Determinants Test

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Determinants Test - 38

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Weekly Quiz Competition

Question 1
1 / -0

If $$a, b, c \ \epsilon \ R$$, find the number of real roots of the equation given by $$ \Delta = 0 $$, where
$$\Delta =\begin{vmatrix}
x & c & -b\\
-c & x & a\\
b & -a & x
\end{vmatrix}$$.

A
$$0$$

B
$$1$$

C
$$2$$

D
$$3$$

Solution

$$\triangle =\left| \begin{matrix} x & c & -b \\ -c & x & a \\ b & -a & x \end{matrix} \right| \\ \triangle =\quad x({ x }^{ 2 }\quad +\quad { a }^{ 2 })\quad -c(-cx-ab)\quad -b(ac-xb)\\ \because \quad \triangle \quad =\quad 0\\ \therefore \quad x({ x }^{ 2 }\quad +\quad { a }^{ 2 })\quad -c(-cx-ab)\quad -b(ac-xb)\quad =\quad 0\\ \Rightarrow \quad { x }^{ 3 }\quad +\quad { a }^{ 2 }x\quad +\quad { c }^{ 2 }x\quad +\quad abc\quad -\quad abc\quad +\quad x{ b }^{ 2 }\quad =\quad 0\\ \Rightarrow \quad { x }^{ 3 }\quad +\quad { a }^{ 2 }x\quad +\quad { c }^{ 2 }x\quad +\quad x{ b }^{ 2 }\quad =\quad 0\\ \Rightarrow \quad { x }^{ 3 }\quad +\quad ({ a }^{ 2 }\quad +\quad { c }^{ 2 }\quad +\quad { b }^{ 2 })x\quad =\quad 0$$
As $$a^{2} + b^{2}+c^{2} $$ is positive, so $$ x\ =\ 0 $$ is the only real solution
Question 2
1 / -0

If $$A=\begin{bmatrix} 0&\sin \alpha & \sin \alpha\sin \beta \\-\sin \alpha &0 &\cos \alpha \sin \beta \\-\sin \alpha \sin \beta &-\cos \alpha \cos \beta &0 \end{bmatrix}$$, then which of the following is true?

A
$$|A|$$ is independent of $$\alpha $$ and $$\beta $$.

B
$$A^{-1}$$ depends only on $$\alpha $$.

C
$$A^{-1}$$ depends only on $$\beta $$.

D
None of these

Solution

$$A=\begin{bmatrix} 0 & \sin { \alpha } & \sin { \alpha } \sin { \beta } \\ -\sin { \alpha } & 0 & \cos { \alpha } \cos { \beta } \\ -\sin { \alpha } \sin { \beta } & -\cos { \alpha } \cos { \beta } & 0 \end{bmatrix}$$
A is a skew symmetric matrix
$$\Rightarrow \left| A \right| =0$$
$$\Rightarrow { A }^{ -1 }$$ does not exist and
$$\left| A \right| $$ is independent of $$\alpha \& \beta $$
Option A
Question 3
1 / -0

For which value of '$$k$$' the points $$(7, -2), (5, 1), (3, k)$$ are collinear?

A
$$-4$$

B
$$4$$

C
$$8$$

D
$$-8$$

Solution

Since points are collinear,

so, $$x_1(y_2 - y_3)+x_2(y_3 - y_1) + x_3(y_1 - y_2)= 0$$

Take ($$7, -2$$) as $$(x_1 , y_1)$$, ($$5, 1$$) as $$(x_2 , y_2)$$ and ($$3 , k$$) as $$(x_3 , y_3)$$, we have

$$7(1 -k ) + 5(k + 2) + 3(-2 -1) = 0$$

$$\Rightarrow$$ $$7- 7k + 5k + 10 -9 =0$$

$$\Rightarrow$$ $$2k =8$$

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

If $$(3,2)$$, $$(4,k)$$ and $$(5,3)$$ are collinear, then $$k$$ is equal to:

A
$$\dfrac { 3 }{ 2 }$$

B
$$\dfrac { 2 }{ 5 }$$

C
$$\dfrac { 5 }{ 2 }$$

D
$$\dfrac { 3 }{ 5 }$$

Solution

As the points are collinear, the slope of the line joining any two points, should be same as the slope of the line joining two other points.
Slope of the line passing through points $$\left( { x }_{ 1 },{ y }_{ 1 } \right) $$ and $$\left( { x }_{ 2 },{ y }_{ 2 } \right)$$ is $$\dfrac { { y }_{ 2 }-{y }_{ 1 } }{ { x }_{ 2 }-x_{ 1 } } $$
So, Slope of the line joining $$ (3,2) , (4,k) = $$ Slope of the line joining $$ (3,2) $$ and $$ (5,3) $$
$$ \dfrac { k - 2 }{ 4 - 3 } = \dfrac { 3- 2 }{ 5 - 3 } $$
$$ k - 2 = \dfrac { 1 }{ 2 } $$
$$ k = \dfrac { 5 }{ 2 } $$
Hence, option C.
Question 5
1 / -0

Let $$A$$ be a non-singular matrix. Then $$\left| adjA \right| $$ is equal to

A
$${ \left| A \right| }^{ n }$$

B
$${ \left| A \right| }^{ n-1 }$$

C
$${ \left| A \right| }^{ n-2 }$$

D
None of these

Solution

Since A is non-singular
$$\therefore { A }^{ -1 }$$ exists
Now ,$$A$$ $$\left( adjA \right) $$=$$\displaystyle\left| A \right| I=$$$$(adjA)A$$
$$\displaystyle \Rightarrow \left| A \right| \left| adjA \right| ={ \left| A \right| }^{ n }=\left| adjA \right| \left| A \right| $$
$$\displaystyle \therefore \left| adjA \right| ={ \left| A \right| }^{ n-1 }$$
Question 6
1 / -0

The value of the determinant $$\begin{vmatrix} -a& b & c\\ a & -b &c \\ a & b &-c \end{vmatrix}$$ is equal to

A
$$0$$

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

C
$$(a+b)(b+c)(c+a)4abc$$

D
$$4abc$$

Solution

$$\begin{vmatrix} -a & b & c \\ a & -b & c \\ a & b & -c \end{vmatrix}=\left( -a \right) \left( bc-bc \right) -b\left( -ac-ac \right) +c\left( ab+ab \right) \\ =2abc+2abc=4abc$$
Question 7
1 / -0

If $$D_p=\begin{vmatrix} p& 15 & 8\\ p^2 & 35 & 9\\ p^3 & 25 & 10\end{vmatrix}$$, then $$D_1+D_2+D_3+D_4+D_5$$ is equal to

A
0

B
25

C
625

D
none of these

Solution

$$D_{ p }=\begin{vmatrix} p & 15 & 8 \\ p^{ 2 } & 35 & 9 \\ p^{ 3 } & 25 & 10 \end{vmatrix}=5\begin{vmatrix} p & 3 & 8 \\ p^{ 2 } & 7 & 9 \\ p^{ 3 } & 5 & 10 \end{vmatrix}\\ =5\left( p\left( 70-45 \right) -{ p }^{ 2 }\left( 30-40 \right) +{ p }^{ 3 }\left( 27-56 \right) \right) \\ =5\left( 25p+10{ p }^{ 2 }-29{ p }^{ 3 } \right) $$
$$D_{ 1 }+D_{ 2 }+D_{ 3 }+D_{ 4 }+D_{ 5 }\\ =5\left( 25+10-29 \right) +5\left( 25.2+10.2^{ 2 }-29.{ 2 }^{ 3 } \right) +5\left( 25.3+10{ .3 }^{ 2 }-29{ .3 }^{ 3 } \right) \\ +5\left( 25.4+10{ .4 }^{ 2 }-29{ .4 }^{ 3 } \right) +5\left( 25.5+10{ .5 }^{ 2 }-29{ .5 }^{ 3 } \right) \\ =28000$$
Question 8
1 / -0

Directions For Questions

$$\displaystyle \Delta (x)=\begin{vmatrix}
2x^{3}-3x^{2} &5x+7 &2 \\
4x^{3}-7x &3x+2 & 1 \\
7x^{3}-8x^{2} &x-1 &3
\end{vmatrix}$$$$\displaystyle =a_{0}+a_{1}x+...+a_{4}x^{4}$$

To evaluate $$a_j$$ we differentiate $$\displaystyle \Delta (x), j$$ times w.r.t. x and put $$x=0$$ or
divide $$\displaystyle \Delta (x)$$ by $$x^{4}$$ put $$ \dfrac{1}{x}=t$$ differentiate $$(4-j)$$ time w.r.t $$t$$ and put $$t=0$$

...view full instructions

$$\displaystyle a_{0}$$ equals

A
0

B
1

C
2

D
3

Solution

$$\triangle \left( x \right) =\begin{vmatrix} 2{ x }^{ 3 }-3{ x }^{ 2 } & 5x+7 & 2 \\ 4{ x }^{ 3 }-7x & 3x+2 & 1 \\ 7{ x }^{ 3 }-8{ x }^{ 2 } & x-1 & 3 \end{vmatrix}={ a }_{ 0 }+{ a }_{ 1 }x+....{ a }_{ 4 }{ x }^{ 4 }$$
Putting $$x=0$$ we get $${ a }_{ 0 }=?$$
$$\triangle \left( 0 \right) ={ a }_{ 0 }=\begin{vmatrix} 0 & 7 & 2 \\ 0 & 2 & 1 \\ 0 & -1 & 3 \end{vmatrix}=0$$
$$\therefore { a }_{ 0 }=0$$
Option A
Question 9
1 / -0

If $$A=\begin{bmatrix} 4 & 2 \\ 3 & 3 \end{bmatrix}$$, then adj (adj $$A$$) is equal to

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

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

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

D
None of these

Solution

For $$A=\begin{bmatrix} 4 & 2 \\ 3 & 3 \end{bmatrix}$$
$$adj\left( A \right) =\begin{bmatrix} 3 & -3 \\ -2 & 4 \end{bmatrix}$$
$$adj\left( adj\left( A \right) \right) =\begin{bmatrix} 4 & 2 \\ 3 & 3 \end{bmatrix}$$
Question 10
1 / -0

A set of points which lie on same line are called as

A
collinear

B
non-collinear

C
concurrent

D
none of these

Solution

A set of points which lie on same line are called as collinear.

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Determinants Test - 38

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Determinants Test - 38

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SHARING IS CARING

Question 1

$$0$$

$$1$$

$$2$$

$$3$$

Solution

Question 2

$$|A|$$ is independent of $$\alpha $$ and $$\beta $$.

$$A^{-1}$$ depends only on $$\alpha $$.

$$A^{-1}$$ depends only on $$\beta $$.

None of these

Solution

Question 3

$$-4$$

$$4$$

$$8$$

$$-8$$

Solution

Question 4

$$\dfrac { 3 }{ 2 }$$

$$\dfrac { 2 }{ 5 }$$

$$\dfrac { 5 }{ 2 }$$

$$\dfrac { 3 }{ 5 }$$

Solution

Question 5

$${ \left| A \right| }^{ n }$$

$${ \left| A \right| }^{ n-1 }$$

$${ \left| A \right| }^{ n-2 }$$

None of these

Solution

Question 6

$$0$$

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

$$(a+b)(b+c)(c+a)4abc$$

$$4abc$$

Solution

Question 7

0

25

625

none of these

Solution

Question 8

0

1

2

3

Solution

Question 9

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

$$\begin{bmatrix} 4 & 2 \\ 3 & 3 \end{bmatrix}$$

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

None of these

Solution

Question 10

collinear

non-collinear

concurrent

none of these

Solution

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