Determinants Test

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Determinants Test - 44

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Weekly Quiz Competition

Question 1
1 / -0

If $$P = \begin{bmatrix} 1& 2 & 1\\ 1 & 3 & 1\end{bmatrix}, Q = PP^{T}$$, then the value of the determinant of $$Q$$ is

A
$$2$$

B
$$-2$$

C
$$1$$

D
$$0$$

Solution

Given,
$$P = \begin{bmatrix} 1& 2 & 1\\ 1 & 3 & 1\end{bmatrix}$$

$$\therefore Q = PP^{T} = \begin{bmatrix} 1& 2 & 1\\ 1 & 3 & 1\end{bmatrix} \begin{bmatrix} 1& 1\\ 2 & 3\\ 1 & 1\end{bmatrix}$$

$$= \begin{bmatrix} 1\times 1 + 2\times 2 + 1\times 1& 1\times 1 + 2\times 3 + 1\times 1\\ 1\times 1 + 3\times 2 + 1\times 1 & 1\times 1 + 3\times 3 + 1\times 1\end{bmatrix}$$

$$= \begin{bmatrix} 1 + 4 + 1& 1 + 6 + 1\\ 1 + 6 + 1 & 1 + 9 + 1\end{bmatrix} = \begin{bmatrix}6 & 8\\ 8 & 11\end{bmatrix}$$

$$\therefore$$ The determinant of $$Q = \begin{bmatrix} 6& 8\\ 8 & 11\end{bmatrix}$$

$$= 66 - 64 = 2$$
Question 2
1 / -0

Evaluate $$\begin{vmatrix} \cos 15^o& \sin 15^o\\ \sin 75^o & \cos 75^o\end{vmatrix}$$

A
$$1$$

B
$$0$$

C
$$2$$

D
$$3$$

Solution

Using fundamental theorem of determinant,
$$\begin{vmatrix} \cos 15^o& \sin 15^o\\ \sin 75^o & \cos 75^o\end{vmatrix}=\cos15^o\cos75^o-\sin15^0\sin75^o$$
$$=\cos(75^o+15^o)=\cos90^o=0$$, [using sum to product formula]

We have used the Fundamental theorem of determinant is $$\begin{vmatrix}a&b\\c&d\end{vmatrix}=ad-bc$$
Question 3
1 / -0

Let $$A$$ be a $$3 \times 3$$ matrix and $$B$$ be its adjoint matrix. If $$|B| = 64$$, then $$|A| =$$

A
$$\pm 2$$

B
$$\pm 4$$

C
$$\pm 8$$

D
$$\pm 12$$

Solution

$$|\text{adj}\,A|=|A|^{n-1}$$ where $$n$$ is the order of the square matirx.
Hence $$|B|=|\text{adj}\,A|=|A|^{3-1}$$
$$=|A|^{2}$$
Therefore
$$|A|=\sqrt{|B|}=\sqrt{64}=\pm8$$.
Question 4
1 / -0

The points $$(-a, -b), (a, b), (0, 0)$$ and $$(a^2, ab), a\ne 0, b\ne 0$$ are

A
Collinear

B
Vertices of a parallelogram

C
Vertices of rectangle

D
Lie on a circle

Solution

Let the points are $$A(-a,-b), B(a,b), C(0,0)$$ and $$D(a^2, ab)$$, then

slope of line $$AB =\dfrac{b}{a}=$$ slope of line $$BC =$$ slope of line $$CD =$$ slope of line $$DA$$

Hence all the given $$4$$ points are collinear.

Note: Slope of line joining the points $$(x_1,y_1)$$ and $$(x_2, y_2)$$ is given by $$\dfrac{y_2-y_1}{x_2-x_1}$$
Question 5
1 / -0

If a, b and c are in A, P., then the value of $$\begin{vmatrix}
x+2 & x+3 &x+a \\
x+4& x+5 &x+b \\
x+6 & x+7 & x+c
\end{vmatrix}$$ is.

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

B
$$9x^2+a+b+c$$

C
0

D
$$a+b+c$$

Solution

$$\begin{vmatrix}x+2&x+3&x+a\\x+4&x+5&x+b\\x+6&x+7&x+c \end{vmatrix}$$

$$=\begin{vmatrix}2x+8&2x+10&2x+(a+c)\\x+4&x+5&x+b\\x+6&x+7&x+c \end{vmatrix}$$, applying $$R_1\to R_1+R_3$$

$$=\begin{vmatrix}2(x+4)&2(x+5)&2(x+b)\\x+4&x+5&x+b\\x+6&x+7&x+c \end{vmatrix}$$, since a,b,c are in A.P. $$ a+c=2b$$

$$= 2\begin{vmatrix}x+4&x+5&x+b\\x+4&x+5&x+b\\x+6&x+7&x+c \end{vmatrix}$$, take 2 common from first row

$$=2(0)=0$$, since first and second row of the determinant are same
Question 6
1 / -0

If $$\Delta_r = \begin{vmatrix}2r - 1 & ^mC_r & 1\\ m^2 - 1 & 2^m & m+1\\ sin^2(m^2) & sin^2(m) & sin^2(m+1)\end{vmatrix}$$, then the value of $$\displaystyle \sum_{r = 0}^m \Delta_r$$, is

A
1

B
0

C
2

D
None of these

Solution

$$\Delta_r = \begin{vmatrix}2r -1 & ^mC_r & 1\\m^2 - 1 & 2^m & m+1\\ sin^2(m^2) & sin^2(m) & sin^2(m + 1)\end{vmatrix}$$

$$\therefore \sum_{r = 0}^m \Delta_r = \begin{vmatrix}\sum_{r = 0}^m (2r - 1) & \sum_{r = 0}^m { }^{m} C_r& \sum_{r = 0}^m 1\\ m^2 - 1 & 2^m & m+1\\ sin^2 (m^2) & sin^2(m) & sin^2(m + 1)\end{vmatrix}$$

$$= \begin{vmatrix} m^2 -1& 2^m & m+1\\ m^2 - 1 & 2^m & m+1\\ sin^2 (m^2) & sin^2(m) & sin^2(m + 1)\end{vmatrix}$$

$$=0$$ ($$\because$$ two rows are identical)
Hence, option B is correct.
Question 7
1 / -0

The value of the determinant $$\begin{vmatrix}1 & \cos (\alpha - \beta) & \cos \alpha\\ \cos(\alpha - \beta) & 1 & \cos \beta\\ \cos \alpha & \cos \beta & 1\end{vmatrix}$$ is

A
$$\alpha^{2} + \beta^{2}$$

B
$$\alpha^{2} - \beta^{2}$$

C
$$1$$

D
$$0$$

Solution

On solving the determinant, we have
$$1(1 - \cos^{2}\beta) - \cos (\alpha - \beta) [\cos (\alpha - \beta) - \cos \alpha \cdot \cos \beta] + \cos \alpha [\cos \beta \cdot \cos (\alpha - \beta) - \cos \alpha]$$

$$= 1 - \cos^{2}\beta - \cos^{2} \alpha - \cos^{2}(\alpha - \beta) + 2\cos \alpha \cdot \cos \beta . \cos (\alpha - \beta)$$

$$= 1 - \cos^{2}\beta - \cos^{2} \alpha + \cos (\alpha - \beta) [2\cos \alpha. \cos \beta - \cos (\alpha - \beta)]$$

$$= 1 - \cos^{2}\beta - \cos^{2} \alpha + \cos(\alpha - \beta) [\cos (\alpha + \beta) + \cos (\alpha - \beta) -\cos (\alpha - \beta)]$$

$$= 1 - \cos^{2}\beta - \cos^{2} \alpha - \cos^{2} \alpha . \cos^{2}\beta - \sin^{2} \alpha . \sin^{2}\beta$$

$$= 1 - \cos^{2} \beta - \cos^{2} \alpha (1 - \cos^{2}\beta) - \sin^{2} \alpha . \sin^{2}\beta$$

$$= 1 - \cos^{2} \beta - \cos^{2} \alpha . \sin^{2}\beta - \sin^{2}\alpha. \sin^{2}\beta$$

$$= (1 - \cos^{2}\beta) - \sin^{2} \beta(\sin^{2}\alpha + \cos^{2} \alpha)$$
$$= \sin^{2}\beta - \sin^{2}\beta = 0$$

Hence, determinant is $$0$$.
Question 8
1 / -0

The value of the determinant $$\begin{vmatrix}cos\, \alpha & -sin \,\alpha &1 \\ sin \, \alpha & cos \, \alpha & 1\\ cos (\alpha +\beta) & -sin (\alpha +\beta ) & 1\end{vmatrix} $$ is

A
Independent of $$\alpha$$

B
Independent of $$\beta$$

C
Independent of $$\alpha$$ and $$\beta$$

D
None of the above

Solution

Given, $$\begin{vmatrix}cos\, \alpha & sin \,\alpha &1 \\ sin \, \alpha & cos \, \alpha & 1\\ cos (\alpha +\beta) & -sin (\alpha +\beta ) & 1\end{vmatrix} $$

[Applying $$R_3\rightarrow R_3-R_1(cos \,\beta )+R_2(sin \, \beta )$$]

$$\begin{vmatrix}cos\, \alpha & sin \,\alpha &1 \\ sin \, \alpha & cos \, \alpha & 1\\ 0 & 0 & 1+sin \,\beta -cos \,\beta \end{vmatrix} $$

$$=(1 + sin \,\beta -cos\,\beta )(cos^2\alpha +sin^2\alpha )$$

$$=1+sin\,\beta -cos \,\beta $$, which is independent of $$\alpha $$.
Hence, option A is correct.
Question 9
1 / -0

The centre of a circle is $$(-6,4)$$. If one end of the diameter of the circle is at $$(-12,8)$$, then the other end is at

A
$$(-18,12)$$

B
$$(-9,6)$$

C
$$(-3,2)$$

D
$$(0,0)$$

Solution

The centre is the midpont of the line joining the two points of the diameter.
Using Midpoint-formula $$\left(\alpha,\beta\right)\equiv\left(\dfrac{{x}_{1}+{x}_{2}}{2},\dfrac{{y}_{1}+{y}_{2}}{2}\right)$$
From above we have $$\alpha=-6,\beta=4$$
Let $${x}_{1}=x,{y}_{1}=y$$
Given$${x}_{2}=-12,{y}_{2}=8$$
We have $$\alpha=\dfrac{{x}_{1}+{x}_{2}}{2}$$ and $$\beta=\dfrac{{y}_{1}+{y}_{2}}{2}$$
$$x+\dfrac{\left(-12\right)}{2} = -6$$
$$\Rightarrow x-6=-6$$
$$\therefore x = 0$$
$$\Rightarrow y+\dfrac{8}{2} = 4$$
$$\Rightarrow y+4=4$$
$$\therefore y = 0$$
Hence, the other point is at $$\left(0,0\right)$$
Question 10
1 / -0

If $$A(x)=\begin{vmatrix} x+1 & 2x+1 & 3x+1 \\ 2x+1 & 3x+1 & x+1 \\ 3x+1 & x+1 & 2x+1 \end{vmatrix}$$
then $$\displaystyle \int _{ 0 }^{ 1 }{ A(x) } dx=$$

A
$$-15$$

B
$$\cfrac { -15 }{ 2 } $$

C
$$-30$$

D
$$-5$$

Solution

$$ \\ A(x)=\begin{vmatrix} x+1 & 2x+1 & 3x+1 \\ 2x+1 & 3x+1 & x+1 \\ 3x+1 & x+1 & 2x+1 \end{vmatrix}\\ =(x+1)((3x+1)(2x+1)-(x+1)(x+1))-(2x+1)((2x+1)(2x+1)-(x+1)(3x+1)) +\\\,\,\,\,(3x+1)((x+1)(2x+1)-(3x+1)(3x+1))\\ A(x)=-18{ x }^{ 3 }-9{ x }^{ 2 }\\\displaystyle \int _{ 0 }^{ 1 }{ A(x) } dx=\cfrac { -18 }{ 4 } { x }^{ 4 }-3{ x }^{ 3 }=\cfrac { -18 }{ 4 } -3=\cfrac {- 30 }{ 4 } =\cfrac { -15 }{ 2 } \\ $$

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Re-Attempt Weekly Quiz Competition

Determinants Test - 44

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Determinants Test - 44

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SHARING IS CARING

Question 1

$$2$$

$$-2$$

$$1$$

$$0$$

Solution

Question 2

$$1$$

$$0$$

$$2$$

$$3$$

Solution

Question 3

$$\pm 2$$

$$\pm 4$$

$$\pm 8$$

$$\pm 12$$

Solution

Question 4

Collinear

Vertices of a parallelogram

Vertices of rectangle

Lie on a circle

Solution

Question 5

$$x-(a+b+c)$$

$$9x^2+a+b+c$$

0

$$a+b+c$$

Solution

Question 6

1

0

2

None of these

Solution

Question 7

$$\alpha^{2} + \beta^{2}$$

$$\alpha^{2} - \beta^{2}$$

$$1$$

$$0$$

Solution

Question 8

Independent of $$\alpha$$

Independent of $$\beta$$

Independent of $$\alpha$$ and $$\beta$$

None of the above

Solution

Question 9

$$(-18,12)$$

$$(-9,6)$$

$$(-3,2)$$

$$(0,0)$$

Solution

Question 10

$$-15$$

$$\cfrac { -15 }{ 2 } $$

$$-30$$

$$-5$$

Solution

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