Properties of Determinants MCQ Quiz in मराठी - Objective Question with Answer for Properties of Determinants - मोफत PDF डाउनलोड करा
Last updated on Mar 9, 2025
Latest Properties of Determinants MCQ Objective Questions
Properties of Determinants Question 1:
जर ω हे एकतेचे घनमूळ असेल, तर समीकरणाचे मूळ काय आहे?\(\left| {\begin{array}{*{20}{c}} {x + 1}&\omega &{{\omega ^2}}\\ \omega &{x + {\omega ^2}}&1\\ {{\omega ^2}}&1&{x + \omega } \end{array}} \right| = 0\)
Answer (Detailed Solution Below)
Properties of Determinants Question 1 Detailed Solution
संकल्पना:
जर ω हे एकतेचे घनमूळ असेल, म्हणजेच ω3 = 1.
तर 1 + ω + ω2 = 0.
ω4 = ω3ω = ω [∵ ω3 =1]
गणना:
दिलेले आहे:
\(\left| {\begin{array}{*{20}{c}} {x + 1}&ω &{{ω ^2}}\\ ω &{x + {ω ^2}}&1\\ {{ω ^2}}&1&{x + ω } \end{array}} \right| = 0\)
C'1 = C1 + C2
\(\left| {\begin{array}{*{20}{c}} {{x+1}+{ω}+{ω^2}}&ω &{{ω ^2}}\\ {{x+1}+{ω}+{ω^2}} &{x + {ω ^2}}&1\\ {{{x+1}+{ω}+{ω^2}}}&1&{x + ω } \end{array}} \right| = 0\)
\(\left| {\begin{array}{*{20}{c}} {x}&ω &{{ω ^2}}\\ {x} &{x + {ω ^2}}&1\\ {{x}}&1&{x + ω } \end{array}} \right| = 0\;\;\;(∵\;1+ω+ω^2=0)\)
R'2 = R2 - R1 आणि R'3 = R3 - R1
\(\left| {\begin{array}{*{20}{c}} {x}&ω &{{ω ^2}}\\ {0} &{x + {ω ^2}-ω}&{1-ω^2}\\ {{0}}&1-ω&{x + ω -ω^2 } \end{array}} \right| = 0\)
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.
Top Properties of Determinants MCQ Objective Questions
जर ω हे एकतेचे घनमूळ असेल, तर समीकरणाचे मूळ काय आहे?\(\left| {\begin{array}{*{20}{c}} {x + 1}&\omega &{{\omega ^2}}\\ \omega &{x + {\omega ^2}}&1\\ {{\omega ^2}}&1&{x + \omega } \end{array}} \right| = 0\)
Answer (Detailed Solution Below)
Properties of Determinants Question 2 Detailed Solution
Download Solution PDFसंकल्पना:
जर ω हे एकतेचे घनमूळ असेल, म्हणजेच ω3 = 1.
तर 1 + ω + ω2 = 0.
ω4 = ω3ω = ω [∵ ω3 =1]
गणना:
दिलेले आहे:
\(\left| {\begin{array}{*{20}{c}} {x + 1}&ω &{{ω ^2}}\\ ω &{x + {ω ^2}}&1\\ {{ω ^2}}&1&{x + ω } \end{array}} \right| = 0\)
C'1 = C1 + C2
\(\left| {\begin{array}{*{20}{c}} {{x+1}+{ω}+{ω^2}}&ω &{{ω ^2}}\\ {{x+1}+{ω}+{ω^2}} &{x + {ω ^2}}&1\\ {{{x+1}+{ω}+{ω^2}}}&1&{x + ω } \end{array}} \right| = 0\)
\(\left| {\begin{array}{*{20}{c}} {x}&ω &{{ω ^2}}\\ {x} &{x + {ω ^2}}&1\\ {{x}}&1&{x + ω } \end{array}} \right| = 0\;\;\;(∵\;1+ω+ω^2=0)\)
R'2 = R2 - R1 आणि R'3 = R3 - R1
\(\left| {\begin{array}{*{20}{c}} {x}&ω &{{ω ^2}}\\ {0} &{x + {ω ^2}-ω}&{1-ω^2}\\ {{0}}&1-ω&{x + ω -ω^2 } \end{array}} \right| = 0\)
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.
Properties of Determinants Question 3:
जर ω हे एकतेचे घनमूळ असेल, तर समीकरणाचे मूळ काय आहे?\(\left| {\begin{array}{*{20}{c}} {x + 1}&\omega &{{\omega ^2}}\\ \omega &{x + {\omega ^2}}&1\\ {{\omega ^2}}&1&{x + \omega } \end{array}} \right| = 0\)
Answer (Detailed Solution Below)
Properties of Determinants Question 3 Detailed Solution
संकल्पना:
जर ω हे एकतेचे घनमूळ असेल, म्हणजेच ω3 = 1.
तर 1 + ω + ω2 = 0.
ω4 = ω3ω = ω [∵ ω3 =1]
गणना:
दिलेले आहे:
\(\left| {\begin{array}{*{20}{c}} {x + 1}&ω &{{ω ^2}}\\ ω &{x + {ω ^2}}&1\\ {{ω ^2}}&1&{x + ω } \end{array}} \right| = 0\)
C'1 = C1 + C2
\(\left| {\begin{array}{*{20}{c}} {{x+1}+{ω}+{ω^2}}&ω &{{ω ^2}}\\ {{x+1}+{ω}+{ω^2}} &{x + {ω ^2}}&1\\ {{{x+1}+{ω}+{ω^2}}}&1&{x + ω } \end{array}} \right| = 0\)
\(\left| {\begin{array}{*{20}{c}} {x}&ω &{{ω ^2}}\\ {x} &{x + {ω ^2}}&1\\ {{x}}&1&{x + ω } \end{array}} \right| = 0\;\;\;(∵\;1+ω+ω^2=0)\)
R'2 = R2 - R1 आणि R'3 = R3 - R1
\(\left| {\begin{array}{*{20}{c}} {x}&ω &{{ω ^2}}\\ {0} &{x + {ω ^2}-ω}&{1-ω^2}\\ {{0}}&1-ω&{x + ω -ω^2 } \end{array}} \right| = 0\)
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.