Determinants(12th Grade > Mathematics ) Questions and answers for exam Preparation

12th Grade > Mathematics

DETERMINANTS MCQs

Total Questions : 60 | Page 6 of 6 pages

A + B + C = π

, then

∣ ∣∣∣ ∣ \begin{matrix} s i n (A + B + C) & s i n B & c o s C - s i n B & 0 & t a n A c o s (A + B) & - t n a A & 0 \end{matrix} ∣ ∣∣∣ ∣

is equal to

1
0
-1
2

Discuss Question

Answer: Option B. -> 0
:
B

Δ = ∣ ∣∣ ∣ \begin{matrix} s i n π & s i n B & c o s C - s i n B & 0 & t a n A c o s (π - C) & - t n a A & 0 \end{matrix} ∣ ∣∣ ∣ = ∣ ∣∣ ∣ \begin{matrix} 0 & s i n B & c o s C - s i n B & 0 & t a n A - c o s C & - t a n A & 0 \end{matrix} ∣ ∣∣ ∣

=0 (

∵ Δ

is skew symmetric)

Question 52. Let

f (x) = ∣ ∣∣ ∣ \begin{matrix} c o s x & s i n x & c o s x c o x 2 x & s i n 2 x & 2 c o s 2 x c o s 3 x & s i n 3 x & 3 c o s 3 x \end{matrix} ∣ ∣∣ ∣

, then

f^{'} (\frac{π}{2})

is equal to

Discuss Question

Answer: Option C. -> 4
:
C

f^{'} (x) = ∣ ∣∣ ∣ \begin{matrix} - s i n x & s i n x & c o s x - 2 s i n 2 x & s i n 2 x & 2 c o s 2 x - 3 s i n 3 x & s i n 3 x & 3 c o s 3 x \end{matrix} ∣ ∣∣ ∣ + ∣ ∣∣ ∣ \begin{matrix} c o s x & c o s x & c o s x c o s 2 x & 2 c o s 2 x & 2 c o s 2 x c o s 3 x & 3 c o s 3 x & 3 c o s 3 x \end{matrix} ∣ ∣∣ ∣ + ∣ ∣∣ ∣ \begin{matrix} c o s x & s i n x & - s i n x c o s 2 x & s i n 2 x & - 4 s i n 2 x c o s 3 x & s i n 3 x & - 9 s i n 3 x \end{matrix} ∣ ∣∣ ∣

f^{'} (\frac{π}{2}) = ∣ ∣∣ ∣ \begin{matrix} - 1 & 1 & 0 0 & 0 & - 2 3 & - 1 & 0 \end{matrix} ∣ ∣∣ ∣ + 0 + ∣ ∣∣ ∣ \begin{matrix} 0 & 1 & - 1 - 1 & 0 & 0 0 & - 1 & 9 \end{matrix} ∣ ∣∣ ∣ = 2 (1 - 3) + 0 + 1 (9 - 1) = - 4 + 8 = 4

Question 53.

∣ ∣∣ ∣ \begin{matrix} 1 & a & b - a & 1 & c - b & - c & 1 \end{matrix} ∣ ∣∣ ∣ =

1+a2+b2+c2
1−a2+b2+c2
1+a2+b2−c2
1+a2−b2+c2

Discuss Question

Answer: Option A. -> 1+a2+b2+c2
:
A

∣ ∣∣ ∣ \begin{matrix} 1 & a & b - a & 1 & c - b & - c & 1 \end{matrix} ∣ ∣∣ ∣ = 1 (1 + c^{2}) - a (- a + b c) + b (a c + b) = 1 + a^{2} + b^{2} + c^{2}

Question 54. The number of values of k for which the system of equations

(k + 1) x + 8 y = 4 k, k x + (k + 3) y = 3 k - 1

has infinitely many solutions, is

0
1
2
Infinite

Discuss Question

Answer: Option B. -> 1
:
B
For infinitely many solutions, the two equations must be identical

\Rightarrow \frac{k + 1}{k} = \frac{8}{k + 3} = \frac{4 k}{3 k - 1} \Rightarrow (k + 1) (k + 3) = 8 k

and 8(3k-1) = 4k(k+3)

\Rightarrow k^{2} - 4 k + 3 = 0

and

k^{2} - 3 k + 2 = 0 .

By cross multiplication,

\frac{k^{2}}{- 8 + 9} = \frac{k}{3 - 2} = \frac{1}{- 3 + 4}

k^{2} = 1

and k=1 ;

∴

k=1.

Question 55. If

D_{k} = ∣ ∣∣∣ ∣ \begin{matrix} 1 & n & n 2 k & n^{2} + n + 1 & n^{2} + n 2 k - 1 & n^{2} & n^{2} + n + 1 \end{matrix} ∣ ∣∣∣ ∣

and

\sum_{k = 1}^{n} D_{k} = 56

, then n equals

4
6
8
None of these

Discuss Question

Answer: Option D. -> None of these
:
D

∴ \sum_{k = 1}^{n} D_{k} = ∣ ∣∣∣ ∣ \begin{matrix} \sum_{k = 1}^{n} 1 & n & n 2 \sum_{k = 1}^{n} k & n^{2} + n + 1 & n^{2} + n 2 \sum_{k = 1}^{n} k - \sum_{k = 1}^{n} 1 & n^{2} & n^{2} + n + 1 \end{matrix} ∣ ∣∣∣ ∣

= ∣ ∣∣∣ ∣ \begin{matrix} n & n & n n^{2} + n & n^{2} + n + 1 & n^{2} + n n^{2} & n^{2} & n^{2} + n + 1 \end{matrix} ∣ ∣∣∣ ∣ = 56

=Applying

C_{2} \to C_{2} \to C_{1} \to C_{3} \to C_{3} - C_{1}

, then

∣ ∣∣ ∣ \begin{matrix} n & 0 & 0 n^{2} + n & 1 & 0 n & 0 & n + 1 \end{matrix} ∣ ∣∣ ∣ = 56 \Rightarrow n (n + 1) = 56 = 7 \times 8 \Rightarrow n = 7

Question 56. If

Δ_{1} = ∣ ∣∣ ∣ \begin{matrix} 1 & 1 & 1 a & b & c a^{2} & b^{2} & c^{2} \end{matrix} ∣ ∣∣ ∣, Δ_{2} = ∣ ∣∣ ∣ \begin{matrix} 1 & b c & a 1 & c a & b 1 & a b & c \end{matrix} ∣ ∣∣ ∣

, then

Δ1+Δ2=0
Δ1+2Δ2=0
Δ1=Δ2
Δ1=2Δ2

Discuss Question

Answer: Option A. -> Δ1+Δ2=0
:
A

Δ_{1} = ∣ ∣∣ ∣ \begin{matrix} 1 & 1 & 1 a & b & c a^{2} & b^{2} & c^{2} \end{matrix} ∣ ∣∣ ∣,

Δ_{2} = ∣ ∣∣ ∣ \begin{matrix} 1 & b c & a 1 & c a & b 1 & a b & c \end{matrix} ∣ ∣∣ ∣

Δ_{2}

,Transposing the determinant

Δ_{2} = ∣ ∣∣ ∣ \begin{matrix} 1 & 1 & 1 b c & c a & a b a & a & c \end{matrix} ∣ ∣∣ ∣

C_{1} \to a C_{1}, C_{2} \to b C_{2}, C_{3} \to C C_{3}

Δ_{2} = \frac{1}{a b c} ∣ ∣∣ ∣ \begin{matrix} a & b & c a b c & a b c & a b c a^{2} & b^{2} & c^{2} \end{matrix} ∣ ∣∣ ∣

Taking abc common from

R_{2}

Δ_{2} = ∣ ∣∣ ∣ \begin{matrix} a & b & c 1 & 1 & 1 a^{2} & b^{2} & c^{2} \end{matrix} ∣ ∣∣ ∣

R_{1} \leftrightarrow R_{2}

Δ_{2} = - ∣ ∣∣ ∣ \begin{matrix} 1 & 1 & 1 a & b & c a^{2} & b^{2} & c^{2} \end{matrix} ∣ ∣∣ ∣

\Rightarrow Δ_{2} = - Δ_{1} \Rightarrow Δ_{2} + Δ_{1} = 0

Question 57. The cofactor of the element '4' in the determinant

∣ ∣∣∣∣ ∣ \begin{matrix} 1 & 3 & 5 & 1 2 & 3 & 4 & 2 8 & 0 & 1 & 1 0 & 2 & 1 & 1 \end{matrix} ∣ ∣∣∣∣ ∣

4
-10
10
-4

Discuss Question

Answer: Option C. -> 10
:
C
The cofactor of element 4, in the

2^{n d}

row and

3^{r d}

column is

= (- 1)^{2 + 3} ∣ ∣∣ ∣ \begin{matrix} 1 & 3 & 1 8 & 0 & 1 0 & 2 & 1 \end{matrix} ∣ ∣∣ ∣

= - {1(-2)-3(8-0)+1.16}
=10.

Question 58. If a,b,c are non – zero real numbers and if the equations (a-1) x = y + z, (b -1)y = z + x, (c - 1)z = x + y has a non trivial solution, then ab + bc + ca is equal to

a +b + c
abc
1
None of these

Discuss Question

Answer: Option B. -> abc
:
B
For non trivial solution

∣ ∣∣ ∣ \begin{matrix} a - 1 & - 1 & - 1 1 & 1 - b & 1 1 & 1 & 1 - c \end{matrix} ∣ ∣∣ ∣ = 0

Applying

C_{1} \to C_{1} - C_{3}

and

C_{2} \to C_{2} - C_{3},

then

∣ ∣∣ ∣ \begin{matrix} a & 0 & 1 0 & - b & 1 c & c & 1 - c \end{matrix} ∣ ∣∣ ∣ = 0 \Rightarrow a (- b + b c - c) - 0 + c (0 - b) = 0 \Rightarrow a b + b c + c a = a b c

Question 59. If

f (n) = α^{n} + β^{n}

and

∣ ∣∣∣ ∣ \begin{matrix} 3 & 1 + f (1) & 1 + f (2) 1 + f (1) & 1 + f (2) & 1 + f (3) 1 + f (2) & 1 + f (3) & 1 + f (4) \end{matrix} ∣ ∣∣∣ ∣ = k (1 - α)^{2} (1 - β)^{2} (α - β)^{2}

, then k is equal to

1
-1
αβ
αβγ

Discuss Question

Answer: Option A. -> 1
:
A

Δ = ∣ ∣∣∣ ∣ \begin{matrix} 3 & 1 + α + β & 1 + α^{2} + β^{2} 1 + α + β & 1 + α^{2} + β^{2} & 1 + α^{3} + β^{3} 1 + α^{2} + β^{2} & 1 + α^{3} + β^{3} & 1 + α^{4} + β^{4} \end{matrix} ∣ ∣∣∣ ∣ = ∣ ∣∣ ∣ \begin{matrix} 1 & 1 & 1 1 & α & β 1 & α^{2} & β^{2} \end{matrix} ∣ ∣∣ ∣ \times ∣ ∣∣ ∣ \begin{matrix} 1 & 1 & 1 1 & α & β 1 & α^{2} & β^{2} \end{matrix} ∣ ∣∣ ∣

Applying

C_{2} \to - C_{2} - C_{3} \to C_{3} - C_{1}

{∣ ∣∣ ∣ \begin{matrix} 1 & 0 & 0 1 & α - 1 & β - 1 1 & α^{2} - 1 & β^{2} - 1 \end{matrix} ∣ ∣∣ ∣}^{2} = (α - 1)^{2} (β - 1)^{2} (β - α)^{2} = (1 - α)^{2} (1 - β)^{2} (α - β)^{2}

Hence, k=1

Question 60. If

α, β

are non real numbers satisfying

x^{3} - 1 = 0

then the value of

∣ ∣∣ ∣ \begin{matrix} λ + 1 & α & β α & λ + β & 1 β & 1 & λ + α \end{matrix} ∣ ∣∣ ∣

is equal to

0
λ3
λ3+1
λ3−1

Discuss Question

Answer: Option B. -> λ3
:
B

x^{3} - 1 = 0 ∴ x = 1, ω, ω^{2}

Here,

α = ω, β = ω^{2} ∣ ∣∣∣ ∣ \begin{matrix} λ + 1 & ω & ω^{2} ω & λ + ω^{2} & 1 ω^{2} & 1 & λ + ω \end{matrix} ∣ ∣∣∣ ∣

Applying

C_{1} \to C_{1} + C_{2} + C_{3}

, then

∣ ∣∣∣ ∣ \begin{matrix} λ & ω & ω^{2} λ & λ + ω^{2} & 1 λ & 1 & λ + ω \end{matrix} ∣ ∣∣∣ ∣

Applying

R - 2 \to R_{2} - R_{1}

and

R_{3} \to R_{3} - R_{1}

, then we get

∣ ∣∣∣ ∣ \begin{matrix} λ & ω & ω^{2} 0 & λ + ω^{2} - ω & 1 - ω^{2} 0 & 1 - ω & λ + ω - ω^{2} \end{matrix} ∣ ∣∣∣ ∣ = λ ((λ + ω^{2} - ω) (λ + ω - ω^{2}) - (1 - ω) (1 - ω^{2})) = λ (λ^{2}) = λ^{3}

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