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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

GSEB Solutions Class 12 Maths Chapter 4 નિશ્ચાયક Ex 4.1

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

GSEB Solutions Class 12 Maths Chapter 4 નિશ્ચાયક Ex 4.1

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

GSEB Solutions Class 12 Maths Chapter 4 નિશ્ચાયક Ex 4.1

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

GSEB Solutions Class 12 Maths Chapter 4 નિશ્ચાયક Ex 4.1

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

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