CBSE Class 12 Maths Previous Year Question Paper 2014
This CBSE Class 12 Maths Previous Year Question Paper from 2014 focuses on the topic of finding the inverse of a matrix using elementary operations. It includes questions requiring the application of elementary column operations and elementary row transformations (ERT). For instance, one question involves using column operations on a matrix equation, while others demonstrate the step-by-step process of finding the inverse of given 2x2 matrices like [[6, 5], [5, 4]], [[3, 2], [7, 5]], and [[2, 5], [1, 3]]. Solving these previous year questions helps students understand the application of matrix operations and prepare effectively for their board examinations.
Explore more subjects
Quick info
| Board | CBSE |
|---|---|
| Class | 12 |
| Subject | Maths |
| Session | 2014 |
| Language | English |
| Type | Previous Year Question Paper |
| Exam type | Board Exam |
Paper pattern
Questions in this paper are worth 4 marks each and focus on applying elementary row and column operations to find the inverse of matrices.
Topics covered
Paper topics
- Inverse of a Matrix
- Elementary Operations
- Elementary Row Transformation
- Elementary Column Operations
- Matrix Equations
Important topics
- Inverse of a Matrix by Elementary Operations
PDF preview
Read page by page below. PDF is streamed from the official NCERT website — no download button on this page.
Question paper text
Inverse of a Matrix by Elementary Operations
4 Marks Questions
- Use elementary column operations <math>C_2 \rightarrow C_2 - 2C_1</math> in the matrix equation
<math>\begin{vmatrix} 4 & 2 \\ 3 & 3 \end{vmatrix} = \begin{vmatrix} 1 & 2 & 2 & 0 \\ 0 & 3 & 1 & 1 \end{vmatrix}</math>
Foreign 2014
We write the matrix A as <math>A = AI</math> for applying elementary column operations. So, apply column operation on the matrix of LHS and on the second matrix of RHS.
Given matrix equation is
<math display="block">\begin{bmatrix} 4 & 2 \\ 3 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 2 & 0 \\ 0 & 3 & 1 & 1 \end{bmatrix}</math>
On applying <math>C_2 \rightarrow C_2 - C_1</math>, we get
<math display="block">\begin{bmatrix} 4 & 2 - 8 \\ 3 & 3 - 6 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 2 & 0 - 4 \\ 0 & 3 & 1 & 1 - 2 \end{bmatrix}</math> <math display="block">\Rightarrow \begin{vmatrix} 4 & -6 \\ 3 & -3 \end{vmatrix} = \begin{vmatrix} 1 & 2 \\ 0 & 3 \end{vmatrix} \begin{vmatrix} 2 & -4 \\ 1 & -1 \end{vmatrix}</math>
which is the required answer.
Using elementary row transformation (ERT), find inverse of matrix <math>A = \begin{bmatrix} 6 & 5 \\ 5 & 4 \end{bmatrix}</math>. Foreign 2010; HOTS
Firstly, put <math>A = IA</math>. Then, by applying elementary row transformation on A of LHS and I of RHS, convert this matrix in the form <math>I = BA</math>, where B gives the inverse of A.
Given matrix is <math>A = \begin{bmatrix} 6 & 5 \\ 5 & 4 \end{bmatrix}</math>.
Let <math>A = IA</math> ...
<math>\Rightarrow</math>
<math>\begin{vmatrix} 6 & 5 \\ 5 & 4 \end{vmatrix} = \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} A</math>
(1/2)
Applying <math>R_1 \rightarrow R_1 - R_2</math>, we get
<math display="block">\begin{bmatrix} 1 & 1 \\ 5 & 4 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix} A</math> (1) Applying <math>R_2 \rightarrow R_2 - 5R_1</math>, we get
<math>\begin{vmatrix} 1 & 1 \\ 0 & -1 \end{vmatrix} = \begin{vmatrix} 1 & -1 \\ -5 & 6 \end{vmatrix} A</math> (1) Applying <math>R_1 \rightarrow R_1 + R_2</math>, we get
<math>\begin{vmatrix} 1 & 0 \\ 0 & -1 \end{vmatrix} = \begin{vmatrix} -4 & 5 \\ -5 & 6 \end{vmatrix} A</math>
(1/2)
Now, applying <math>R_2 \rightarrow (-1)R_2</math>, we get
<math display="block">\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} -4 & 5 \\ 5 & -6 \end{bmatrix} A</math>
(1/2)
Hence, <math>A^{-1} = \begin{bmatrix} -4 & 5 \\ 5 & -6 \end{bmatrix} [:: A^{-1}A = I]</math> (1/2)
- Find <math>A^{-1}</math>, by using elementary row transformation for matrix <math>A = \begin{bmatrix} 3 & 2 \\ 7 & 5 \end{bmatrix}</math>. Foreign 2010
Do same as Que 2. Ans. <math>A^{-1} = \begin{bmatrix} 5 & -2 \\ -7 & 3 \end{bmatrix}</math>
- Using elementary row transformation, find
inverse of matrix <math>A = \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}</math>. Delhi 2010
Do same as Que 2. Ans. <math>A^{-1} = \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix}</math>
Frequently asked questions
What is this document?
This is a CBSE Class 12 Maths Previous Year Question Paper from 2014, focusing on matrix operations.
What is the main topic covered?
The main topic is finding the inverse of a matrix using elementary row and column operations.
How does this paper help students?
Solving this previous year question paper helps students understand the board exam pattern and practice specific matrix operations for better scores.
What is the marking scheme for these questions?
The questions related to finding the inverse of a matrix using elementary operations are marked for 4 marks.
How can students best use this paper?
Students should solve these problems step-by-step, similar to the provided examples, to master the technique of finding matrix inverses using elementary transformations.
Content reviewed by the NCERT Help team. Editorial Team and update policy
Question Papers PDF on NCERT Help. URL unchanged for search indexing.