12th Grade > Mathematics
DETERMINANTS MCQs
Total Questions : 60
| Page 4 of 6 pages
Answer: Option B. -> bd
:
B
Δ2Δ1=∣∣∣10cd∣∣∣∣∣∣10ab∣∣∣=∣∣∣10c+adbd∣∣∣=bd.
:
B
Δ2Δ1=∣∣∣10cd∣∣∣∣∣∣10ab∣∣∣=∣∣∣10c+adbd∣∣∣=bd.
Answer: Option A. -> ∣∣
∣
∣∣fghf'g'h'(x3f'')'(x3h'')'(x3h'')'∣∣
∣
∣∣
:
A
∣∣
∣
∣∣fghxf'+fxg'+gxh'+h4xf'+2f+x2f''4xg'+2g+x2g''4xh'+2h+x2h''∣∣
∣
∣∣OperatingR2→R1−R1;R3→R3−4R2+2R1andshiftingxofR2toR3△⎛⎜
⎜⎝x⎞⎟
⎟⎠=∣∣
∣
∣∣fghf'g'h'x3f''x3g''x3h''∣∣
∣
∣∣⇒△'⎛⎜
⎜⎝x⎞⎟
⎟⎠=0+0+∣∣
∣
∣∣fghf'g'h'(x3f'')'(x3g'')'(x3h'')'∣∣
∣
∣∣
:
A
∣∣
∣
∣∣fghxf'+fxg'+gxh'+h4xf'+2f+x2f''4xg'+2g+x2g''4xh'+2h+x2h''∣∣
∣
∣∣OperatingR2→R1−R1;R3→R3−4R2+2R1andshiftingxofR2toR3△⎛⎜
⎜⎝x⎞⎟
⎟⎠=∣∣
∣
∣∣fghf'g'h'x3f''x3g''x3h''∣∣
∣
∣∣⇒△'⎛⎜
⎜⎝x⎞⎟
⎟⎠=0+0+∣∣
∣
∣∣fghf'g'h'(x3f'')'(x3g'')'(x3h'')'∣∣
∣
∣∣
Answer: Option C. -> -39, 27, 11
:
C
C21=(−1)2+1(18+21)=−39C22=(−1)2+2(15+12)=27C23=(−1)2+3(−35+24)=11.
:
C
C21=(−1)2+1(18+21)=−39C22=(−1)2+2(15+12)=27C23=(−1)2+3(−35+24)=11.
Answer: Option C. -> 0
:
C
∣∣
∣∣0a−b−a0cb−c0∣∣
∣∣=0 (Since value of determinant of skew-symmetric matrix of odd orders is 0).
:
C
∣∣
∣∣0a−b−a0cb−c0∣∣
∣∣=0 (Since value of determinant of skew-symmetric matrix of odd orders is 0).
Answer: Option C. -> ΔABC is an isosceles triangle
:
C
Δ=∣∣
∣
∣∣a2b2c2(a+1)2(b+1)2(c+1)2(a−1)2(b−1)2(c−1)2∣∣
∣
∣∣
Applying R2→R2−R3
=4∣∣
∣
∣∣a2b2c2abc(a−1)2(b−1)2(c−1)2∣∣
∣
∣∣
Applying R3→R3−R1+2R2
∴Δ=4∣∣
∣∣a2b2c2abc111∣∣
∣∣=−4(a−b)(b−c)(c−a)=0
If a – b = 0 or b – c = 0 or c – a = 0
∴a = b or b = c or c = a
∴ΔABC is an isosceles triangle.
:
C
Δ=∣∣
∣
∣∣a2b2c2(a+1)2(b+1)2(c+1)2(a−1)2(b−1)2(c−1)2∣∣
∣
∣∣
Applying R2→R2−R3
=4∣∣
∣
∣∣a2b2c2abc(a−1)2(b−1)2(c−1)2∣∣
∣
∣∣
Applying R3→R3−R1+2R2
∴Δ=4∣∣
∣∣a2b2c2abc111∣∣
∣∣=−4(a−b)(b−c)(c−a)=0
If a – b = 0 or b – c = 0 or c – a = 0
∴a = b or b = c or c = a
∴ΔABC is an isosceles triangle.
Answer: Option D. -> None of these
:
D
Multiplying R1 by a,R2 by b and R3 by c, we have
Δ=1abc∣∣
∣
∣∣ab2c2abcab+aca2bc2abcbc+aba2b2cabcac+bc∣∣
∣
∣∣=a2b2c2abc∣∣
∣∣bc1ab+acac1bc+abab1ac+bc∣∣
∣∣=abc∣∣
∣
∣∣bc1∑abac1∑abab1∑ab∣∣
∣
∣∣{byC3→C3+C1}=abc.∑ab∣∣
∣∣bc11ca11ab11∣∣
∣∣=0,[SinceC2≡C3].
Trick : Put a=1, b=2, c=3 and check it.
:
D
Multiplying R1 by a,R2 by b and R3 by c, we have
Δ=1abc∣∣
∣
∣∣ab2c2abcab+aca2bc2abcbc+aba2b2cabcac+bc∣∣
∣
∣∣=a2b2c2abc∣∣
∣∣bc1ab+acac1bc+abab1ac+bc∣∣
∣∣=abc∣∣
∣
∣∣bc1∑abac1∑abab1∑ab∣∣
∣
∣∣{byC3→C3+C1}=abc.∑ab∣∣
∣∣bc11ca11ab11∣∣
∣∣=0,[SinceC2≡C3].
Trick : Put a=1, b=2, c=3 and check it.
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