Determinants Test

Result

Determinants Test - 52

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Let $$\Delta =$$$$\begin{vmatrix} sin\theta cos \phi & sin\theta sin\phi & cos\theta \\ cos\theta cos\phi & cos\theta sin\phi & -sin\theta \\ -sin\theta sin\phi & sin\theta cos\phi & 0\end{vmatrix}$$, then

A
$$\Delta$$ is independent of $$\theta$$

B
$$\Delta$$ is independent of $$\phi$$

C
$$\Delta$$ is a constant

D
none of these

Solution

$$\triangle = \begin{vmatrix} \sin \theta . \cos \phi & \sin \theta . \sin \phi & \cos \theta \\ \cos \theta . \cos \phi & \cos \theta . \sin \theta & -\sin \theta \\ -\sin \theta. \sin \theta & \sin \theta . \cos \phi & 0 \end{vmatrix}$$
$$= \sin \theta . \sin \phi (\sin^{2} \theta . \cos \theta) - \cos \theta (- \sin \theta . \cos \theta . \cos \theta)- \sin \theta . \sin \phi (- \sin^{2} \theta. \sin \phi - \cos^{2} \theta- \sin \theta)$$
$$=\sin^{3} \theta . \sin \phi. \cos \phi+ \cos^{2} \theta. \sin \theta. \cos^{2} \phi+ \sin^{3} \theta. \sin^{2} \phi + \sin \theta . \cos^{2} \theta . \sin^{2} \phi$$
So, We can say that $$\triangle $$ is depend $$d$$ on both.
$$\theta $$ and $$\phi$$
Question 2
1 / -0

$$\begin{vmatrix} 1 & bc & a(b+c) \\ 1 & ca & b(c+a) \\ 1 & ab & c(a+b) \end{vmatrix}=$$

A
$$0$$

B
$$1$$

C
$$abc$$

D
$$ab+bc+ca$$

Solution

$$\begin{vmatrix} 1 & bc & ab+ac \\ 1 & ca & ba+bc \\ 1 & ab & ca+cb \end{vmatrix}$$
$$\begin{vmatrix} 1 & bc & ab+bc+ac \\ 1 & ca & ab+bc+ca \\ 1 & ab & ab+bc+ca \end{vmatrix}$$
$$=(ab+bc+ca)\begin{vmatrix} 1 & bc & 1 \\ 1 & ca & 1 \\ 1 & ab & 1 \end{vmatrix}$$
$$=0$$
Question 3
1 / -0

If the system of equations of $$3x-2y+z=0, kx-14y+15z=0,x+2y+3z=0$$ has non trivial solution then $$k=$$

A
$$29$$

B
$$26$$

C
$$23$$

D
$$19$$

Solution

According to the given information

$$\Rightarrow \begin{vmatrix} 3&-2&1\\k&-14&15\\1&2&3 \end{vmatrix}=0$$

$$\Rightarrow 3(-42-30)+2(3k-15)+1(2k+14)=0$$

$$\Rightarrow -216+6k-30+2k+14=0$$

$$\Rightarrow 8k=232$$

$$\Rightarrow k=29$$
Question 4
1 / -0

$$A=\left[ \begin{matrix} 5 & 5\alpha & \alpha \\ 0 & \alpha & 5\alpha \\ 0 & 0 & 5 \end{matrix} \right] $$; If $$\left| { A }^{ 2 } \right| =25$$, then $$\left| \alpha \right| =$$

A
$$5$$

B
$$5^{2}$$

C
$$1$$

D
$$\dfrac{1}{5}$$

Solution

$$A=\begin{bmatrix} 5 & 5\alpha & \alpha\\ 0 & \alpha & 5\alpha\\ 0 & 0 & 5\end{bmatrix}$$
$$A^2=\begin{bmatrix} 5 & 5\alpha & \alpha\\ 0 & \alpha & 5\alpha\\ 0 & 0 & 5\end{bmatrix}\begin{bmatrix} 5 & 5\alpha & \alpha\\ 0 & \alpha & 5\alpha\\ 0 & 0 & 5\end{bmatrix}=\begin{bmatrix} 25 & 30\alpha & 25\alpha^2+10\alpha\\ 0 & \alpha^2 & 5\alpha^2+25\alpha\\ 0 & \alpha^2 & 5\alpha^2 +25\alpha\\ 0 & 0 & 25\end{bmatrix}$$
$$|A^2|=25$$
$$25\begin{vmatrix} \alpha^2 & 5\alpha^2+25\alpha\\ 0 & 25\end{vmatrix} -30\alpha \begin{vmatrix} 0 & 5\alpha^2+25\alpha\\ 0 & 25\end{vmatrix} +25\alpha^2+10\alpha\begin{vmatrix} 0 & \alpha^2\\ 0 & 0\end{vmatrix}=25$$
$$25\times 25\alpha^2=25$$
$$25\alpha^2=1$$
$$\alpha =\dfrac{1}{5}$$.
Question 5
1 / -0

If $$f(x) = \left|
\begin{array}{111}
x-3 & 2x^2 -18 & 3x^3 -81\\
x-5 & 2x^2-50 & 4x^3-500\\
1 & 2 & 3 \\
\end {array}
\right|
$$ then $$f(1). f(3) + f(3) .f(5) + f(5) .f(1)$$ is equal to-

A
$$f(1)$$

B
$$f(3)$$

C
$$f(1) + f(3)$$

D
$$f(1) + f(5)$$

Solution

$$f(x)=\begin{vmatrix} x-3 & \quad \quad \quad { 2x }^{ 2 }-18 & \quad \quad \quad { 3x }^{ 3 }-81 \\ x-5 & { \quad \quad \quad 2x }^{ 2 }-50 & \quad \quad \quad { 4x }^{ 2 }-500 \end{vmatrix}$$

Now $$f(3)=\begin{vmatrix} 3-3 & \quad \quad { 2x3 }^{ 2 }-18 & \quad \quad { 3x3 }^{ 2 }-81 \\ 3-5 & \quad \quad { 2x3 }^{ 2 }-50 & \quad \ { 3 }^{ 3 }-500 \\ 1 & 2 & 3 \end{vmatrix}$$

$$=0$$
Also , $$f(5)=0$$
Hence, $$f(1), f(3) + f(3), f(3)+f(3)+f(5), f(1)$$
$$=0$$
$$f(3)$$
Question 6
1 / -0

The points $$({X}_{1},{Y}_{1})$$, $$({X}_{2},{Y}_{2})$$, $$({X}_{1},{Y}_{2})$$ and $$({X}_{2},{Y}_{1})$$ are always

A
Collinear

B
Concyclic

C
Vertices of a square

D
Vertices of rectangle

Solution

As all the points lie on the same line this implies that area of the triangle formed by the points is zero.
So,using area formula in determinant form

$${ X }_{ 1 }∗({ Y }_{ 2 }−{ Y }_{ 3 })−{ X }_{ 2 }∗({ Y }_{ 1 }−{ Y }_{ 3 })+{ X }_{ 3 }∗({ Y }_{ 1 }−{ Y }_{ 2 })=0$$
$${ X }_{ 1 }∗({ Y }_{ 2 }−{ Y }_{ 3 })+{ X }_{ 2 }∗({ Y }_{ 3 }−{ Y }_{ 1 })+{ X }_{ 3 }∗({ Y }_{ 1 }−{ Y }_{ 2 })=0$$
Now,divide both sides by $$ { X }_{ 1 }∗{ X }_{ 2 }∗{ X }_{ 3 }$$
$$({ Y }_{ 2 }−{ Y }_{ 3 })+{ X }_{ 2 }∗({ Y }_{ 3 }−{ Y }_{ 1 })+{ X }_{ 3 }∗({ Y }_{ 1 }−{ Y }_{ 2 })=0$$
$$\dfrac { { Y }_{ 2 }−{ Y }_{ 3 } }{ { X }_{ 2 }−{ X }_{ 3 } } +\dfrac { { Y }_{ 3 }−{ Y }_{ 1 } }{ { X }_{ 3 }−{ X }_{ 1 } } +\dfrac { { Y }_{ 1 }−{ Y }_{ 2 } }{ { X }_{ 1 }−{ X }_{ 2 } } =0$$
Its collinear.
Question 7
1 / -0

$$\begin{vmatrix}\dfrac{1}{a} &bc &a^2 \\ \dfrac{1}{b} &ca & b^2\\ \dfrac{1}{c} & ab & c^2\end{vmatrix}$$ is equal to -

A
$$(a - b)(b-c) ( c-a)$$

B
$$abc(a-b)(b-c)(c-a)$$

C
$$0$$

D
1

Solution
Question 8
1 / -0

$$\begin{vmatrix} log e & log e^2 & log e^3\\ log e^2 & log e^3 & log e^4\\ log e^3 & log e^4 & log e^5\end{vmatrix}=$$?

A
$$0$$

B
$$1$$

C
$$4$$ log e

D
$$5$$ log e

Solution

$$\begin{vmatrix} \log { e } & \log { { e }^{ 2 } } & \log { e^{ 3 } } \\ \log { { e }^{ 2 } } & \log { { e }^{ 3 } } & \log { e^{ 4 } } \\ \log { e^{ 3 } } & \log { { e }^{ 4 } } & \log { { e }^{ 3 } } \end{vmatrix}$$

$$\begin{vmatrix} \log { e } & 2\log { e } & 3\log { e } \\ 2\log { e } & 3\log { e } & 4\log { e } \\ 3\log { e } & 4\log { e } & 5\log { e } \end{vmatrix}$$

$${ \left( \log { e } \right) }^{ 3 }\begin{vmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix}\xrightarrow [ { R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 3 } ]{ { R }_{ 1 }\rightarrow { R }_{ 1 }-{ R }_{ 2 } } $$

$${ \left( \log { e } \right) }^{ 3 }\begin{vmatrix} 1- & -1 & -1 \\ -1 & -1 & -1 \\ 3 & 4 & 5 \end{vmatrix}=0$$ ($$\because $$ two rowa are same)
Question 9
1 / -0

If $$ \begin{vmatrix} \lambda^2 + 3 \lambda & \lambda -1 & \lambda +3 \\ \lambda + 1 & 2- \lambda & \lambda - 4 \\ \lambda-3 & \lambda + 4 & 3 \lambda \end{vmatrix} = p \lambda^4 + q \lambda^3 + r \lambda^2 + s \lambda + t $$ then $$ t = $$

A
$$16$$

B
$$17$$

C
$$18$$

D
$$19$$

Solution

To get $$t$$,

Put $$\lambda=0$$

We get, value of determinant as,

$$[0(0+16)+1(0-12)+3(4+6)]=t$$

=> $$t=18$$
Question 10
1 / -0

$$\begin{vmatrix} a + x& a - x & a - x\\ a - x & a + x & a - x\\ a - x & a - x & a + x\end{vmatrix} = 0$$ then $$x$$ is

A
$$0$$

B
$$a$$

C
$$2a$$

D
$$3a$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Determinants Test - 52

Result

Determinants Test - 52

Score

-

Rank

-

SHARING IS CARING

Question 1

$$\Delta$$ is independent of $$\theta$$

$$\Delta$$ is independent of $$\phi$$

$$\Delta$$ is a constant

none of these

Solution

Question 2

$$0$$

$$1$$

$$abc$$

$$ab+bc+ca$$

Solution

Question 3

$$29$$

$$26$$

$$23$$

$$19$$

Solution

Question 4

$$5$$

$$5^{2}$$

$$1$$

$$\dfrac{1}{5}$$

Solution

Question 5

$$f(1)$$

$$f(3)$$

$$f(1) + f(3)$$

$$f(1) + f(5)$$

Solution

Question 6

Collinear

Concyclic

Vertices of a square

Vertices of rectangle

Solution

Question 7

$$(a - b)(b-c) ( c-a)$$

$$abc(a-b)(b-c)(c-a)$$

$$0$$

1

Solution

Question 8

$$0$$

$$1$$

$$4$$ log e

$$5$$ log e

Solution

Question 9

$$16$$

$$17$$

$$18$$

$$19$$

Solution

Question 10

$$0$$

$$a$$

$$2a$$

$$3a$$

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.