Class 12 CBSE Mathematics (NCERT) — Study Material
Based on Latest Curriculum
Matrices
Introduction
Matrices are rectangular arrangements of numbers, symbols, or expressions in rows and columns. They are fundamental in mathematics, physics, computer graphics, economics, and engineering.
- Computer graphics: 3D transformations (rotation, scaling, translation)
- Cryptography: Encoding and decoding secret messages
- Economics: Input-output analysis, Leontief models
- Robotics: Representing joint movements and positions
- Machine learning: Storing datasets and weights in neural networks
Topic 1: Elementary Transformations (Operations) of a Matrix
Definition
Elementary transformations are operations that can be performed on rows or columns of a matrix. They are of three types:
- Interchange any two rows (columns): \( R_i \leftrightarrow R_j \) (or \( C_i \leftrightarrow C_j \))
- Multiply a row (column) by a non-zero scalar: \( R_i \rightarrow k R_i \), \( k \neq 0 \) (or \( C_i \rightarrow k C_i \))
- Add a multiple of one row (column) to another: \( R_i \rightarrow R_i + k R_j \) (or \( C_i \rightarrow C_i + k C_j \))
Inverse of a Matrix using Elementary Transformations
To find the inverse of a square matrix \(A\) using elementary transformations, we use the concept: \[ A = I A \quad \text{or} \quad A I = A \] By applying the same row transformations to \(A\) and \(I\), we convert \(A\) into \(I\), and \(I\) becomes \(A^{-1}\).
Column method: Write \(A = A I\). Apply column operations to make \(A\) into \(I\). The same operations on \(I\) yield \(A^{-1}\).
Algorithm for Inverse using Row Operations
- Write the augmented matrix \([A | I]\).
- Apply row operations to transform \(A\) into \(I\).
- The right side becomes \(A^{-1}\).
- If \(A\) cannot be transformed to \(I\), then \(A\) is singular (no inverse).
Solved Examples (NCERT & Additional)
NCERT Example:
Find the inverse of \(A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\) using elementary row operations.
Solution:
Write \([A|I] = \left[\begin{array}{cc|cc} 1 & 2 & 1 & 0 \\ 3 & 4 & 0 & 1 \end{array}\right]\)
\(R_2 \rightarrow R_2 – 3R_1\): \(\left[\begin{array}{cc|cc} 1 & 2 & 1 & 0 \\ 0 & -2 & -3 & 1 \end{array}\right]\)
\(R_2 \rightarrow -\frac{1}{2} R_2\): \(\left[\begin{array}{cc|cc} 1 & 2 & 1 & 0 \\ 0 & 1 & \frac{3}{2} & -\frac{1}{2} \end{array}\right]\)
\(R_1 \rightarrow R_1 – 2R_2\): \(\left[\begin{array}{cc|cc} 1 & 0 & -2 & 1 \\ 0 & 1 & \frac{3}{2} & -\frac{1}{2} \end{array}\right]\)
Thus \(A^{-1} = \begin{bmatrix} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \end{bmatrix}\)
Additional Example 1:
Find inverse of \(A = \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}\) using column operations.
Solution:
Write \(A = A I\): \(\begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\).
Apply column operations:
\(C_1 \rightarrow C_1 – C_2\): \(\begin{bmatrix} -3 & 5 \\ -2 & 3 \end{bmatrix} = A \begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix}\)
\(C_1 \rightarrow -\frac{1}{3} C_1\): \(\begin{bmatrix} 1 & 5 \\ \frac{2}{3} & 3 \end{bmatrix} = A \begin{bmatrix} -\frac{1}{3} & 0 \\ \frac{1}{3} & 1 \end{bmatrix}\)
\(C_2 \rightarrow C_2 – 5C_1\): \(\begin{bmatrix} 1 & 0 \\ \frac{2}{3} & -\frac{1}{3} \end{bmatrix} = A \begin{bmatrix} -\frac{1}{3} & \frac{5}{3} \\ \frac{1}{3} & -\frac{2}{3} \end{bmatrix}\)
So \(A^{-1} = \begin{bmatrix} -\frac{1}{3} & \frac{5}{3} \\ \frac{1}{3} & -\frac{2}{3} \end{bmatrix} = \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix}\). Check: \(\begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix} \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\).
Additional Example 2:
Find \(A^{-1}\) for \(A = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix}\) using row operations.
Solution:
\([A|I] = \left[\begin{array}{ccc|ccc} 0 & 1 & 2 & 1 & 0 & 0 \\ 1 & 2 & 3 & 0 & 1 & 0 \\ 3 & 1 & 1 & 0 & 0 & 1 \end{array}\right]\)
Interchange \(R_1 \leftrightarrow R_2\): \(\left[\begin{array}{ccc|ccc} 1 & 2 & 3 & 0 & 1 & 0 \\ 0 & 1 & 2 & 1 & 0 & 0 \\ 3 & 1 & 1 & 0 & 0 & 1 \end{array}\right]\)
\(R_3 \rightarrow R_3 – 3R_1\): \(\left[\begin{array}{ccc|ccc} 1 & 2 & 3 & 0 & 1 & 0 \\ 0 & 1 & 2 & 1 & 0 & 0 \\ 0 & -5 & -8 & 0 & -3 & 1 \end{array}\right]\)
\(R_3 \rightarrow R_3 + 5R_2\): \(\left[\begin{array}{ccc|ccc} 1 & 2 & 3 & 0 & 1 & 0 \\ 0 & 1 & 2 & 1 & 0 & 0 \\ 0 & 0 & 2 & 5 & -3 & 1 \end{array}\right]\)
\(R_3 \rightarrow \frac{1}{2} R_3\): \(\left[\begin{array}{ccc|ccc} 1 & 2 & 3 & 0 & 1 & 0 \\ 0 & 1 & 2 & 1 & 0 & 0 \\ 0 & 0 & 1 & \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \end{array}\right]\)
\(R_2 \rightarrow R_2 – 2R_3\): \(\left[\begin{array}{ccc|ccc} 1 & 2 & 3 & 0 & 1 & 0 \\ 0 & 1 & 0 & -4 & 3 & -1 \\ 0 & 0 & 1 & \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \end{array}\right]\)
\(R_1 \rightarrow R_1 – 2R_2 – 3R_3\): After simplification, \(R_1 \rightarrow \begin{bmatrix} 1 & 0 & 0 & \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \end{bmatrix}\)
Thus \(A^{-1} = \begin{bmatrix} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & -1 \\ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \end{bmatrix}\).
Practice Problems
Solve the following problems:
- Find the inverse of \(\begin{bmatrix} 1 & 3 \\ 2 & 7 \end{bmatrix}\) using row operations.
- Find \(A^{-1}\) for \(A = \begin{bmatrix} 3 & 10 \\ 2 & 7 \end{bmatrix}\) using column operations.
- Using elementary row transformations, find the inverse of \(\begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix}\).
- Check if \(A = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}\) is invertible. If yes, find its inverse.
- Find the inverse of \(\begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}\) using row operations.
- Using column operations, find \(A^{-1}\) for \(A = \begin{bmatrix} 4 & 7 \\ 2 & 6 \end{bmatrix}\).
- Find the inverse of \(\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{bmatrix}\).
- If \(A = \begin{bmatrix} 2 & 3 \\ 1 & -4 \end{bmatrix}\), find \(A^{-1}\) using elementary transformations.
- Find the inverse of \(\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}\) using row operations.
- Determine the inverse of \(\begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix}\).
- Apply column operations to find \(A^{-1}\) for \(A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}\).
- Find \(k\) such that \(\begin{bmatrix} 1 & 2 \\ k & 4 \end{bmatrix}\) is singular.
- Using row operations, find the inverse of \(\begin{bmatrix} 0 & 2 & 1 \\ 1 & 0 & 0 \\ 3 & 1 & 2 \end{bmatrix}\).
- Verify that \(A^{-1}A = I\) for \(A = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{bmatrix}\).
- Find the inverse of \(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\) where \(ad-bc \neq 0\) using elementary transformations.
Chapter-End MCQ Test
Multiple Choice Questions (Choose the correct option)
- Which of the following is not an elementary row operation?
(A) \(R_i \leftrightarrow R_j\) (B) \(R_i \rightarrow R_i + kR_j\) (C) \(R_i \rightarrow k R_i\) where \(k=0\) (D) \(R_i \rightarrow k R_i\) where \(k \neq 0\)Answer: C - The inverse of \(\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\) using row operations is:
(A) \(\begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}\) (B) \(\begin{bmatrix} -1 & 1 \\ 0 & 1 \end{bmatrix}\) (C) \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\) (D) \(\begin{bmatrix} 1 & 1 \\ 0 & -1 \end{bmatrix}\)Answer: A - If a matrix \(A\) is singular, then:
(A) \(A^{-1}\) exists (B) \(A^{-1}\) does not exist (C) \(A\) is square (D) \(A\) is symmetricAnswer: B - Which operation is used to convert a diagonal element to 1?
(A) Interchange (B) Scaling \(R_i \rightarrow kR_i\), \(k \neq 0\) (C) \(R_i \rightarrow R_i + R_j\) (D) None of theseAnswer: B - The inverse of \(\begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}\) is:
(A) \(\begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix}\) (B) \(\begin{bmatrix} -3 & 5 \\ 1 & -2 \end{bmatrix}\) (C) \(\begin{bmatrix} 3 & 5 \\ 1 & 2 \end{bmatrix}\) (D) \(\begin{bmatrix} 2 & -5 \\ -1 & 3 \end{bmatrix}\)Answer: A - While finding inverse using column operations, we write:
(A) \(A = IA\) (B) \(A = AI\) (C) \(A = A^{-1}I\) (D) \(A = I A^{-1}\)Answer: B - If \(A = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix}\), then \(A^{-1}\) is:
(A) \(\begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 1 & -2 & 1 \end{bmatrix}\) (B) \(\begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix}\) (C) \(\begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 1 & 2 & 1 \end{bmatrix}\) (D) \(\begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ -3 & 2 & 1 \end{bmatrix}\)Answer: A - How many elementary operations are possible on a matrix?
(A) 2 (B) 3 (C) 4 (D) 6Answer: B - The elementary operation \(C_2 \rightarrow C_2 + 5C_1\) means:
(A) Add 5 times column 2 to column 1 (B) Add 5 times column 1 to column 2 (C) Multiply column 2 by 5 and add to column 1 (D) None of theseAnswer: B - A square matrix is invertible if and only if:
(A) It is symmetric (B) It is diagonal (C) Its determinant is non-zero (D) All entries are non-zeroAnswer: C - Find \(A^{-1}\) for \(A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\) using row operations:
(A) \(\begin{bmatrix} -2 & 1 \\ 1.5 & -0.5 \end{bmatrix}\) (B) \(\begin{bmatrix} 2 & -1 \\ -1.5 & 0.5 \end{bmatrix}\) (C) \(\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\) (D) \(\begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}\)Answer: A - The inverse of identity matrix \(I_n\) is:
(A) \(I_n\) (B) \(-I_n\) (C) \(0\) (D) Does not existAnswer: A - If \(A\) and \(B\) are invertible matrices of same order, then \((AB)^{-1} =\)
(A) \(A^{-1}B^{-1}\) (B) \(B^{-1}A^{-1}\) (C) \(AB\) (D) \(BA\)Answer: B - Which operation is used to make elements above and below a pivot zero?
(A) \(R_i \rightarrow kR_i\) (B) \(R_i \leftrightarrow R_j\) (C) \(R_i \rightarrow R_i + kR_j\) (D) NoneAnswer: C - The elementary row operation \(R_1 \rightarrow R_1 – 3R_2\) on \([A|I]\):
(A) Does not change the matrix (B) Changes \(A\) and \(I\) accordingly (C) Changes only \(A\) (D) Changes only \(I\)Answer: B
© CBSE Class 12 Mathematics — Matrices (NCERT based) | Study Material