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

12th Grade > Mathematics

DETERMINANTS MCQs

Total Questions : 60 | Page 5 of 6 pages

Question 41. The system of equations

x + y + z = 2, 3 x - y + 2 z = 6

and

3 x + y + z = - 18

has

A unique solution
No solutions
An infinite number of solutions
Zero solution as the only solution

Discuss Question

Answer: Option A. -> A unique solution
:
A
Given system of equation can be written as

⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 3 & - 1 & 2 3 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} x y z \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 26 - 18 \end{matrix} ⎤ ⎥ ⎦

On solving the above system we get the unique solution x = -10, y = -4, z = 16.

Question 42. The value of the determinant

∣ ∣∣∣∣ ∣ \begin{matrix} l o g_{a} (\frac{x}{y}) & l o g_{a} (\frac{y}{z}) & l o g_{a} (\frac{z}{x}) l o g_{b} (\frac{y}{z}) & l o g_{b} (\frac{z}{x}) & l o g_{b} (\frac{x}{y}) l o g_{c} (\frac{z}{x}) & l o g_{c} (\frac{x}{y}) & l o g_{c} (\frac{y}{z}) \end{matrix} ∣ ∣∣∣∣ ∣

1
-1
logcxyz
None of these

Discuss Question

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

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

Then,

∣ ∣∣∣∣ ∣ \begin{matrix} 0 & l o g_{a} (\frac{y}{z}) & l o g_{a} (\frac{z}{x}) 0 & l o g_{b} (\frac{z}{x}) & l o g_{b} (\frac{x}{y}) 0 & l o g_{c} (\frac{x}{y}) & l o g_{c} (\frac{y}{z}) \end{matrix} ∣ ∣∣∣∣ ∣ = 0

Question 43.

If f, g and h are differentiable functions of x and △ ⎛ ⎜⎜ ⎝ x ⎞ ⎟⎟ ⎠ = ∣ ∣∣∣ ∣ \begin{matrix} f & g & h (x f)' & (x g)' & (x h)' (x^{2} f)'' & (x^{2} g)'' & (x^{2} h)'' \end{matrix} ∣ ∣∣∣ ∣, then △' ⎛ ⎜⎜ ⎝ x ⎞ ⎟⎟ ⎠ is

∣∣ ∣ ∣∣fghf'g'h'(x3f'')'(x3h'')'(x3h'')'∣∣ ∣ ∣∣
∣∣ ∣ ∣∣fghf'g'h'(x3f'')(x3h'')(x3h'')∣∣ ∣ ∣∣
∣∣ ∣ ∣∣fghf'g'h'(x3f'')''(x3h'')''(x3h'')''∣∣ ∣ ∣∣
0

Discuss Question

Answer: Option A. -> ∣∣ ∣ ∣∣fghf'g'h'(x3f'')'(x3h'')'(x3h'')'∣∣ ∣ ∣∣
:
A

∣ ∣∣∣ ∣ \begin{matrix} f & g & h x f' + f & x g' + g & x h' + h 4 x f' + 2 f + x^{2} f'' & 4 x g' + 2 g + x^{2} g'' & 4 x h' + 2 h + x^{2} h'' \end{matrix} ∣ ∣∣∣ ∣ Operating R_{2} \to R_{1} - R_{1}; R_{3} \to R_{3} - 4 R_{2} + 2 R_{1} and shifting x of R_{2} to R_{3} △ ⎛ ⎜⎜ ⎝ x ⎞ ⎟⎟ ⎠ = ∣ ∣∣∣ ∣ \begin{matrix} f & g & h f' & g' & h' x^{3} f'' & x^{3} g'' & x^{3} h'' \end{matrix} ∣ ∣∣∣ ∣ \Rightarrow △' ⎛ ⎜⎜ ⎝ x ⎞ ⎟⎟ ⎠ = 0 + 0 + ∣ ∣∣∣ ∣ \begin{matrix} f & g & h f' & g' & h' (x^{3} f'')' & (x^{3} g'')' & (x^{3} h'')' \end{matrix} ∣ ∣∣∣ ∣

Question 44. If

A_{1}, B_{1}, C_{1} . . . .

are respectively the co-factors of the elements

a_{1}, b_{1}, c_{1} . . . .

of the determinant

Δ = ∣ ∣∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣∣ ∣,

then

∣ ∣ ∣ \begin{matrix} B_{2} & C_{2} B_{3} & C_{3} \end{matrix} ∣ ∣ ∣ =

a1Δ
a1a3Δ
(a1+b1)Δ
None of these

Discuss Question

Answer: Option A. -> a1Δ
:
A

B_{2} = ∣ ∣ ∣ \begin{matrix} a_{1} & c_{1} a_{3} & c_{3} \end{matrix} ∣ ∣ ∣ = a_{1} c_{3} - c_{1} a_{3} C_{2} = - ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} a_{3} & b_{3} \end{matrix} ∣ ∣ ∣ = - (a_{1} b_{3} - a_{3} b_{1}) B_{3} = - ∣ ∣ ∣ \begin{matrix} a_{1} & c_{1} a_{2} & c_{2} \end{matrix} ∣ ∣ ∣ = - (a_{1} c_{2} - a_{2} c_{1}) C_{3} = ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} a_{2} & b_{2} \end{matrix} ∣ ∣ ∣ = (a_{1} b_{2} - a_{2} b_{1}) ∣ ∣ ∣ \begin{matrix} B_{2} & C_{2} B_{3} & C_{3} \end{matrix} ∣ ∣ ∣ = ∣ ∣ ∣ \begin{matrix} a_{1} c_{3} - a_{3} c_{1} & - (a_{1} b_{3} - a_{3} b_{1}) - (a_{1} c_{2} - a_{2} c_{1}) & a_{1} b_{2} - a_{2} b_{1} \end{matrix} ∣ ∣ ∣ = ∣ ∣ ∣ \begin{matrix} a_{1} c_{3} & - a_{1} b_{3} - a_{1} c_{2} & a_{1} b_{2} \end{matrix} ∣ ∣ ∣ + ∣ ∣ ∣ \begin{matrix} a_{1} c_{3} & a_{3} b_{1} - a_{1} c_{2} & - a_{2} b_{1} \end{matrix} ∣ ∣ ∣ + ∣ ∣ ∣ \begin{matrix} - a_{3} c_{1} & - a_{1} b_{3} a_{2} c_{1} & a_{1} b_{2} \end{matrix} ∣ ∣ ∣ + ∣ ∣ ∣ \begin{matrix} - a_{3} c_{1} & a_{3} b_{1} a_{2} c_{1} & - a_{2} b_{1} \end{matrix} ∣ ∣ ∣ = {a_{1}}^{2} (b_{2} c_{3} - b_{3} c_{2}) + a_{1} b_{1} (- c_{3} a_{2} + a_{3} c_{2}) + a_{1} c_{1} (- a_{3} b_{2} + a_{2} b_{3}) + c_{1} b_{1} (a_{3} a_{2} - a_{2} a_{3}) = a_{1} Δ .

Question 45. If A =

∣ ∣∣ ∣ \begin{matrix} 5 & 6 & 3 - 4 & 3 & 2 - 4 & - 7 & 3 \end{matrix} ∣ ∣∣ ∣,

then cofactors of the elements of

2^{n d}

row are

39, -3, 11
-39, 3, 11
-39, 27, 11
-39, -3, 11

Discuss Question

Answer: Option C. -> -39, 27, 11
:
C

C_{21} = (- 1)^{2 + 1} (18 + 21) = - 39 C_{22} = (- 1)^{2 + 2} (15 + 12) = 27 C_{23} = (- 1)^{2 + 3} (- 35 + 24) = 11 .

Question 46. The system of equations
X+2y+3z=4
2x+3y+4z=5
3x+4y+5z=6 has

Many solutions
No solution
Unique solution
Atmost two solutions

Discuss Question

Answer: Option A. -> Many solutions
:
A

The System Of EquationsX+2y+3z=42x+3y+4z=53x+4y+5z=6 Has...

Question 47. If

ω

is a cube root of unity and

Δ = ∣ ∣ ∣ \begin{matrix} 1 & 2 ω ω & ω^{2} \end{matrix} ∣ ∣ ∣

, then

Δ^{2}

is equal to

−ω
ω2
1
ω

Discuss Question

Answer: Option D. -> ω
:
D
Since

Δ = ω^{2} - 2 ω^{2} = - ω^{2} .

Therefore

Δ^{2} = ω^{4} = ω .

Question 48. If

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

, then f(200) is equal to

1
0
200
-200

Discuss Question

Answer: Option B. -> 0
:
B

f (X) = x (x + 1) ∣ ∣∣ ∣ \begin{matrix} 1 & 1 & 1 2 x & x - 1 & x 3 x (x - 1) & (x - 1) (x - 2) & x (x - 1) \end{matrix} ∣ ∣∣ ∣ = x (x + 1) (x - 1) ∣ ∣∣ ∣ \begin{matrix} 1 & 1 & 1 2 x & x - 1 & x 3 x & x - 2 & x \end{matrix} ∣ ∣∣ ∣

Applying

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

and

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

then

f (x) = x (x + 1) (x - 1) ∣ ∣∣ ∣ \begin{matrix} 1 & 0 & 0 2 x & - x - 1 & - x 3 x & - 2 x - 2 & - 2 x \end{matrix} ∣ ∣∣ ∣ = x^{2} (x + 1)^{2} (x - 1) ∣ ∣∣ ∣ \begin{matrix} 1 & 0 & 0 2 x & - 1 & - 1 3 x & - 2 & - 2 \end{matrix} ∣ ∣∣ ∣ = 0 ∴ f (200) = 0

Question 49. If sin 2x = 1, then

{∣ ∣∣ ∣ \begin{matrix} 0 & c o s x & - s i n x s i n x & 0 & c o s x c o s x & s i n x & 0 \end{matrix} ∣ ∣∣ ∣}^{2}

equals

3
0
1
None of these

Discuss Question

Answer: Option B. -> 0
:
B

∴ s i n 2 x = 1 t h e n X = \frac{π}{4}

Then,

{∣ ∣∣ ∣ \begin{matrix} 0 & c o s x & - s i n x s i n x & 0 & c o s x c o s x & s i n x & 0 \end{matrix} ∣ ∣∣ ∣}^{2} = {∣ ∣∣∣∣∣ ∣ \begin{matrix} 0 & \frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \end{matrix} ∣ ∣∣∣∣∣ ∣}^{2}

= (\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}) {∣ ∣∣ ∣ \begin{matrix} 0 & 1 & - 1 1 & 0 & 1 1 & 1 & 0 \end{matrix} ∣ ∣∣ ∣}^{2}

= \frac{1}{8} {0 - 1 (0 - 1) - 1 (1)}^{2}

Question 50.

x + k y - z = 0, 3 x - k y - z = 0

and

x - 3 y + z = 0

has non-zero solution for k =

-1
0
1
2

Discuss Question

Answer: Option C. -> 1
:
C
It has a non-zero solution if

∣ ∣∣ ∣ \begin{matrix} 1 & k & - 1 3 & - k & - 1 1 & - 3 & 1 \end{matrix} ∣ ∣∣ ∣ = 0 \Rightarrow - 6 k + 6 = 0 \Rightarrow k = 1 .

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