Chapter 4 of Class 12 NCERT Mathematics Part I introduces the concept of determinants, a crucial topic in linear algebra. Exercise 4.1 focuses on the fundamentals of determinants, including their definition, calculation for 2x2 and 3x3 matrices, and their basic properties. This exercise lays the groundwork for understanding determinants' role in solving systems of linear equations, finding inverse matrices, and their applications in various fields of mathematics and science. Students learn to evaluate determinants, understand their geometric interpretation, and apply their properties to solve problems and prove mathematical statements.
Evaluate the determinants from the following Questions.
Question 1. {\begin{vmatrix}2&4\\-5&-1\end{vmatrix}}
Solution:
The determinant of a 2 x 2 matrix
{\begin{vmatrix}a&b\\c&d\end{vmatrix}} = ad - bc\\ Hence,
{\begin{vmatrix}2&4\\-5&-1\end{vmatrix}} \\ = (2)(-1) - (4)(-5)\\ = -2 + 20\\ = 18\\
Question 2. (i) {\begin{vmatrix}cosθ&-sinθ\\sinθ&cosθ\end{vmatrix}}
Solution:
{\begin{vmatrix}cosθ&-sinθ\\sinθ&cosθ\end{vmatrix}}
= (cosθ)(cosθ) - (-sinθ)(sinθ)\\ = cos^2θ + sin^2θ\\ = 1 from trigonometric identities
(ii) {\begin{vmatrix}x^2-x+1&x-1\\x+1&x+1\end{vmatrix}}
Solution:
{\begin{vmatrix}x^2-x+1&x-1\\x+1&x+1\end{vmatrix}}
= (x^2-x+1)(x+1) - (x-1)(x+1)\\ = (x^3-x^2+x+x^2-x+1) - (x^2-1)\\ = (x^3+1) - (x^2-1)\\ = x^3 - x^2 + 2\\
Question 3. If A = {\begin{vmatrix}1&2\\4&2\end{vmatrix}} show that |2A| = 4|A|
Solution:
LHS=>
Matrix,
2A\\ = 2{\begin{pmatrix}1&2\\4&2\end{pmatrix}}\\ = {\begin{pmatrix}2*1&2*2\\2*4&2*2\end{pmatrix}}\\ = {\begin{pmatrix}2&4\\8&4\end{pmatrix}}\\ Hence, determinant,
|2A|\\ = {\begin{vmatrix}2&4\\8&4\end{vmatrix}}\\ = (2)(4) - (4)(8)\\ = -24 RHS=>
Determinant,
|A|\\ = {\begin{vmatrix}1&2\\4&2\end{vmatrix}}\\ = (1)(2) - (2)(4)\\ = -6 Now,
4|A|\\ = 4(-6)\\ = -24 Hence, proved, LHS = RHS
Question 4. If A = {\begin{vmatrix}1&0&1\\0&1&2\\0&0&4\end{vmatrix}} then show that ||3A| = 27|A|
Solution:
LHS=>
Matrix,
3A\\ = 3{\begin{pmatrix}1&0&1\\0&1&2\\0&0&4\end{pmatrix}}\\ = {\begin{pmatrix}3*1&3*0&3*1\\3*0&3*1&3*2\\3*0&3*0&3*4\end{pmatrix}}\\ = {\begin{pmatrix}3&0&3\\0&3&6\\0&0&12\end{pmatrix}}\\ Hence, determinant,
|3A|\\ = {\begin{vmatrix}3&0&3\\0&3&6\\0&0&12\end{vmatrix}}\\ = 3{\begin{vmatrix}3&6\\0&12\end{vmatrix}} - 0{\begin{vmatrix}0&6\\0&12\end{vmatrix}} + 3{\begin{vmatrix}0&3\\0&0\end{vmatrix}} \\ = 3(36) - 0(0) + 3(0)\\ = 108 RHS =>
Determinant,
|A|\\ = {\begin{vmatrix}1&0&1\\0&1&2\\0&0&4\end{vmatrix}}\\ = 1{\begin{vmatrix}1&2\\0&4\end{vmatrix}} - 0{\begin{vmatrix}0&2\\0&4\end{vmatrix}} + 1{\begin{vmatrix}0&1\\0&0\end{vmatrix}} \\ = 1(4) - 0(0) + 1(0) Now,
27|A|\\ = 27(4)\\ = 108 Hence, proved, LHS = RHS
Question 5. Evaluate the determinants
(i) {\begin{vmatrix}3&-1&-2\\0&0&-1\\3&-5&0\end{vmatrix}}\\
Solution:
Since the maximum number of zeroes are in the second row, we will expand the determinant along row 2.
= -0{\begin{vmatrix}-1&-2\\-5&0\end{vmatrix}} + 0{\begin{vmatrix}3&-2\\3&0\end{vmatrix}} -(-1){\begin{vmatrix}3&-1\\3&-5\end{vmatrix}} \\ = -(-1)(-15 + 3)\\ = -12
(ii) {\begin{vmatrix}3&-4&5\\1&1&-2\\2&3&1\end{vmatrix}}\\
Solution:
= 3{\begin{vmatrix}1&-2\\3&1\end{vmatrix}} -(-4){\begin{vmatrix}1&-2\\2&1\end{vmatrix}} +5{\begin{vmatrix}1&1\\2&3\end{vmatrix}} \\ = 3(7) + 4(5) + 5(1)\\ = 46
(iii) {\begin{vmatrix}0&1&2\\-1&0&-3\\-2&3&0\end{vmatrix}}\\
Solution:
= 0{\begin{vmatrix}0&-3\\3&0\end{vmatrix}} -1{\begin{vmatrix}-1&-3\\-2&0\end{vmatrix}} +2{\begin{vmatrix}-1&0\\-2&3\end{vmatrix}} \\ = 0 - 1(-6) + 2(-3)\\ = 0 Note: This matrix is skew symmetric i.e.
A^T = -A For every skew symmetric matrix of "odd dimension", the determinant vanishes i.e. determinant is zero.
(iv) {\begin{vmatrix}2&-1&-2\\0&2&-1\\3&-5&0\end{vmatrix}}\\
Solution:
Since the maximum number of zeroes are in the second row, we will expand the determinant along row 2.
= -0{\begin{vmatrix}-1&-2\\-5&0\end{vmatrix}} + 2{\begin{vmatrix}2&-2\\3&0\end{vmatrix}} - (-1){\begin{vmatrix}2&-1\\3&-5\end{vmatrix}} \\ = 0 + 2(6) +1(-7)\\ = 5
Question 6. If A = {\begin{pmatrix}1&1&-2\\2&1&-3\\5&4&-9\end{pmatrix}} find |A|
Solution:
|A| \\ = 1{\begin{vmatrix}1&-3\\4&-9\end{vmatrix}} -1{\begin{vmatrix}2&-3\\5&-9\end{vmatrix}} + (-2){\begin{vmatrix}2&1\\5&4\end{vmatrix}} \\ = 1(3) - 1(-3) - 2(3)\\ = 0
Question 7. Find the values of x if
(i) {\begin{vmatrix}2&4\\5&1\end{vmatrix}} = {\begin{vmatrix}2x&4\\6&x\end{vmatrix}}
Solution:
Solving determinants on both sides,
(2)(1) - (4)(5) = (2x)(x) - (4)(6)\\ -18 = 2x^2 - 24\\ x^2 = 3\\ x = ±√3
(ii) {\begin{vmatrix}2&3\\4&5\end{vmatrix}} = {\begin{vmatrix}x&3\\2x&5\end{vmatrix}}
Solution:
Solving determinants on both sides
(2)(5) - (3)(4) = (x)(5) - (3)(2x)\\ x = 2
Question 8. If {\begin{vmatrix}x&2\\18&x\end{vmatrix}} = {\begin{vmatrix}6&2\\18&6\end{vmatrix}} then x is equal to
(A) 6 (B) ±6 (C) -6 (D) 0
Solution:
Solving determinants on both sides
(x)(x) - (2)(18) = (6)(6) - (2)(18)\\ x^2 -36 = 0\\ x = ±6\\ Hence, Option (B)
Summary
Exercise 4.1 of Chapter 4 on Determinants provides a comprehensive set of problems to help students master the fundamental concepts of determinants. It covers the calculation of determinants for both 2x2 and 3x3 matrices, explores their properties, and introduces applications in solving equations. The exercise includes problems that require students to prove certain properties of determinants, enhancing their analytical and proof-writing skills. By working through these problems, students develop a strong foundation in determinant theory, preparing them for more advanced topics in linear algebra and its applications in various fields of mathematics and science.
