Properties of Determinants MCQ Quiz in मराठी - Objective Question with Answer for Properties of Determinants - मोफत PDF डाउनलोड करा

संकल्पना:

जर ω हे एकतेचे घनमूळ असेल, म्हणजेच ω³ = 1.

तर 1 + ω + ω² = 0.

ω⁴ = ω³ω = ω [∵ ω³ =1]

गणना:

दिलेले आहे:

$\left|\begin{array}{ccc}x+1& \omega & {\omega }^2\\ \omega & x+{\omega }^2& 1\\ {\omega }^2& 1& x+\omega \end{array}\right|=0$

C'₁ = C₁ + C₂

$\left|\begin{array}{ccc}x+1+\omega +{\omega }^2& \omega & {\omega }^2\\ x+1+\omega +{\omega }^2& x+{\omega }^2& 1\\ x+1+\omega +{\omega }^2& 1& x+\omega \end{array}\right|=0$

$\left|\begin{array}{ccc}x& \omega & {\omega }^2\\ x& x+{\omega }^2& 1\\ x& 1& x+\omega \end{array}\right|=0(\because 1+\omega +{\omega }^2=0)$

R'₂ = R₂ - R₁ आणि R'₃ = R₃ - R₁

$\left|\begin{array}{ccc}x& \omega & {\omega }^2\\ 0& x+{\omega }^2-\omega & 1-{\omega }^2\\ 0& 1-\omega & x+\omega -{\omega }^2\end{array}\right|=0$

पहिल्या स्तंभासह विस्तारत आहे:

∴ x[(x + ω² - ω)(x + ω - ω²) - (1 - ω)(1 - ω²)]

∴ x[x² + ωx - ω²x + ω²x + ω³ - ω⁴ - ωx - ω² + ω³ - 1 + ω² + ω - ω³]

∴ x³ = 0      (∵ ω³ = 1 आणि ω⁴ = ω³ω ⇒ ω)

∴ x = 0.

संकल्पना:

जर ω हे एकतेचे घनमूळ असेल, म्हणजेच ω³ = 1.

तर 1 + ω + ω² = 0.

ω⁴ = ω³ω = ω [∵ ω³ =1]

गणना:

दिलेले आहे:

$\left|\begin{array}{ccc}x+1& \omega & {\omega }^2\\ \omega & x+{\omega }^2& 1\\ {\omega }^2& 1& x+\omega \end{array}\right|=0$

C'₁ = C₁ + C₂

$\left|\begin{array}{ccc}x+1+\omega +{\omega }^2& \omega & {\omega }^2\\ x+1+\omega +{\omega }^2& x+{\omega }^2& 1\\ x+1+\omega +{\omega }^2& 1& x+\omega \end{array}\right|=0$

$\left|\begin{array}{ccc}x& \omega & {\omega }^2\\ x& x+{\omega }^2& 1\\ x& 1& x+\omega \end{array}\right|=0(\because 1+\omega +{\omega }^2=0)$

R'₂ = R₂ - R₁ आणि R'₃ = R₃ - R₁

$\left|\begin{array}{ccc}x& \omega & {\omega }^2\\ 0& x+{\omega }^2-\omega & 1-{\omega }^2\\ 0& 1-\omega & x+\omega -{\omega }^2\end{array}\right|=0$

पहिल्या स्तंभासह विस्तारत आहे:

∴ x[(x + ω² - ω)(x + ω - ω²) - (1 - ω)(1 - ω²)]

∴ x[x² + ωx - ω²x + ω²x + ω³ - ω⁴ - ωx - ω² + ω³ - 1 + ω² + ω - ω³]

∴ x³ = 0      (∵ ω³ = 1 आणि ω⁴ = ω³ω ⇒ ω)

∴ x = 0.

संकल्पना:

जर ω हे एकतेचे घनमूळ असेल, म्हणजेच ω³ = 1.

तर 1 + ω + ω² = 0.

ω⁴ = ω³ω = ω [∵ ω³ =1]

गणना:

दिलेले आहे:

$\left|\begin{array}{ccc}x+1& \omega & {\omega }^2\\ \omega & x+{\omega }^2& 1\\ {\omega }^2& 1& x+\omega \end{array}\right|=0$

C'₁ = C₁ + C₂

$\left|\begin{array}{ccc}x+1+\omega +{\omega }^2& \omega & {\omega }^2\\ x+1+\omega +{\omega }^2& x+{\omega }^2& 1\\ x+1+\omega +{\omega }^2& 1& x+\omega \end{array}\right|=0$

$\left|\begin{array}{ccc}x& \omega & {\omega }^2\\ x& x+{\omega }^2& 1\\ x& 1& x+\omega \end{array}\right|=0(\because 1+\omega +{\omega }^2=0)$

R'₂ = R₂ - R₁ आणि R'₃ = R₃ - R₁

$\left|\begin{array}{ccc}x& \omega & {\omega }^2\\ 0& x+{\omega }^2-\omega & 1-{\omega }^2\\ 0& 1-\omega & x+\omega -{\omega }^2\end{array}\right|=0$

पहिल्या स्तंभासह विस्तारत आहे:

∴ x[(x + ω² - ω)(x + ω - ω²) - (1 - ω)(1 - ω²)]

∴ x[x² + ωx - ω²x + ω²x + ω³ - ω⁴ - ωx - ω² + ω³ - 1 + ω² + ω - ω³]

∴ x³ = 0      (∵ ω³ = 1 आणि ω⁴ = ω³ω ⇒ ω)

∴ x = 0.