Table of Contents
Ans. (d) 40
Explanation:
According to the property of transpose of a matrix, (kA′) = kA′
Also, from the property of determinant of a matrix, |A′| = |A|
and |kA| = k^n|A|, where n is the order of matrix A.
Thus, |2A′| = 2^3|A| {since A is a square matrix of order 3} = 8 × 5 = 40
Ans. (c) Determinant is a number associated with a square matrix.
Explanation:
Given that A is a square matrix of order 3 and |A| = -4.
We know that |adj A| = |A|^{n−1}, where n is the order of matrix A.
So, |adj A| = (−4)^{3-1} = (-4)^2 = 16
Explanation:
|A| = -7
Order of matrix A is 3×3.
Now, \displaystyle{\sum_{i = 1}^3} a_{i2}\space A_{i2} = \space a_{12}\space A_{12} + a_{22}\space A_{22}\space + \space a_{32}\space A_{32}\space \\[4.5 bp]
= |A|
= -7
Explanation:
Given that the A is an invertible matrix of order 2.
If the matrix is invertible, then its determinant is not equal to 0.
We know that,
AA^{-1} = I, where I is the identity matrix
Taking determinant on both sides,
\\[4.5 bp]|AA^{-1}| = |I|\\[4.4 bp]
|A| |A^{-1}| = 1\\[4.5 bp]
|A^{-1}| = \dfrac{1}{|A|} {since A is non-singular, |A| ≠ 0}
Chapter No. | Chapter Name |
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Chapter 1 | Relations and Functions |
Chapter 2 | Inverse Trigonometric Functions |
Chapter 3 | Matrices |
Chapter 4 | Determinants |
Chapter 5 | Continuity and Differentiability |
Chapter 6 | Applications of Derivatives |
Chapter 7 | Integrals |
Chapter 8 | Applications of the Integrals |
Chapter 9 | Differential Equations |
Chapter 10 | Vectors |
Chapter 11 | Three - dimensional Geometry |
Chapter 12 | Linear Programming |
Chapter 13 | Probability |
Chapter Wise Important Questions for CBSE Board Class 12 Maths |
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Relations and Functions |
Inverse Trigonometric Functions |
Matrices |
Determinants |
Continuity and Differentiability |
Applications of Derivatives |
Integrals |
Applications of the Integrals |
Differential Equations |
Vectors |
Three - dimensional Geometry |
Linear Programming |
Probability |
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