Free Determinants 03 Practice Test - 12th Grade - Commerce

Question 1

SOLUTION

Solution : D

Question 2

The cofactor of the element '4' in the determinant $∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 3 & 5 & 1 2 & 3 & 4 & 2 8 & 0 & 1 & 1 0 & 2 & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣$ is

A. 4

B. -10

C. 10

D. -4

SOLUTION

Solution : 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 3

$x + k y - z = 0, 3 x - k y - z = 0$ and $x - 3 y + z = 0$ has non-zero solution for k =

A. -1

B. 0

C. 1

D. 2

SOLUTION

Solution : C

It has a non-zero solution if
$∣ ∣ ∣ ∣ \begin{matrix} 1 & k & - 1 3 & - k & - 1 1 & - 3 & 1 \end{matrix} ∣ ∣ ∣ ∣ = 0 \Rightarrow - 6 k + 6 = 0 \Rightarrow k = 1 .$

Question 4

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

A. 0

B. 1

C. 2

D. Infinite

SOLUTION

Solution : 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 5

If $Δ (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{n} & s i n x & c o s x n! & s i n \frac{n π}{2} & c o s \frac{n π}{2} a & a^{2} & a^{3} \end{matrix} ∣ ∣ ∣ ∣ ∣$ , then the value of $\frac{d^{n}}{d x^{n}} [Δ (x)] a t x = 0$ is

A. -1

B. 0

C. 1

D. Dependent of a

SOLUTION

Solution : B

$\frac{d^{n}}{d x^{n}} [Δ (x)] = ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{d^{n}}{d x^{n}} x^{n} & \frac{d^{n}}{d x^{n}} s i n x & \frac{d^{n}}{d x^{n}} c o s x n! & s i n (\frac{n π}{2}) & c o s (\frac{n π}{2}) a & a^{2} & a^{3} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} n! & s i n (x + \frac{n π}{2}) & c o s (x + \frac{n π}{2}) n! & s i n (\frac{n π}{2}) & c o s (\frac{n π}{2}) a & a^{2} & a^{3} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ \Rightarrow [Δ^{n} (x)]_{x = 0} = ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} n! & s i n (0 + \frac{n π}{2}) & c o s (0 + \frac{n π}{2}) n! & s i n (\frac{n π}{2}) & c o s (\frac{n π}{2}) a & a^{2} & a^{3} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ = 0 {S i n c e R_{1} \equiv R_{2}} .$

Question 6

The value of the determinant $∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} l o g_{a} (\frac{x}{y}) & l o g_{a} (\frac{y}{z}) & l o g_{a} (\frac{z}{x}) l o g_{b} (\frac{y}{z}) & l o g_{b} (\frac{z}{x}) & l o g_{b} (\frac{x}{y}) l o g_{c} (\frac{z}{x}) & l o g_{c} (\frac{x}{y}) & l o g_{c} (\frac{y}{z}) \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣$ is

A. 1

B. -1

l o g_{c} x y z

D. None of these

SOLUTION

Solution : 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 7

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

A. 1

B. 0

C. 200

D. -200

SOLUTION

Solution : 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 8

If a, b, c are sides of a triangle and $∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} & b^{2} & c^{2} (a + 1)^{2} & (b + 1)^{2} & (c + 1)^{2} (a - 1)^{2} & (b - 1)^{2} & (c - 1)^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0$ , then

Δ A B C

s an equilateral triangle

Δ A B C

is right angled isosceles triangle

Δ A B C

is an isosceles triangle

Δ A B C

None of the above

SOLUTION

Solution : C

$Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} & b^{2} & c^{2} (a + 1)^{2} & (b + 1)^{2} & (c + 1)^{2} (a - 1)^{2} & (b - 1)^{2} & (c - 1)^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$
Applying $R_{2} \to R_{2} - R_{3}$
$= 4 ∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} & b^{2} & c^{2} a & b & c (a - 1)^{2} & (b - 1)^{2} & (c - 1)^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$
Applying $R_{3} \to R_{3} - R_{1} + 2 R_{2}$
$∴ Δ = 4 ∣ ∣ ∣ ∣ \begin{matrix} a^{2} & b^{2} & c^{2} a & b & c 1 & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ = - 4 (a - b) (b - c) (c - a) = 0$
If a – b = 0 or b – c = 0 or c – a = 0
$∴$ a = b or b = c or c = a
$∴ Δ A B C$ is an isosceles triangle.

Question 9

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

A. -2abc

B. abc

C. 0

a^{2} + b^{2} + c^{2}

SOLUTION

Solution : C

$∣ ∣ ∣ ∣ \begin{matrix} 0 & a & - b - a & 0 & c b & - c & 0 \end{matrix} ∣ ∣ ∣ ∣ = 0$ (Since value of determinant of skew-symmetric matrix of odd orders is 0).

Question 10

If a,b and care non zero numbers, then $Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} b^{2} c^{2} & b c & b + c c^{2} a^{2} & c a & c + a a^{2} b^{2} & a b & a + b \end{matrix} ∣ ∣ ∣ ∣ ∣$ is equal to

A. abc

a^{2} b^{2} c^{2}

C. ab+bc+ca

D. None of these

SOLUTION

Solution : D

Multiplying $R_{1}$ by $a, R_{2}$ by b and $R_{3}$ by c, we have
$Δ = \frac{1}{a b c} ∣ ∣ ∣ ∣ ∣ \begin{matrix} a b^{2} c^{2} & a b c & a b + a c a^{2} b c^{2} & a b c & b c + a b a^{2} b^{2} c & a b c & a c + b c \end{matrix} ∣ ∣ ∣ ∣ ∣ = \frac{a^{2} b^{2} c^{2}}{a b c} ∣ ∣ ∣ ∣ \begin{matrix} b c & 1 & a b + a c a c & 1 & b c + a b a b & 1 & a c + b c \end{matrix} ∣ ∣ ∣ ∣ = a b c ∣ ∣ ∣ ∣ ∣ \begin{matrix} b c & 1 & \sum a b a c & 1 & \sum a b a b & 1 & \sum a b \end{matrix} ∣ ∣ ∣ ∣ ∣ {b y C_{3} \to C_{3} + C_{1}} = a b c . \sum a b ∣ ∣ ∣ ∣ \begin{matrix} b c & 1 & 1 c a & 1 & 1 a b & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ = 0, [S i n c e C_{2} \equiv C_{3}]$ .
Trick : Put a=1, b=2, c=3 and check it.

Free Determinants 03 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

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