2nd PUC Maths Question Bank Chapter 4 Determinants Ex 4.1

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Karnataka 2nd PUC Maths Question Bank Chapter 4 Determinants Ex 4.1

2nd PUC Maths Determinants NCERT Text Book Questions and Answers Ex 4.1

Evaluate the determinations

Question 1.
\(\left|\begin{array}{rr}{2} & {4} \\{-5} & {-1}\end{array}\right|\)
Answer:
2 x (-1) -4x (-5)
= -2 + 20 = 18

Question 2.
(i)
\(\left|\begin{array}{rr}{\cos \theta} & {-\sin \theta} \\{\sin \theta} & {\cos \theta}\end{array}\right|\)
Answer:
Cos θ x Cos θ – sin θ x sin θ
= cos² θ + sin² θ = 1

(ii)
\(\left|\begin{array}{cc}{\mathbf{x}^{2}-\mathbf{x}+\mathbf{1}} & {\mathbf{x}-\mathbf{1}} \\{\mathbf{x}+\mathbf{1}} & {\mathbf{x}+\mathbf{1}}\end{array}\right|\)
Answer:
(x² – x + 1) (x + 1) – (x – 1) (x + 1)
= x³ + 1 – (x² – 1)
= x³ – x² + 2

Question 3.
\(\text { If } \mathbf{A}=\left[\begin{array}{ll}{1} & {2} \\{4} & {2}\end{array}\right]\),then show that |2A| = 4 |A|
Answer:

Question 4.
\(\text { If } \mathbf{A}=\left[\begin{array}{lll}{\mathbf{1}} & {\mathbf{0}} & {\mathbf{1}} \\{\mathbf{0}} & {\mathbf{1}} & {\mathbf{2}} \\{\mathbf{0}} & {\mathbf{0}} & {\mathbf{4}}\end{array}\right]\) then show that |3A| = 27 |A|
Answer:

Question 5.
Evaluate the determinant
(i)
\(\left|\begin{array}{rrr}{3} & {-4} & {5} \\{1} & {1} & {-2} \\{2} & {3} &{1}\end{array}\right|\)
Answer:
Expanding along R₂ (Second Row where two elements are zeroes)

(ii)
\(\left|\begin{array}{rrr}{3} & {-4} & {5} \\{1} & {1} & {-2} \\{2} & {3} & {1}\end{array}\right|\)
Answer:
Expanding along R₁

(iii)
\(\left|\begin{array}{ccc}{0} & {1} & {2} \\{-1} & {0} & {-3} \\{-2} & {3} & {0}\end{array}\right|\)
Answer:
Expanding along R₁

(iv)
\(\left|\begin{array}{ccc}{2} & {-1} & {-2} \\{0} & {2} & {-1} \\{3} & {-5} &{0}\end{array}\right|\)
Answer:
Expanding along R₃

Question 6.
\(\text { If } A=\left[\begin{array}{ccc}{1} & {1} & {-2} \\{2} & {1} & {-3} \\{5} & {4} & {-9}\end{array}\right] \text { find }|A|\)
Answer:

Question 7.
find X if
(i)
\(\left|\begin{array}{cc}{2} & {4} \\{5} & {1}\end{array}\right|=\left|\begin{array}{cc}{2 x} & {4} \\{6} & {x}\end{array}\right|\)
Answer:
2 x 1 -4 x 5 = 2 × x × -4 × 6
2 – 20 = 2x² – 24 – 18
= 2x² – 24
2x² = 6,
x² = 3
\(x=\pm \sqrt{3}\)

(ii)
\(\left|\begin{array}{cc}{2} & {3} \\{4} & {5}\end{array}\right|=\left|\begin{array}{cc}{x} & {3} \\{2 x} & {5}\end{array}\right|\)
2 x 5 -3 x 4=5 x x – 6x
10 – 12  = 5x – 6x
-2       =-x
x = 2

Question 8.
\(\text { If }\left|\begin{array}{cc}{x} & {2} \\{18} {x}\end{array}\right|=\left|\begin{array}{cc}{6} & {2} \\{18} & {6}\end{array}\right|\)
(A) 6
(B) ± 6
(C) -6
(D) 0
Answer:
2 x 18 = 6 x 6-18 x 2
x² – 36 = 36 – 36
x² – 36 = 0
x² = 36
x= ±6
hence B is the correct answer.
Note: Two determinants ms equal does not mean corresponding elements are equal so do not equate the elements like we are doing in matrix.