11th And 12th > Mathematics
DETERMINANTS MCQs
Determinants
Total Questions : 60
| Page 1 of 6 pages
Answer: Option A. ->
∣∣
∣
∣∣fghf'g'h'(x3f'')'(x3h'')'(x3h'')'∣∣
∣
∣∣
:
A
:
A
∣∣
∣
∣∣fghxf'+fxg'+gxh'+h4xf'+2f+x2f''4xg'+2g+x2g''4xh'+2h+x2h''∣∣
∣
∣∣Operating R2→R1−R1; R3→R3−4R2+2R1and shifting x of R2 to R3△⎛⎜
⎜⎝x⎞⎟
⎟⎠=∣∣
∣
∣∣fghf'g'h'x3f''x3g''x3h''∣∣
∣
∣∣⇒△'⎛⎜
⎜⎝x⎞⎟
⎟⎠=0+0+∣∣
∣
∣∣fghf'g'h'(x3f'')'(x3g'')'(x3h'')'∣∣
∣
∣∣
Answer: Option D. ->
ω
:
D
Since Δ = ω2−2ω2 = −ω2. Therefore Δ2 = ω4 = ω.
:
D
Since Δ = ω2−2ω2 = −ω2. Therefore Δ2 = ω4 = ω.
Answer: Option B. ->
bd
:
B
Δ2Δ1 = ∣∣∣10cd∣∣∣ ∣∣∣10ab∣∣∣ = ∣∣∣10c+adbd∣∣∣ = bd.
:
B
Δ2Δ1 = ∣∣∣10cd∣∣∣ ∣∣∣10ab∣∣∣ = ∣∣∣10c+adbd∣∣∣ = bd.
Answer: Option A. ->
a1Δ
:
A
B2 = ∣∣∣a1c1a3c3∣∣∣ = a1c3 − c1a3C2 = −∣∣∣a1b1a3b3∣∣∣ = −(a1b3 − a3b1)B3 = −∣∣∣a1c1a2c2∣∣∣ = −(a1c2 − a2c1)C3 = ∣∣∣a1b1a2b2∣∣∣ = (a1b2 − a2b1)∣∣∣B2C2B3C3∣∣∣ = ∣∣∣a1c3 − a3c1−(a1b3 − a3b1)−(a1c2 − a2c1)a1b2 − a2b1∣∣∣ =∣∣∣a1c3−a1b3−a1c2a1b2∣∣∣+∣∣∣a1c3a3b1−a1c2−a2b1∣∣∣+∣∣∣−a3c1−a1b3a2c1a1b2∣∣∣+∣∣∣−a3c1a3b1a2c1−a2b1∣∣∣=a12(b2c3−b3c2)+a1b1(−c3a2+a3c2)+a1c1(−a3b2+a2b3)+c1b1(a3a2−a2a3) = a1Δ .
:
A
B2 = ∣∣∣a1c1a3c3∣∣∣ = a1c3 − c1a3C2 = −∣∣∣a1b1a3b3∣∣∣ = −(a1b3 − a3b1)B3 = −∣∣∣a1c1a2c2∣∣∣ = −(a1c2 − a2c1)C3 = ∣∣∣a1b1a2b2∣∣∣ = (a1b2 − a2b1)∣∣∣B2C2B3C3∣∣∣ = ∣∣∣a1c3 − a3c1−(a1b3 − a3b1)−(a1c2 − a2c1)a1b2 − a2b1∣∣∣ =∣∣∣a1c3−a1b3−a1c2a1b2∣∣∣+∣∣∣a1c3a3b1−a1c2−a2b1∣∣∣+∣∣∣−a3c1−a1b3a2c1a1b2∣∣∣+∣∣∣−a3c1a3b1a2c1−a2b1∣∣∣=a12(b2c3−b3c2)+a1b1(−c3a2+a3c2)+a1c1(−a3b2+a2b3)+c1b1(a3a2−a2a3) = a1Δ .
Answer: Option C. ->
Are in H. P.
:
C
∣∣
∣∣12aa13bb14cc∣∣
∣∣ = 0, [C2 → C2−2C3]⇒ ∣∣
∣∣10a1bb12cc∣∣
∣∣ = 0, [R3 → R3−R2, R2 → R2−R1]⇒ ∣∣
∣∣10a0bb−a02c−bc−b∣∣
∣∣ = 0 ; b(c−b)−(b−a)(2c−b) = 0
On simplification, 2b = 1a+1c
∴ a, b, c are in Harmonic progression.
:
C
∣∣
∣∣12aa13bb14cc∣∣
∣∣ = 0, [C2 → C2−2C3]⇒ ∣∣
∣∣10a1bb12cc∣∣
∣∣ = 0, [R3 → R3−R2, R2 → R2−R1]⇒ ∣∣
∣∣10a0bb−a02c−bc−b∣∣
∣∣ = 0 ; b(c−b)−(b−a)(2c−b) = 0
On simplification, 2b = 1a+1c
∴ a, b, c are in Harmonic progression.
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 A. ->
A unique solution
:
A
Given system of equation can be written as ⎡⎢⎣1113−12311⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦ = ⎡⎢⎣26−18⎤⎥⎦
On solving the above system we get the unique solution x = -10, y = -4, z = 16.
:
A
Given system of equation can be written as ⎡⎢⎣1113−12311⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦ = ⎡⎢⎣26−18⎤⎥⎦
On solving the above system we get the unique solution x = -10, y = -4, z = 16.