3/2 | 1/2, 0], [5, 7 | 0, 1]] Step 3: R_2 → R_2 - 5R_1 [[1, 3/2 | 1/2, 0], [0, 7 - 5(3/2) | 0 - 5(1/2), 1 - 5(0)]] [[1, 3/2 | 1/2, 0], [0, 7 - 15/2 | -5/2, 1]] [[1, 3/2 | 1/2, 0], [0, -1/2 | -5/2, 1]] Step 4: R_2 → (-2)R_2 [[1, 3/2 | 1/2, 0], [0, 1 | 5, -2]] Step 5: R_1 → R_1 - (3/2)R_2 [[1, 0 | 1/2 - (3/2)(5), 0 - (3/2)(-2)], [0, 1 | 5, -2]] [[1, 0 | 1/2 - 15/2, 0 + 3], [0, 1 | 5, -2]] [[1, 0 | -14/2, 3], [0, 1 | 5, -2]] [[1, 0 | -7, 3], [0, 1 | 5, -2]] Final answer: D⁻¹ = [[-7, 3], [5, -2]]

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