Determinants Test

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Determinants Test - 49

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Weekly Quiz Competition

Question 1
1 / -0

If maximum and minimum values of $$D = \begin{vmatrix}1 & -\cos \theta & 1\\ \cos \theta & 1 & -\cos \theta\\ 1 & \cos \theta & 1\end{vmatrix}$$ are $$p$$ and $$q$$ respectively, then the value of $$2p + 3q$$ is _________.

A
$$16$$

B
$$6$$

C
$$14$$

D
$$8$$

Solution

$$D = \begin{vmatrix}1 & -\cos \theta & 1\\ \cos \theta & 1 & -\cos \theta\\ 1 & \cos \theta & 1\end{vmatrix}$$
$$\Rightarrow D = 1(1 + \cos^{2}\theta) + \cos \theta (\cos \theta + \cos \theta) - 1 (\cos^{2}\theta - 1)$$
$$= 1 + \cos^{2}\theta + 2\cos^{2}\theta -\cos^{2} \theta + 1$$
$$= 2 (1 + \cos^{2} \theta)$$
Now, $$-1\le \cos \theta\le 1\Rightarrow 0\le \cos^{2}\theta\le 1$$
$$\therefore p = 4, q = 2$$
$$\therefore 2p + 3q = 14$$.
Question 2
1 / -0

If $$A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$ satisfies the equation $$x^2 - (a + d) x + k = 0$$, then ?

A
$$k = bc$$

B
$$k = ad$$

C
$$k = a^2 + b^2 + c^2 + d^2$$

D
$$ad - bc$$

Solution

If a matrix $$A$$ satisfies a polynomial equation, then the product of the roots of that equation is the determinant of $$A$$.

Here, product of roots of the equation = $$k$$, also $$det(A)=ad-bc$$.
Therefore, $$k=ad-bc$$
Question 3
1 / -0

For a positive numbers $$x, y$$ and $$z$$ the numerical value of the determinant $$\begin{bmatrix}1 & \log_{x} y & \log_{x} z \\ \log_{y} x & 1 & \log_{y} z\\ \log_{z} x & \log_{z} y & 1\end{bmatrix}$$ is:

A
$$0$$

B
$$1$$

C
$$\log_{e} xyz$$

D
$$-\log_{e} xyz$$

Solution

$$D=\left| \begin{matrix} 1 & \log _{ x }{ y } & \log _{ x }{ z } \\ \log _{ y }{ x } & 1 & \log _{ y }{ z } \\ \log _{ z }{ x } & \log _{ z }{ y } & 1 \end{matrix} \right| =(1-\log _{ z }{ y } .\log _{ y }{ z } )-\log _{ x }{ y } \left( \log _{ y }{ x } -\log _{ z }{ x } .\log _{ y }{ z } \right) +\log _{ x }{ z } \left( \log _{ y }{ x } .\log _{ z }{ y } -\log _{ z }{ x } \right) \\ =\left( 1-1 \right) -\log _{ x }{ y } \left( \log _{ y }{ x } -\log _{ y }{ x } \right) +\log _{ x }{ z } \left( \log _{ z }{ x } -\log _{ z }{ x } \right) =0-0+0=0$$
Question 4
1 / -0

The value of the determinant
$$\begin{vmatrix} \cos^2 \dfrac{\theta}{2}&\sin^2\dfrac{\theta}{2}\\ \sin^2\dfrac{\theta}{2} &\cos^2\dfrac{\theta}{2} \end{vmatrix}$$
for all values of $$\theta $$, is

A
$$1$$

B
$$\cos\theta $$

C
$$\sin\theta $$

D
$$\cos2\theta $$

Solution

$$\begin{vmatrix}\cos ^2\dfrac{\theta}{2}&\sin ^2\dfrac{\theta}{2}\\ \sin ^2\dfrac{\theta}{2} & {\cos }^2 \dfrac{\theta}{2}\end{vmatrix}=\cos ^4\dfrac{\theta}{2}-\sin ^4\dfrac{\theta}{2}$$

$$=\left({\cos }^2 \dfrac{\theta}2+{\sin }^2 \dfrac{\theta}2\right)\left({\cos }^2 \dfrac{\theta}2-{\sin }^2 \dfrac{\theta}2\right)$$

$$=1\times \cos \theta=\cos \theta$$

Hence, the answer is $$\cos \theta$$.
Question 5
1 / -0

The adjoint of the matric $$A = \begin{bmatrix}1 & 0 & 2\\ 2 & 1 & 0\\ 0 & 3 & 1\end{bmatrix}$$ is

A
$$\begin{bmatrix}-1 & 6 & 2\\ -2 & 1 & -4\\ 6 & 3 & 1\end{bmatrix}$$

B
$$\begin{bmatrix}1 & 6 & -2\\ -2 & 1 & 4\\ 6 & -3 & 1\end{bmatrix}$$

C
$$\begin{bmatrix}6 & 1 & 2\\ 4 & -1 & 2\\ 6 & 3 & -1\end{bmatrix}$$

D
$$\begin{bmatrix}-6 & 2 & 1\\ 4 & -2 & 1\\ 3 & 1 & -6\end{bmatrix}$$

Solution

$$adj(A)=cofactor(A)^T=\begin{bmatrix} A_{11} &A_{21}&A_{31}\\A_{12} &A_{22} &A_{32}\\A_{13} &A_{23} &A_{33}\end{bmatrix}=\begin{bmatrix} 1 &6&-2\\-2 &1 &4\\6 &-3 &1\end{bmatrix}$$
Question 6
1 / -0

If $$A + B + C = \pi$$, then $$\begin{vmatrix} \sin (A + B + C)& \sin B & \cos C\\ -\sin B & 0 & \tan A\\ \cos (A + B) & -\tan A & 0\end{vmatrix}$$ is equal to

A
$$0$$

B
$$2\sin B \tan A \cos C$$

C
$$1$$

D
None of these

Solution

Given $$A+B+C= \pi$$

$$=\begin{vmatrix} \sin (A+B+C) & \sin B & \cos C \\ -\sin B & 0 & \tan A \\ \cos A+B & -\tan A & 0 \end{vmatrix}$$

$$=\begin{vmatrix} \sin \pi & \sin B & \cos C \\ -\sin B & 0 & \tan A \\ \cos (\pi-C) & -\tan A & 0 \end{vmatrix}$$

$$=\sin\pi [\tan^2A]-\sin B[-\tan A \cos (\pi-C)]+\cos C[\sin B\tan A]$$

$$=0+\sin B\, \tan A\, \cos(\pi-C)+\cos C \sin B \tan A$$

$$=\sin B\, \tan A[\cos\pi \cos C+ \sin\pi \sin C]+\cos C\sin B\tan A$$

$$=-\sin B \tan A\cos C +\cos C \sin B \tan A$$

$$=0$$ $$\left[\therefore \sin \pi=0 ,\cos \pi =-1\right]$$
Question 7
1 / -0

If the points $$A(-2,1),B(a,b)$$ and $$C(4,-1)$$ are collinear and $$a-b=1$$, find the values of $$a$$ and $$b$$.

A
$$a=1,b=5$$

B
$$a=1,b=0$$

C
$$a=2,b=0$$

D
None of these

Solution

Three points $$A(-2,1),\ B(a,b)$$ and $$C(4,-1)$$ are collinear so the area of the triangle is equal to zero
Here the equation $$(1)$$ is $$\to a-b=1$$
Here $$x_1=-2\ y_1=1$$
$$x_1 =a\ y_1=b$$
$$x_3=4\ y_3=-1$$
Area of the triangle $$\triangle ABC=\dfrac {1}{2} [x_1 (y_2 -y_3)+x_2 (y_3 -y_1)+x_3 (y_1 -y_2)]=0$$
$$\Rightarrow \ \dfrac {1}{2} [-2(b+1)+a(-1 -1)+4(1-b)]=0$$
$$\Rightarrow \ \dfrac {1}{2}[-2b-2-2b+4-4b]=0$$
$$\Rightarrow \ \dfrac {1}{2} [-6b+2-2b]=0$$
$$\Rightarrow \ -3b+1-a=0$$
$$\Rightarrow \ -a-3b=-1$$......Equation $$2$$
Equation $$(1)\times 1\longrightarrow a-b=1$$
Equation $$(2)\times 1\to \dfrac {-a-2b=-1}{-4b=0\\ \Rightarrow b=0}$$
Putting the value of $$b=0$$ in Equation $$1:-$$
$$a-b=1$$
$$\Rightarrow \ a=1$$
So the value of $$a=1, b=0$$
Question 8
1 / -0

The determinant $$\begin{vmatrix} xp+y & x & y\\ yp+z & y & z\\ 0 & xp+y & yp+z\end{vmatrix}=0$$. If.

A
x, y, z are in A.P.

B
x, y, z are in G.P.

C
x, y, z are in H.P.

D
xy, yz, zx are in A.P.

Solution
Question 9
1 / -0

If $$\triangle$$ =$$
\left |
\begin{array}{111}
x_1+y_1\omega & x_1\omega^2+y_1 & x_1+y_1\omega+z_1\omega^2 \\
x_2+y_2\omega & x_2\omega^2+y_2 & x_2+y_2\omega+z_2\omega^2 \\
x_3+y_3\omega & x_3\omega^2+y_3 & x_3+y_3\omega+z_3\omega^2 \\
\end {array}
\right |
$$
where $$1,\omega,\omega^2$$ are cube roots of unity then $$\triangle$$ is equal to

A
$$0$$

B
$$1$$

C
$$-1$$

D
None of these

Solution
Question 10
1 / -0

The value of determinant $$\begin{vmatrix} x+1 & x+2 & x+4\\ x+3 & x+5 & x+8\\ x+7 & x+10 & x+14\end{vmatrix}$$ is?

A
$$-2$$

B
$$x^2+2$$

C
$$2$$

D
None of these

Solution

Operating $$C_3-C_2$$ and $$C_2-C_1$$, we get $$\Delta =\begin{vmatrix} x+1 & 1 & 2\\ x+3 & 2 & 3\\ x+7 & 3 & 4\end{vmatrix}$$
Apply $$R_3-R_2, R_2-R_1$$
$$\Delta =\begin{vmatrix} x+1 & 1 & 2\\ 2 & 1 & 1\\ 4 & 1 & 1\end{vmatrix}$$ apply $$C_3-C_2$$
$$\begin{vmatrix} x+1 & 1 & 1\\ 2 & 1 & 0 \\ 4 & 1 & 0\end{vmatrix}=1.(2-4)=-2$$

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Re-Attempt Weekly Quiz Competition

Determinants Test - 49

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Determinants Test - 49

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SHARING IS CARING

Question 1

$$16$$

$$6$$

$$14$$

$$8$$

Solution

Question 2

$$k = bc$$

$$k = ad$$

$$k = a^2 + b^2 + c^2 + d^2$$

$$ad - bc$$

Solution

Question 3

$$0$$

$$1$$

$$\log_{e} xyz$$

$$-\log_{e} xyz$$

Solution

Question 4

$$1$$

$$\cos\theta $$

$$\sin\theta $$

$$\cos2\theta $$

Solution

Question 5

$$\begin{bmatrix}-1 & 6 & 2\\ -2 & 1 & -4\\ 6 & 3 & 1\end{bmatrix}$$

$$\begin{bmatrix}1 & 6 & -2\\ -2 & 1 & 4\\ 6 & -3 & 1\end{bmatrix}$$

$$\begin{bmatrix}6 & 1 & 2\\ 4 & -1 & 2\\ 6 & 3 & -1\end{bmatrix}$$

$$\begin{bmatrix}-6 & 2 & 1\\ 4 & -2 & 1\\ 3 & 1 & -6\end{bmatrix}$$

Solution

Question 6

$$0$$

$$2\sin B \tan A \cos C$$

$$1$$

None of these

Solution

Question 7

$$a=1,b=5$$

$$a=1,b=0$$

$$a=2,b=0$$

None of these

Solution

Question 8

x, y, z are in A.P.

x, y, z are in G.P.

x, y, z are in H.P.

xy, yz, zx are in A.P.

Solution

Question 9

$$0$$

$$1$$

$$-1$$

None of these

Solution

Question 10

$$-2$$

$$x^2+2$$

$$2$$

None of these

Solution

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