Determinants Test

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Determinants Test - 63

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Weekly Quiz Competition

Question 1
1 / -0

If $$a_{1}, a_{2}, ...., a_{n}$$, ..... are in G.P. then $$\begin{vmatrix}\log a_{n} & \log a_{n + 1} & \log a_{n + 2}\\ \log a_{n + 3} & \log a_{n + 4} & \log a_{n + 5}\\ \log a_{n + 6} & \log a_{n + 7} & \log a_{n + 8}\end{vmatrix}$$ is

A
$$0$$

B
$$1$$

C
$$-1$$

D
None of these

Solution

We know that $$\dfrac{a_2}{a_1}=\dfrac{a_3}{a_2}=.......\dfrac{a_{n}}{a_{n-1}}=r$$
which means that $$a_n,a_{n+1}$$ and $$a_{n+2}$$ are in G.P
So,
$$a^{2}_{n+1}=a_n \times a_{n+2}$$
Taking log on both the sides, we get
$$2 log\ {a_{n+1}}=log\ {a_{n}}+log{a_{n+2}}$$
$$2 log\ {a_{n+1}}-log\ {a_{n}}-log\ {a_{n+2}}=0$$ --------(1)
Similarly,
$$2 log\ {a_{n+4}}-log\ {a_{n+3}}-log\ {a_{n-5}}=0$$ -------(2)
$$2 log\ {a_{n+7}}-log\ {a_{n+6}}-log\ {a_{n+8}}=0$$ ---------(3)
Apply the operation on column, we get
$$C_{1}\rightarrow 2C_2-C_1-C_3$$
$$det(A)=0$$
Question 2
1 / -0

The coefficient of $${x}^{2}$$ in the expansion of the determinant
$$\begin{vmatrix} { x }^{ 2 } & { x }^{ 3 }+1 & { x }^{ 5 }+2 \\ { x }^{ 3 }+3 & { x }^{ 2 }+x & { x }^{ 3 }+{ x }^{ 4 } \\ x+4 & { x }^{ 3 }+{ x }^{ 4 } & { 2 }^{ 3 } \end{vmatrix}$$ is

A
$$-10$$

B
$$-8$$

C
$$-2$$

D
$$-6$$

E
$$8$$

Solution

For the coefficient of $${x}^{2}$$ on expanding along $$R$$ we get
$$\Delta ={ x }^{ 2 }\left[ 8{ x }^{ 2 }+8x-{ x }^{ 6 }-2{ x }^{ 7 }-{ x }^{ 8 } \right]-\left( { x }^{ 2 }+1 \right)$$
$$\left[ 8{ x }^{ 3 }+24-{ x }^{ 4 }-{ x }^{ 6 }-4{ x }^{ 3 }-4{ x }^{ 4 } \right] +\left( { x }^{ 5 }+2 \right)$$
$$\left[ { x }^{ 6 }+{ x }^{ 7 }+3{ { x }^{ 3 }-3{ x }^{ 4 }-{ x }^{ 3 }-x }^{ 2 }-4{ x }^{ 2 }-4x \right] $$
$$= 8x^4+8x^3-x^5-2x^9-x^{10}-8x^6-24x^3+x^7+x^6+4x^6+4x^7-8x^3-24$$
$$-x^4+x^5+4x^3+4x^4+x^{11}+x^{12}+3x^8+3x^9-x^8-x^7-4x^7-4x^7-4x^6+2x^6$$
$$+2x^7+6x^3+6x^4-2x^3-2x^2-8x^2-8x$$
Coefficient of $${x}^{2}=-2-8=-10$$
Question 3
1 / -0

The value of $$x$$ satisfying the equation $$\begin{vmatrix}\cos 2x & \sin 2x & \sin 2x\\ \sin 2x & \cos 2x & \sin 2x\\ \sin 2x & \sin 2x & \cos 2x\end{vmatrix} = 0$$ and $$x\epsilon \left [0, \dfrac {\pi}{4}\right ]$$ is

A
$$\dfrac {\pi}{4}$$

B
$$\dfrac {\pi}{2}$$

C
$$\dfrac {\pi}{16}$$

D
$$\dfrac {\pi}{3}$$

E
$$\dfrac {\pi}{8}$$

Solution

Given, $$\begin{vmatrix}\cos 2x & \sin 2x & \sin 2x\\ \sin 2x & \cos 2x & \sin 2x\\ \sin 2x & \sin 2x & \cos 2x\end{vmatrix} = 0; x\epsilon \left [0, \dfrac {\pi}{4}\right ]$$
Applying operation: $$R_{1}\rightarrow R_{1} + R_{2} + R_{3}$$, we get
$$\begin{vmatrix}\cos 2x + 2\sin 2x & 2 \sin 2x + \cos 2x & 2 \sin 2x + \cos 2x \\ \sin 2x & \cos 2x & \sin 2x\\ \sin 2x & \sin 2x & \cos 2x\end{vmatrix} = 0$$
$$\Rightarrow (2\sin 2x + \cos 2x) \begin{vmatrix}1 & 1 & 1\\ \sin 2x & \cos 2x & \sin 2x\\ \sin 2x & \sin 2x & \cos 2x\end{vmatrix} = 0$$
Now, apply operation
$$C_{2}\rightarrow C_{2} - C_{1}$$ and $$C_{3} \rightarrow C_{3} - C_{1}$$, we get
$$(2\sin 2x + \cos 2x)$$
$$\begin{vmatrix}1 & 0 & 0\\ \sin 2x & \cos 2x - \sin 2x & 0\\ \sin 2x & 0 & \cos 2x - \sin 2x\end{vmatrix} = 0$$
Expanding along $$R_{1}$$, we get
$$(2\sin 2x + \cos 2x) (\cos 2x - \sin 2x)^{2} = 0$$
$$\Rightarrow 2\sin 2x + \cos 2x = 0$$
or $$\cos 2x - \sin 2x = 0$$
$$\Rightarrow \tan 2x = -\dfrac {1}{2}$$
or $$\tan 2x = 1 = \tan \dfrac {\pi}{4}$$
$$\because x\neq \dfrac {1}{2} \tan^{-1}\left (\dfrac {-1}{2}\right )$$
$$\therefore x = \dfrac {\pi}{8}\epsilon \left [0, \dfrac {\pi}{4}\right ]$$.
Question 4
1 / -0

If the determinant $$\begin{vmatrix} a+p & 1+x & u+f \\ b+q & m+y & v+g \\ c+r & n+z & w+h \end{vmatrix}$$ splits into exactly K determinants of order 3, each element of which contains only one term, then the value of K is

A
9

B
8

C
24

D
12
Question 5
1 / -0

Points (a, 0), (0, b) and (1, 1)are collinear, if:

A
$$\displaystyle \frac{1}{a} + \frac{1}{b} = 1$$

B
$$\displaystyle \frac{1}{a} - \frac{1}{b} = 1$$

C
$$\displaystyle a+b = 1$$

D
$$\displaystyle a-b = ab$$

Solution

$$\textbf{Step 1: Calculating}$$
$$\text{The given points are }(a,0),(0,b),(1,1)$$
$$\implies \text{Slope of line connecting }(a,0)\text{ and }(0,b)=\text{Slope of line connecting }(0,b) \text{ and }(1,1)$$
$$\textbf{Step 2: Calculating}$$
$$\text{Slope }=\dfrac{y_2-y_1}{x_2-x_1}$$
$$\text{Consider the points }(a,0)\text{ and }(0,b)$$
$$\implies \text{Slope}=\dfrac{0-b}{a-0}=\dfrac{-b}{a}$$
$$\text{Consider the points }(0,b)\text{ and }(1,1)$$
$$\implies \text{Slope}=\dfrac{b-1}{0-1}=-(b-1)$$
$$\implies \dfrac{-b}{a}=-(b-1)$$
$$\implies ab-a=b$$
$$\implies a+b=ab$$
$$\implies \dfrac1a+\dfrac1b=1$$
$$\textbf{Hence, The expression }\mathbf{\dfrac1a+\dfrac1b=1 \text{ is true for the given conditions}} $$
Question 6
1 / -0

If planes $$x-cy-bz=0, cx-y+az=0$$ and $$bx+ay-z=0$$ pass through a straight line then $${ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }=$$

A
$$1-abc$$

B
$$abc-1$$

C
$$1-2abc$$

D
$$2abc-1$$

Solution

Since, the given planes are concurrent, hence:-

$$\begin{vmatrix} 1 &-c & -b \\ c & -1 & a \\ b & a & -1 \end{vmatrix}=0$$

$$\implies 1(1-a^2)+c(-c-ab)-b(ac+b)=0$$

$$\implies1-a^2-c^2-abc-abc-b^2=0$$

$$\implies a^2+b^2+c^2=1-2abc$$

Hence, answer is option-(C).
Question 7
1 / -0

The value of the determinant $$\begin{vmatrix} b^2-ab & b-c & bc-ac \\ ab -a^2 & a-b & b^2-ab \\ bc-ac & c-a & ab -a^2 \end{vmatrix}$$ =

A
abc

B
a+b+c

C
0

D
ab+bc+ca
Question 8
1 / -0

If $$B$$ is a square matrix of order 4 such that $$ |B|= 24$$ ,then the value of $$|adj B|$$ is equal to

A
$$24$$

B
$${24}^{2}$$

C
$${24}^{3}$$

D
$${24}^{4}$$

Solution
Question 9
1 / -0

If $$A$$ is $$3*3$$ show symmetric matrix then $$|A| = ?: - $$

A
$$0$$

B
$$1$$

C
$$-1$$

D
N.O.T.

Solution
Question 10
1 / -0

$$\begin{vmatrix} a^2
+2a & 2 a + 1 & 1 \\ 2a+1 & a+2 & 1 \\ 3 & 3 & 1
\end{vmatrix} =$$

A
$$(a-1)^2$$

B
$$(a-1)^3$$

C
$$(a-1)^4$$

D
$$2(a-1)$$

Solution

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Re-Attempt Weekly Quiz Competition

Determinants Test - 63

Result

Determinants Test - 63

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SHARING IS CARING

Question 1

$$0$$

$$1$$

$$-1$$

None of these

Solution

Question 2

$$-10$$

$$-8$$

$$-2$$

$$-6$$

$$8$$

Solution

Question 3

$$\dfrac {\pi}{4}$$

$$\dfrac {\pi}{2}$$

$$\dfrac {\pi}{16}$$

$$\dfrac {\pi}{3}$$

$$\dfrac {\pi}{8}$$

Solution

Question 4

9

8

24

12

Question 5

$$\displaystyle \frac{1}{a} + \frac{1}{b} = 1$$

$$\displaystyle \frac{1}{a} - \frac{1}{b} = 1$$

$$\displaystyle a+b = 1$$

$$\displaystyle a-b = ab$$

Solution

Question 6

$$1-abc$$

$$abc-1$$

$$1-2abc$$

$$2abc-1$$

Solution

Question 7

abc

a+b+c

0

ab+bc+ca

Question 8

$$24$$

$${24}^{2}$$

$${24}^{3}$$

$${24}^{4}$$

Solution

Question 9

$$0$$

$$1$$

$$-1$$

N.O.T.

Solution

Question 10

$$(a-1)^2$$

$$(a-1)^3$$

$$(a-1)^4$$

$$2(a-1)$$

Solution

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