Free Determinants 02 Practice Test - 12th Grade - Commerce

Question 1

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

SOLUTION

Solution : A

Question 2

f=3 and g=-5

f=-3 and g=-5

f=-3 and g=-9

f=-5 and g=9

SOLUTION

Solution : D

Question 3

N²

C. Zero

SOLUTION

Solution : C

Question 4

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

$∣ ∣ ∣ ∣ ∣ \begin{matrix} f & g & h f' & g' & h' (x^{3} f'')' & (x^{3} h'')' & (x^{3} h'')' \end{matrix} ∣ ∣ ∣ ∣ ∣$

$∣ ∣ ∣ ∣ ∣ \begin{matrix} f & g & h f' & g' & h' (x^{3} f'') & (x^{3} h'') & (x^{3} h'') \end{matrix} ∣ ∣ ∣ ∣ ∣$

$∣ ∣ ∣ ∣ ∣ \begin{matrix} f & g & h f' & g' & h' (x^{3} f'')'' & (x^{3} h'')'' & (x^{3} h'')'' \end{matrix} ∣ ∣ ∣ ∣ ∣$

0

SOLUTION

Solution : 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 5

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

ω^{2}

C. 1

ω

SOLUTION

Solution : D

Since $Δ = ω^{2} - 2 ω^{2} = - ω^{2} .$ Therefore $Δ^{2} = ω^{4} = ω .$

Question 6

If $Δ_{1} = ∣ ∣ ∣ \begin{matrix} 1 & 0 a & b \end{matrix} ∣ ∣ ∣$ and $Δ_{2} = ∣ ∣ ∣ \begin{matrix} 1 & 0 c & d \end{matrix} ∣ ∣ ∣$ , then $Δ_{2} Δ_{1}$ is equal to

a c

b d

(b - a) (d - c)

a b c

SOLUTION

Solution : B

$Δ_{2} Δ_{1} = ∣ ∣ ∣ \begin{matrix} 1 & 0 c & d \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 a & b \end{matrix} ∣ ∣ ∣ = ∣ ∣ ∣ \begin{matrix} 1 & 0 c + a d & b d \end{matrix} ∣ ∣ ∣ = b d .$

Question 7

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

a_{1} Δ

a_{1} a_{3} Δ

(a_{1} + b_{1}) Δ

D. None of these

SOLUTION

Solution : 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 8

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

A. 39, -3, 11

B. -39, 3, 11

C. -39, 27, 11

D. -39, -3, 11

SOLUTION

Solution : 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 9

If the system of linear equation $x + 2 a y + a z = 0, x + 3 b y + b z = 0, x + 4 c y + c z = 0$ has a non zero solution, then a, b, c

A. Are in A.P.

B. Are in G. P.

C. Are in H. P.

D. Satisfy a +2b + 3c = 0

SOLUTION

Solution : C

$∣ ∣ ∣ ∣ \begin{matrix} 1 & 2 a & a 1 & 3 b & b 1 & 4 c & c \end{matrix} ∣ ∣ ∣ ∣ = 0, [C_{2} \to C_{2} - 2 C_{3}] \Rightarrow ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & a 1 & b & b 1 & 2 c & c \end{matrix} ∣ ∣ ∣ ∣ = 0, [R_{3} \to R_{3} - R_{2}, R_{2} \to R_{2} - R_{1}] \Rightarrow ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & a 0 & b & b - a 0 & 2 c - b & c - b \end{matrix} ∣ ∣ ∣ ∣ = 0; b (c - b) - (b - a) (2 c - b) = 0$
On simplification, $\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$
$∴$ a, b, c are in Harmonic progression.

Question 10

The system of equations $x + y + z = 2, 3 x - y + 2 z = 6$ and $3 x + y + z = - 18$ has

A. A unique solution

B. No solutions

C. An infinite number of solutions

D. Zero solution as the only solution

SOLUTION

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.

Free Determinants 02 Practice Test - 12th Grade - Commerce

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Related ExamsApplication of Derivatives 03Continuity and Differentiability 02Probability 02

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Application of Derivatives 03
Continuity and Differentiability 02
Probability 02