GSEB Solutions Class 12 Maths Chapter 4 નિશ્ચાયક Ex 4.1

Gujarat Board GSEB Solutions Class 12 Maths Chapter 4 નિશ્ચાયક Ex 4.1 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 12 Maths Chapter 4 નિશ્ચાયક Ex 4.1

પ્રશ્ન 1 અને 2 માં આપેલા નિશ્વારાકનું મૂલ્ય શોધો.

પ્રશ્ન 1.
\(\left|\begin{array}{cc}
2 & 4 \\
-5 & -1
\end{array}\right|\)
ઉત્તરઃ
\(\left|\begin{array}{cc}
2 & 4 \\
-5 & -1
\end{array}\right|\)
= 2(-1) – (-5)(4)
= -2 + 20
= 18

પ્રશ્ન 2.
(i) \(\left|\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right|\)
ઉત્તરઃ
\(\left|\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right|\)
= (cosθ) (cosθ) (-sinθ)(sinθ)
= cos²θ + sin²θ
= 1

(ii) \(\left|\begin{array}{cc}
x^2-x+1 & x-1 \\
x+1 & x+1
\end{array}\right|\)
ઉત્તરઃ
\(\left|\begin{array}{cc}
x^2-x+1 & x-1 \\
x+1 & x+1
\end{array}\right|\)
= (x + 1)(x² – x + 1) – (x + 1)(x – 1)
= (x³ + 1³) – (x² – 1)
= x³ + 1 – x² + 1
= x³ – x² + 1

પ્રશ્ન 3.
જો A = \(\left[\begin{array}{ll}
1 & 2 \\
4 & 2
\end{array}\right]\) હોય, તો સાબિત કરો કે |2A| = 4|A|.
ઉત્તરઃ
|A| = \(\left[\begin{array}{ll}
1 & 2 \\
4 & 2
\end{array}\right]\)
= (2)(1) – (2)(4)
= 2 – 8
= -6
A = \(\left[\begin{array}{ll}
1 & 2 \\
4 & 2
\end{array}\right]\) ∴ 2A = \(\left[\begin{array}{ll}
2 & 4 \\
8 & 4
\end{array}\right]\)
∴ |2A| = \(\left|\begin{array}{ll}
2 & 4 \\
8 & 4
\end{array}\right|\)
= (2)(4) – (4)(8)
= 8 – 32.
= -24
∴ 2|A| = 4|A| = -24

પ્રશ્ન 4.
જો A = \(\left[\begin{array}{lll}
1 & 0 & 1 \\
0 & 1 & 2 \\
0 & 0 & 4
\end{array}\right]\), હોય, તો સાબિત કરો કે |3A| = 27|A|.
ઉત્તરઃ
|A| = \(\left[\begin{array}{lll}
1 & 0 & 1 \\
0 & 1 & 2 \\
0 & 0 & 4
\end{array}\right]\) = 1(4 – 0) = 4
27|A| = 27(4) = 108
3A = \(3\left[\begin{array}{lll}
1 & 0 & 1 \\
0 & 1 & 2 \\
0 & 0 & 4
\end{array}\right]=\left[\begin{array}{ccc}
3 & 0 & 3 \\
0 & 3 & 6 \\
0 & 0 & 12
\end{array}\right]\)
∴ |3A| = \(\left|\begin{array}{ccc}
3 & 0 & 3 \\
0 & 3 & 6 \\
0 & 0 & 12
\end{array}\right|\) = 3(36 – 0) = 108
∴ |3A| = 27|A| = 108

પ્રશ્ન 5.
નીચે આપેલાં નિશ્ચાયકનાં મૂલ્યો શોધો :
(i) \(\left|\begin{array}{ccc}
3 & -1 & -2 \\
0 & 0 & -1 \\
3 & -5 & 0
\end{array}\right|\)
ઉત્તરઃ
\(\left|\begin{array}{ccc}
3 & -1 & -2 \\
0 & 0 & -1 \\
3 & -5 & 0
\end{array}\right|\)
= \(3\left|\begin{array}{cc}
0 & -1 \\
-5 & 0
\end{array}\right|-(-1)\left|\begin{array}{cc}
0 & -1 \\
3 & 0
\end{array}\right|+(-2)\left|\begin{array}{cc}
0 & 0 \\
3 & -5
\end{array}\right|\)
= 3(0 – 5) + 1(0 + 3) – 2(0)
= -15 + 3 – 0
= -12

(ii) \(\left|\begin{array}{ccc}
3 & -4 & 5 \\
1 & 1 & -2 \\
2 & 3 & 1
\end{array}\right|\)
ઉત્તરઃ
\(\left|\begin{array}{ccc}
3 & -4 & 5 \\
1 & 1 & -2 \\
2 & 3 & 1
\end{array}\right|\)
= 3(1 + 6) + 4(1 + 4) + 5(3 – 2)
= 21 + 20 + 5
= 46

(iii) \(\left|\begin{array}{ccc}
0 & 1 & 2 \\
-1 & 0 & -3 \\
-2 & 3 & 0
\end{array}\right|\)
ઉત્તરઃ
\(\left|\begin{array}{ccc}
0 & 1 & 2 \\
-1 & 0 & -3 \\
-2 & 3 & 0
\end{array}\right|\)
= \(0\left|\begin{array}{cc}
0 & -3 \\
3 & 0
\end{array}\right|-1\left|\begin{array}{cc}
-1 & -3 \\
-2 & 0
\end{array}\right|+2\left|\begin{array}{cc}
-1 & 0 \\
-2 & 3
\end{array}\right|\)
= 0 – 1(-6) + 2(-3)
= 6 – 6
= 0

(iv) \(\left|\begin{array}{ccc}
2 & -1 & -2 \\
0 & 2 & -1 \\
3 & -5 & 0
\end{array}\right|\)
ઉત્તરઃ
\(\left|\begin{array}{ccc}
2 & -1 & -2 \\
0 & 2 & -1 \\
3 & -5 & 0
\end{array}\right|\)
= \(2\left|\begin{array}{cc}
2 & -1 \\
-5 & 0
\end{array}\right|+1\left|\begin{array}{cc}
0 & -1 \\
3 & 0
\end{array}\right|-2\left|\begin{array}{cc}
0 & 2 \\
3 & -5
\end{array}\right|\)
= 2(0 – 5) + 1(3) – 2(-6)
= – 10 + 3 + 12
= 5

પ્રશ્ન 6.
જો A = \(\left[\begin{array}{lll}
1 & 1 & -2 \\
2 & 1 & -3 \\
5 & 4 & -9
\end{array}\right]\) હોય, તો |A| શોધો.
ઉત્તરઃ
|A| = \(\left|\begin{array}{lll}
1 & 1 & -2 \\
2 & 1 & -3 \\
5 & 4 & -9
\end{array}\right|\)
= \(1\left|\begin{array}{ll}
1 & -3 \\
4 & -9
\end{array}\right|-1\left|\begin{array}{ll}
2 & -3 \\
5 & -9
\end{array}\right|-2\left|\begin{array}{ll}
2 & 1 \\
5 & 4
\end{array}\right|\)
= 1 (-9 + 12) – 1(-18 + 15) – 2(8 – 5)
= 3 + 3 – 6
= 0

પ્રશ્ન 7.
xનું મૂલ્ય શોધો :
(i) \(\left|\begin{array}{ll}
2 & 4 \\
5 & 1
\end{array}\right|=\left|\begin{array}{cc}
2 x & 4 \\
6 & x
\end{array}\right|\)
ઉત્તરઃ
\(\left|\begin{array}{ll}
2 & 4 \\
5 & 1
\end{array}\right|=\left|\begin{array}{cc}
2 x & 4 \\
6 & x
\end{array}\right|\)
∴ 2 – 20 = 2x² – 24
∴ 6 = 2x²
∴ x² = 3
∴ x = ±\(\sqrt{3}\)

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

પ્રશ્ન 8 માં વિધાન સાચું બને તે રીતે આપેલ વિોમાંથી યોગ્ય વિકલ્પ પસંદ કરો :

પ્રશ્ન 8.
જો \(\left|\begin{array}{cc}
x & 2 \\
18 & x
\end{array}\right|=\left|\begin{array}{cc}
6 & 2 \\
18 & 6
\end{array}\right|\) હોય, તો x = …………………
(A) 6
(B) ±6
(C) -6
(D) 0
ઉત્તરઃ
\(\left|\begin{array}{cc}
x & 2 \\
18 & x
\end{array}\right|=\left|\begin{array}{cc}
6 & 2 \\
18 & 6
\end{array}\right|\)
∴ x² – 36 = 36 – 36
∴ x² = 36
x = ±6