Determinants Test

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Determinants Test - 35

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Weekly Quiz Competition

Question 1
1 / -0

In a third order determinant $$a_{ij}$$ denotes the element in the ith row and the jth column If $$ a_{ij} = \left\{\begin{matrix}0, & i = j\\ 1, & i > j\\ -1, & i < j\end{matrix}\right.$$ then the value of the determinant

A
0

B
1

C
-1

D
none of these

Solution

$$ a_{ij} = \left\{\begin{matrix}0, & i = j\\ 1, & i > j\\ -1, & i < j\end{matrix}\right.$$

Let $$A=\begin{bmatrix} 0&-1&-1\\1&0&-1\\1&1&0\end{bmatrix}$$

Now, $$|A|=\begin{vmatrix} 0&-1&-1\\1&0&-1\\1&1&0\end{vmatrix}$$

$$=1-1=0$$

$$\Rightarrow |A|=0$$
Question 2
1 / -0

If $$z = \begin{vmatrix}3 + 3i & 5 - i & 7 - 3i\\ i & 2i & - 3i\\ 3 - 2i & 5 + i & 7 + 3i\end{vmatrix} $$, then

A
z is purely real

B
z is purely imaginary

C
0

D
none of these

Solution

$$z=$$ $$\begin{vmatrix}
3+3i & 5-i & 7-3i\\
i & 2i & -3i \\
3-2i & 5+i & 7+3i
\end{vmatrix}$$

On finding $$\left | z \right |$$ ie determinant value

$$=3(1+i)[2i(7+3i)+3i(5+i)]-(5-i)[i(7+3i)+3i(3-2i)]+(7-3i)[i(5+i)-2i(3-2i)]$$

$$=-3(3i+61)$$

Hence $$\left | z \right |$$ is complex so option $$(d)$$ is correct
Question 3
1 / -0

Evaluate $$\begin{vmatrix}cos (x + x^2) & sin(x + x^2) &- cos (x + x^2) \\sin(x - x^2) & cos (x - x^2) & sin (x - x^2)\\ sin 2x & 0 & sin 2x^2\end{vmatrix}$$

A
$$sin (2x + 2x^2)$$

B
$$-sin (2x + 2x^2)$$

C
$$cos (2x + 2x^2)$$

D
$$-cos (2x + 2x^2)$$

Solution

$$\begin{vmatrix}cos (x + x^2) & sin(x + x^2) &- cos (x + x^2) \\sin(x - x^2) & cos (x - x^2) & sin (x - x^2)\\ sin 2x & 0 & sin 2x^2\end{vmatrix}$$
expanding along $$R_3$$ gives
$$=\sin 2x\left(\sin(x+x^2)\sin(x-x^2)+\cos(x+x^2)\cos(x-x^2)\right)$$ $$+\sin 2x^2\left(\cos(x+x^2)\cos(x-x^2)-\sin(x+x^2)\sin(x-x^2)\right)$$
$$=\sin 2x\cos 2x^2+\sin 2x^2\cos 2x$$
$$=\sin(2x+2x^2)$$
Hence, option A.
Question 4
1 / -0

If $$\displaystyle \left | \begin{matrix}a+x &a &x \\ a-x &a &x \\ a-x &a &-x \end{matrix} \right |=0$$ then $$\displaystyle x$$ is

A
$$0$$

B
$$\displaystyle a$$

C
$$3$$

D
$$\displaystyle 2a$$

Solution
Question 5
1 / -0

The value of the determinant $$\begin{vmatrix}\sqrt{a} + \sqrt{b} & \sqrt{bc} + \sqrt{2a} & b + \sqrt{ca}\\ 2 \sqrt{c} & c & \sqrt{bc}\\ \sqrt{c} & \sqrt{2c} &c \end{vmatrix}$$ is

A
$$c(\sqrt{2} b - c \sqrt{b})$$

B
$$b(\sqrt{2b} - c \sqrt{b})$$

C
$$c(\sqrt{2b} - c \sqrt{b})$$

D
$$b(\sqrt{2} b - c \sqrt{b})$$

Solution

$$\begin{vmatrix} \sqrt { a } +\sqrt { b } & \sqrt { bc } +\sqrt { 2a } & b+\sqrt { ac } \\ 2\sqrt { c } & c & \sqrt { bc } \\ \sqrt { c } & \sqrt { 2c } & c \end{vmatrix}\\ =\left( \sqrt { a } +\sqrt { b } \right) \left( c\sqrt { 2b } -{ c }^{ 2 } \right) -\left( \sqrt { bc } +\sqrt { 2a } \right) \left( 2c\sqrt { c } -c\sqrt { b } \right) +\left( b+\sqrt { ac } \right) \left( c\sqrt { c } -2c\sqrt { c } \right) \\ =c\sqrt { 2ab } +bc\sqrt { 2 } -{ c }^{ 2 }\sqrt { a } -{ c }^{ 2 }\sqrt { b } -2{ c }^{ 2 }\sqrt { b } -2a\sqrt { 2ac } +bc\sqrt { c } \\ +c\sqrt { ab } +bc\sqrt { c } +{ c }^{ 2 }\sqrt { a } -2bc\sqrt { c } -2{ c }^{ 2 }\sqrt { a } \\ =c\sqrt { 2 } b-c\sqrt { b } $$
Question 6
1 / -0

The value of the $$\displaystyle m^{th}$$ order determinant of a matrix $$\displaystyle A$$ is $$\displaystyle 15$$ then the value of determinant formed by the cofactors of $$\displaystyle A$$ will be

A
$$\displaystyle \left ( 15 \right )^{m}$$

B
$$\displaystyle 15^{2m}$$

C
$$\displaystyle \left ( 15 \right )^{m-1}$$

D
$$\displaystyle \left ( 15 \right )^{2m-1}$$

Solution

If A be the matrix of order m and B is matrix of cofactors of a then determinant of B will be equal to $$\displaystyle \left ( m-1 \right )^{th}$$ power of the determinant of A.
$$\displaystyle \left | B \right |= \left | A \right |^{m-1}$$
$$\displaystyle \therefore |B| = \left ( 15 \right )^{m-1}$$
Question 7
1 / -0

If the determinant $$\begin{vmatrix}a & b & at-b\\ b & c & bt-c\\ 2 & 1 & 0\end{vmatrix}=0$$, if $$a, b, c$$ are in

A
$$A.P.$$

B
$$G.P.$$

C
$$H.P.$$

D
$$k=1/2$$

Solution

$$\begin{vmatrix}a & b & at-b\\ b & c & bt-c\\ 2 & 1 & 0\end{vmatrix}=0$$
Expanding it along third row,
$$\Rightarrow 2[b(bt-c)-c(at-b)]-1[a(bt-c)-b(at-b)]=0$$
$$\Rightarrow 2t(b^2-ac)-(b^2-ac)=0$$
$$\Rightarrow (2t-1)(b^2-ac)=0$$
$$\Rightarrow t=\dfrac{1}{2}$$ or $$b^2=ac$$
If $$b^2 =ac$$ then $$a,b,c\in $$ G.P.
Question 8
1 / -0

The points $$\displaystyle(a, b+c),(b, c+a),(c, a+b)$$ are

A
vertices of an equilateral triangle

B
collinear

C
concyclic

D
none of these

Solution

To check for collineauty, we can check the slope of the two points. Suppose the points be A, B and C then

$$(slope)_{AB} = \dfrac {(c+a)-(b+c)}{b-a} =-1$$

$$(slope)_{BC}= \dfrac{(a+b)-(c+a)}{c-b} =-1$$

$$m_{AB}= m_{BC} =$$ points are collinear
Question 9
1 / -0

If $$\begin{vmatrix}x^{2}+3x &x+1 &x-2 \\ x-1 &1-2x &x+4 \\ x+3 &x-4 &3x \end{vmatrix}= Ax^{4}+Bx^{3}+Cx^{2}+Dx+\varrho $$
Then value of $$\varrho$$ equals to,

A
-10

B
10

C
0

D
None of these

Solution

$$\begin{vmatrix}x^{2}+3x &x+1 &x-2 \\ x-1 &1-2x &x+4 \\ x+3 &x-4 &3x \end{vmatrix}= Ax^{4}+Bx^{3}+Cx^{2}+Dx+\varrho $$

Substitute $$x=0$$

$$\Rightarrow \begin{vmatrix} 0 & 1 & -2 \\ -1 & 1 & 4 \\ 3 & -4 & 0 \end{vmatrix}=\varrho $$

$$=12-2=10$$
Question 10
1 / -0

The value of the determinant $$\displaystyle \begin{vmatrix}1 & e^{i \pi/3} &e^{i \pi/4} \\ e^{- i \pi/3} &1 &e^{2 i \pi / 3} \\ e^{- i \pi/4} & e^{- 2 i \pi / 3} & 1\end{vmatrix}$$ is ..................

A
$$(2 + \sqrt{2})$$

B
$$ (2 - \sqrt{2})$$

C
$$- (2 + \sqrt{2})$$

D
$$ -(2 - \sqrt{2})$$

Solution

Let $$\Delta = \begin{vmatrix} 1 & { e }^{ i\pi /3 } & { e }^{ i\pi /4 } \\ { e }^{ -i\pi /3 } & 1 & { e }^{ i2\pi /3 } \\ { e }^{ -i\pi /4 } & { e }^{ -i2\pi /3 } & 1 \end{vmatrix}$$
Expanding along first column we get,
$$\displaystyle \Delta = (1)\begin{vmatrix} 1 & { e }^{ i2\pi /3 } \\ { e }^{ -i2\pi /3 } & 1 \end{vmatrix}-{ e}^{ -i\pi /3 }\begin{vmatrix} { e }^{ i\pi /3 } & { e }^{ i\pi /4 }

\\ { e }^{ -i2\pi /3 } & 1 \end{vmatrix}+{ e }^{ -i\pi /4 } \begin{vmatrix} { e }^{ i\pi /3 } & { e }^{ i\pi /4 } \\ 1 & { e }^{ i2\pi /3 } \end{vmatrix}$$
$$\Rightarrow \Delta = 1(1-1)-{ e }^{ -i\pi /3 }({ e }^{ i\pi /3 }-{ e}^{ -5i\pi /12 })+{ e }^{ -i\pi /4 } ({ e }^{ \pi } -{ e }^{ i\pi /4 } )$$
$$\Rightarrow \Delta =-2+{ e }^{ 3i\pi /4 }+{ e }^{ -3i\pi /4 }=-2+2\cos { (3\pi /4) }=-2-\sqrt{2} =-(2+\sqrt{2})$$

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Determinants Test - 35

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Determinants Test - 35

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Question 1

0

1

-1

none of these

Solution

Question 2

z is purely real

z is purely imaginary

0

none of these

Solution

Question 3

$$sin (2x + 2x^2)$$

$$-sin (2x + 2x^2)$$

$$cos (2x + 2x^2)$$

$$-cos (2x + 2x^2)$$

Solution

Question 4

$$0$$

$$\displaystyle a$$

$$3$$

$$\displaystyle 2a$$

Solution

Question 5

$$c(\sqrt{2} b - c \sqrt{b})$$

$$b(\sqrt{2b} - c \sqrt{b})$$

$$c(\sqrt{2b} - c \sqrt{b})$$

$$b(\sqrt{2} b - c \sqrt{b})$$

Solution

Question 6

$$\displaystyle \left ( 15 \right )^{m}$$

$$\displaystyle 15^{2m}$$

$$\displaystyle \left ( 15 \right )^{m-1}$$

$$\displaystyle \left ( 15 \right )^{2m-1}$$

Solution

Question 7

$$A.P.$$

$$G.P.$$

$$H.P.$$

$$k=1/2$$

Solution

Question 8

vertices of an equilateral triangle

collinear

concyclic

none of these

Solution

Question 9

-10

10

0

None of these

Solution

Question 10

$$(2 + \sqrt{2})$$

$$ (2 - \sqrt{2})$$

$$- (2 + \sqrt{2})$$

$$ -(2 - \sqrt{2})$$

Solution

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