Students can cross-reference their work with OP Malhotra Class 12 Solutions Chapter 5 Determinants Ex 5(a) to ensure accuracy.
S Chand Class 12 ICSE Maths Solutions Chapter 5 Determinants Ex 5(a)
Question 1.
Evaluate the following determinants :
(i) \(\left|\begin{array}{ll}
2 & 5 \\
4 & 1
\end{array}\right|\)
(ii) \(\left|\begin{array}{ll}
3 & 5 \\
1 & 2
\end{array}\right|\)
(iii) \(\left|\begin{array}{rr}
a & b \\
-b & a
\end{array}\right|\)
(iv) \(\left|\begin{array}{rr}
x+2 & 2 x+5 \\
3 x-1 & x-3
\end{array}\right|\)
(v) \(\left|\begin{array}{rr}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right|\)
(vi) \(\left|\begin{array}{cc}
y-x & -x^2+x y-y^2 \\
x+y & x^2+x y+y^2
\end{array}\right|\)
Solution:
Question 2.
Prove that
\(\left|\begin{array}{rr}
\sin 10^{\circ} & -\cos 10^{\circ} \\
\sin 80^{\circ} & \cos 80^{\circ}
\end{array}\right|\) = 1
Solution:
\(\left|\begin{array}{rr}
\sin 10^{\circ} & -\cos 10^{\circ} \\
\sin 80^{\circ} & \cos 80^{\circ}
\end{array}\right|\)
= sin 10° cos80° + cos 10° sin 80°
= sin(10° + 80°) = sin 90° = 1
Question 3.
If \(\left|\begin{array}{rr}
3 & m \\
4 & 5
\end{array}\right|\) = 3, find the value of m.
Solution:
Given
\(\left|\begin{array}{rr}
3 & m \\
4 & 5
\end{array}\right|\) = 3
⇒ 15 – 4m = 3
⇒ 12 = 4 m
⇒ m = 3
Question 4.
(i) Find the value of x, if
\(\left|\begin{array}{rr}
x-1 & x-2 \\
x & x-3
\end{array}\right|\) = 0
\(\left|\begin{array}{ll}
3 x & 7 \\
-2 & 4
\end{array}\right|=\left|\begin{array}{ll}
8 & 7 \\
6 & 4
\end{array}\right|\)
Solution:
(i) Given \(\left|\begin{array}{rr}
x-1 & x-2 \\
x & x-3
\end{array}\right|\) = 0
⇒ (x – 1) (x – 3) – x(x – 2) = 0
⇒ x² – 4x + 3 – x² + 2x = 0
⇒ – 2x + 3 = 0 ⇒ x = \(\frac { 3 }{ 2 }\)
(ii) Given \(\left|\begin{array}{ll}
3 x & 7 \\
-2 & 4
\end{array}\right|=\left|\begin{array}{ll}
8 & 7 \\
6 & 4
\end{array}\right|\)
⇒ 12x + 14 = 32 – 42
⇒ 12x = – 10 – 14 = – 24
⇒ x = – 2
Question 5.
Determine the value of k for which \(\left|\begin{array}{rr}
k & k \\
4 & 2 k
\end{array}\right|\) = 0
Solution:
Given \(\left|\begin{array}{rr}
k & k \\
4 & 2 k
\end{array}\right|\) = 0
⇒ k x 2k – 4 x k = 0
⇒ 2k² – 4k = 0
⇒ 2k(k – 2) = 0
⇒ k = 0, 2