Determinants Test

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Determinants Test - 51

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Weekly Quiz Competition

Question 1
1 / -0

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

A
$$0$$

B
$$1$$

C
$$4 \log e$$

D
$$5 \log e$$

Solution

As we know that, $$loge^n=nloge$$

So, We can write given Determinant as

$$\begin{vmatrix}log e & 2log e & 3log e \\ 2log e & 3log e & 4log e \\ 3log e &\ 4log e & 5log e \end{vmatrix}=(loge)^3\begin{vmatrix}1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix}$$ .... [Taking $$loge$$ common from each row]

Expanding the Given Determinant , we get

$$=(loge)^3|(15-16)-2(10-12)+3(8-9)|=(loge)^3\times 0=0$$
Question 2
1 / -0

If the lines $$p_{ 1 }x+q_{ 1 }y=1, p_{ 2 }x+q_{ 2 }y=1$$ and $$p_{3}x+q_{3}y=1$$ be concurrent, show that the points $$(p_{1},q_{1}), (p_{2}, q_{2})$$ and $$ (p_{3}, q_{3})$$ are collinear.

A
vertices of right angle triangle

B
vertices of an equilateral triangle

C
vertices of an isosceles triangle

D
Collinear

Solution

Given:
$$p_{ 1 }x+q_{ 1 }y=1, p_{ 2 }x+q_{ 2 }y=1$$ and $$p_{3}x+q_{3}y=1$$

The lines will be concurrent if

$$\begin{vmatrix} { p }_{ 1 } & { q }_{ 1 } & -1 \\ { p }_{ 2 } & { q }_{ 2 } & -1 \\ { p }_{ 3 } & { q }_{ 3 } & -1 \end{vmatrix}=0$$

Or

$$\dfrac { 1 }{ 2 } \begin{vmatrix} { p }_{ 1 } & { q }_{ 1 } & 1 \\ { p }_{ 2 } & { q }_{ 2 } & 1 \\ { p }_{ 3 } & { q }_{ 3 } & 1 \end{vmatrix}=0$$

Or

$$\triangle=0$$

i.e., area of a triangle formed by the points $$({p}_{1}, {q}_{1})$$, $$({p}_{2}, {q}_{2})$$, and $$({p}_{3}, {q}_{3})$$ is zero and as such the points are collinear.
Question 3
1 / -0

The value of $$\dfrac { 1 }{ x-y } \begin{vmatrix} 1 & 0 & 0 \\ 3 & { x }^{ 3 } & 1 \\ 5 & { y }^{ 3 } & 1 \end{vmatrix}$$ is-

A
$$x+y$$

B
$${x}^{2}-xy+{y}^{2}$$

C
$${x}^{2}+xy+{y}^{2}$$

D
$${x}^{3}-{y}^{3}$$

Solution

Expanding along $$R_1$$

$$=\dfrac{1}{(x-y)}[1(x^3-y^3)]$$

$$=\dfrac{(x-y)(x^2+y^2+xy)}{x-y}=x^2+y^2+xy$$
Question 4
1 / -0

Find the determinant of given matrix $$\left[ \begin{matrix} a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{matrix} \right] $$

A
$$2(a+b+c)^{3}$$

B
$$(a-b-c)^{3}$$

C
$$2(a-b-c)^{3}$$

D
$$(a+b+c)^{3}$$

Solution

$$\left| \begin{matrix} a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{matrix} \right| $$

$$ =\left| \begin{matrix} a+b+c & a+b+c & a+b+c \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{matrix} \right| { R }_{ 1 }\rightarrow { R }_{ 1 }+{ R }_{ 2 }+{ R }_{ 3 }$$

Taking out common $$a+b+c$$ from $${R_1}$$

$$=\left( a+b+c \right) \left| \begin{matrix} 1 & 1 & 1 \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{matrix} \right|$$

$$ =\left( a+b+c \right) \left| \begin{matrix} 1 & 0 & 0 \\ 2b & -\left( a+b+c \right) & 0 \\ 2c & 0 & -\left( a+b+c \right) \end{matrix} \right| { R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 },{ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 1 }$$

Expanding along $${R_1}$$

$$={ \left( a+b+c \right) }^{ 3 }$$
Question 5
1 / -0

If $$\triangle =\begin{bmatrix} { a }_{ 1 } & { b }_{ 1 } & { c }_{ 1 } \\ { a }_{ 2 } & { b }_{ 2 } & { c }_{ 2 } \\ { a }_{ 3 } & { b }_{ 3 } & { c }_{ 3 } \end{bmatrix}$$ and $${A}_{2},{B}_{2},{C}_{2}$$ are respectively cofactors of $${a}_{2},{b}_{2},{c}_{2}$$ then $${a}_{1}{A}_{2}+{b}_{1}{B}_{2}+{c}_{1}{C}_{2}$$ is equal to ?

A
$$-\triangle$$

B
$$0$$

C
$$\triangle$$

D
$$none\ of\ these$$

Solution

Co-factor of $$ \displaystyle a_2 = (-1)^{i+j}\left|\begin{matrix} b_1 & c_1 \\ b_3 & c_3 \end{matrix}\right| = - [b_1 \, c_3 - c_1 \, b_3] = A_2 $$
i=2 j=1

of $$ \displaystyle b_2 = (-1)^{2+2} \left|\begin{matrix} a_1 & c_1 \\ a_3 & c_3 \end{matrix}\right| = [a_1 \, c_3 - c_1 \, a_3] = B_2 $$

of $$ \displaystyle c_2 = (-1)^{2+3} \left|\begin{matrix} a_1 & b_1 \\ a_3 & b_3 \end{matrix}\right| = -[a_1 \, b_3 \, - b_1 \, a_3 ] = C_2 $$

$$ a_1 \, A_2 + b_1 \, B_2 +c_1 \, C_2 $$

$$ \displaystyle -a_1 \, b_1 \, c_3 + a_1 \, c_1 \, b_3 + b_1 \, a_1 \, c_3 - b_1 c_1 \, a_3 -c_1 \, a_1 \, b_3 + c_1 \, b_1 \, a_3 $$

$$ \displaystyle = 0 $$
Question 6
1 / -0

If $${x}^{a}{y}^{b}={e}^{m},{x}^{c}{y}^{d}={e}^{n}, { \triangle }_{ 1 }=\begin{vmatrix} m & b \\ n & d \end{vmatrix},{ \triangle }_{ 2 }=\begin{vmatrix} a & m \\ c & n \end{vmatrix}$$ and $${ \triangle }_{ 3 }=\begin{vmatrix} a & b \\ c & d \end{vmatrix}$$ the value of $$x$$ and $$y$$ are respectively

A
$$\dfrac { { \triangle }_{ 1 } }{ { \triangle }_{ 3 } }$$ and $$\dfrac { { \triangle }_{ 2 } }{ { \triangle }_{ 3 } }$$

B
$$\dfrac { { \triangle }_{ 2 } }{ { \triangle }_{ 1 } }$$ and $$\dfrac { { \triangle }_{ 3 } }{ { \triangle }_{ 1 } }$$

C
$$\log { \left( \dfrac { { \triangle }_{ 1 } }{ { \triangle }_{ 3 } } \right) } and\log { \left( \dfrac { { \triangle }_{ 2 } }{ { \triangle }_{ 3 } } \right) }$$

D
$${ e }^{ { \triangle }_{ 1 }/{ \triangle }_{ 3 } }\,\ and\,\ { e }^{ { \triangle }_{ 2 }/{ \triangle }_{ 3 } }$$

Solution

$${ x }^{ a }{ y }^{ b }={ e }^{ m }$$

Taking $${ \log }_{ e }$$ on both sides,

$$\log _{ e }{ \left( { x }^{ a }{ y }^{ b } \right) } =\log _{ e }{ { e }^{ m } } $$

$$a\log _{ e }{ x } +b\log _{ e }{ y } =m\rightarrow 1$$

Similarly,

$$c\log _{ e }{ x } +d\log _{ e }{ y } =n\rightarrow 2$$

Solving $$1$$ and $$2$$ by Cramer's rule,

$$\log _{ e }{ x } =\cfrac { \begin{vmatrix} m & b \\ n & d \end{vmatrix} }{ \begin{vmatrix} a & b \\ c & d \end{vmatrix} } =\cfrac { { \triangle }_{ 1 } }{ { \triangle }_{ 3 } } $$

$$\log _{ e }{ y } =\cfrac { \begin{vmatrix} a & m \\ c & n \end{vmatrix} }{ \begin{vmatrix} a & b \\ c & d \end{vmatrix} } =\cfrac { { \triangle }_{ 2 } }{ { \triangle }_{ 3 } } $$

$$\therefore x={ e }^{ \left( { { \triangle }_{ 1 } }/{ { \triangle }_{ 3 } } \right) },y={ e }^{ \left( { { \triangle }_{ 2 } }/{ { \triangle }_{ 3 } } \right) }$$

Answer: D
Question 7
1 / -0

For distinct numbers $$a,b,c,x,y,z\ \epsilon R$$ if $${ \Delta }_{ 1 }\left| \begin{matrix} \left( a-x \right) ^{ 2 } & \left( b-x \right) ^{ 2 } & \left( c-x \right) ^{ 2 } \\ \left( a-y \right) ^{ 2 } & \left( b-y \right) ^{ 2 } & \left( c-y \right) ^{ 2 } \\ \left( a-z \right) ^{ 2 } & \left( b-z \right) ^{ 2 } & \left( c-z \right) ^{ 2 } \end{matrix} \right| { \Delta }_{ 2 }\left| \begin{matrix} (ax+1)^{ 2 } & (bx+1)^{ 2 } & (cx+1)^{ 2 } \\ (ay+1)^{ 2 } & (by+1)^{ 2 } & (cy+1)^{ 2 } \\ (az+1)^{ 2 } & (bz+1)^{ 2 } & (cz+1)^{ 2 } \end{matrix} \right| $$ then $$\frac { { \Delta }_{ 1 }^{ 2 } }{ { \Delta }_{ 2 }^{ 2 } } +\frac { { \Delta }_{ 2 }^{ 2 } }{ { \Delta }_{ 1 }^{ 2 } } =$$

A
$$\dfrac {5}{4}$$

B
$$\dfrac {10}{3}$$

C
$$\dfrac {1}{4}$$

D
$$None\ of\ these$$
Question 8
1 / -0

Three straight lines $$2x+11y-5=0, 24x+7y-20=0$$ and $$4x-3y-2=0$$

A
form a triangle

B
are only congruent

C
are concurrent with one line bisecting the angle between the other two

D
None of above
Question 9
1 / -0

$$\Delta =\left| \begin{matrix} 0 & i-100 & i-500 \\ 100-i & 0 & 1000-i \\ 500-i & i-1000 & 0 \end{matrix} \right|$$ is equal to

A
$$100$$

B
$$500$$

C
$$1000$$

D
$$0$$

Solution
Question 10
1 / -0

If $$p{\lambda ^4} + p{\lambda ^3} + p{\lambda ^2} + s\lambda + t = $$ $$\left| {\begin{array}{*{20}{c}}{{\lambda ^2} + 3\lambda } & {\lambda + 1} & {\lambda + 3}\\{\lambda + 1} & {2 - \lambda } & {\lambda - 4}\\{\lambda - 3} & {\lambda + 4} & {3\lambda }\end{array}} \right|$$, then value of t is

A
$$16$$

B
$$18$$

C
$$17$$

D
$$19$$

Solution

first we solve RHS then compare with LHS
$$=>(\lambda^2+3\lambda)[(2-\lambda)(3\lambda)-(\lambda+4)(\lambda-4)]-(\lambda+1)[(\lambda+1)(3\lambda)-(\lambda-3)(\lambda-4)]$$ $$+(\lambda+3)$$
$$[(\lambda+1)(\lambda+4-(\lambda-3)(2-\lambda))]$$

$$=>[-4\lambda^2+6\lambda^3+16\lambda^2-12\lambda^3+18\lambda^2+48\lambda-2\lambda^3-10\lambda^2+12\lambda-2\lambda^2-10\lambda+12+2\lambda^3+10\lambda+6\lambda^2+30]$$

$$=>[-4\lambda^4-6\lambda^3+28\lambda^2+60\lambda+18]$$

Now compare both side

$$=>[p\lambda^4+p\lambda^3+p\lambda^2+s\lambda+t=-4\lambda^4-6\lambda^3+28\lambda^2+60\lambda+18]$$

hence,$$t=18$$

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Re-Attempt Weekly Quiz Competition

Determinants Test - 51

Result

Determinants Test - 51

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SHARING IS CARING

Question 1

$$0$$

$$1$$

$$4 \log e$$

$$5 \log e$$

Solution

Question 2

vertices of right angle triangle

vertices of an equilateral triangle

vertices of an isosceles triangle

Collinear

Solution

Question 3

$$x+y$$

$${x}^{2}-xy+{y}^{2}$$

$${x}^{2}+xy+{y}^{2}$$

$${x}^{3}-{y}^{3}$$

Solution

Question 4

$$2(a+b+c)^{3}$$

$$(a-b-c)^{3}$$

$$2(a-b-c)^{3}$$

$$(a+b+c)^{3}$$

Solution

Question 5

$$-\triangle$$

$$0$$

$$\triangle$$

$$none\ of\ these$$

Solution

Question 6

$$\dfrac { { \triangle }_{ 1 } }{ { \triangle }_{ 3 } }$$ and $$\dfrac { { \triangle }_{ 2 } }{ { \triangle }_{ 3 } }$$

$$\dfrac { { \triangle }_{ 2 } }{ { \triangle }_{ 1 } }$$ and $$\dfrac { { \triangle }_{ 3 } }{ { \triangle }_{ 1 } }$$

$$\log { \left( \dfrac { { \triangle }_{ 1 } }{ { \triangle }_{ 3 } } \right) } and\log { \left( \dfrac { { \triangle }_{ 2 } }{ { \triangle }_{ 3 } } \right) }$$

$${ e }^{ { \triangle }_{ 1 }/{ \triangle }_{ 3 } }\,\ and\,\ { e }^{ { \triangle }_{ 2 }/{ \triangle }_{ 3 } }$$

Solution

Question 7

$$\dfrac {5}{4}$$

$$\dfrac {10}{3}$$

$$\dfrac {1}{4}$$

$$None\ of\ these$$

Question 8

form a triangle

are only congruent

are concurrent with one line bisecting the angle between the other two

None of above

Question 9

$$100$$

$$500$$

$$1000$$

$$0$$

Solution

Question 10

$$16$$

$$18$$

$$17$$

$$19$$

Solution

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