Determinants Test

Result

Determinants Test - 23

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The point $$(-a,-b),(0,0),(a,b)$$ and $$(a^{2},ab)$$ are-

A
collinear

B
concyclic

C
vertices of a rectangle

D
vertices of a parallelogram

Solution

The equation of line passing through points $$\left( {0,0} \right)$$ and $$\left( {a,b} \right)$$ is given by

$$\begin{array}{l} L:\left( { y-0 } \right) =\frac { { b-0 } }{ { a-0 } } \left( { x-0 } \right) \\ \Rightarrow y=\frac { b }{ a } x \\ \Rightarrow ay=bx \\ \Rightarrow L:bx-ay=0 \end{array}$$

Putting $$\left( { - a, - b} \right)$$ in $$R.H.S$$

we get

$$L: - ab - a\left( { - b} \right)$$ Satisfied $$L$$ $$\left( { - a, - b} \right)$$ lies on $$L$$

Now Putting  $$\left( { { a^{ 2 } },ab } \right) $$ in $$R.H.S$$

we get

$$L:b\left( { { a^{ 2 } } } \right) -a\left( { ab } \right) -{ a^{ 2 } }b=0$$

Therefore, $$\left( { { a^{ 2 } },ab } \right) $$ Satisfies $$L$$

So $$\left( { { a^{ 2 } },ab } \right) $$ lies on $$L$$

Hence The points  $$\left( { 0,0 } \right) ,\left( { a,b } \right) ,\left( { -a,-b } \right) ,\left( { { a^{ 2 } },ab } \right) $$ are Collinear. So the option $$(A)$$ is the correct answer.
Question 2
1 / -0

$$A=\begin{bmatrix} 5 & 5a & a \\ 0 & a & 5a \\ 0 & 0 & 5 \end{bmatrix}$$ If $$\left| A^{ 2 } \right| =25$$ then $$|a|=$$

A
$$5$$

B
$$5^{2}$$

C
$$1$$

D
$$\dfrac{1}{5}$$

Solution

$$A=\left[ { \begin{array} { *{ 20 }{ c } }5 & { 5a } & a \\ 0 & a & { 5a } \\ 0 & 0 & 5 \end{array} } \right] $$
$$\left| A \right| =\left[ { \begin{array} { *{ 20 }{ c } }5 & { 5a } & a \\ 0 & a & { 5a } \\ 0 & 0 & 5 \end{array} } \right] =25a$$
$$|A^2|=|A|^2=25$$
$$(25a)^2=25$$
$$a=\pm \frac{1}{5}$$
$$|a|=\frac{1}{5}$$
Question 3
1 / -0

The value of (adj $$A$$) is equal to

A
$$2A$$

B
$$4A$$

C
$$8A$$

D
$$16A$$

Solution

The value of (adj A) is equal to $$2A$$.
Option $$A$$ is correct answer.
Question 4
1 / -0

Two points $$(a, 0)$$ and $$(0, b)$$ are joined by a straight line. Another point on this line is

A
$$(3a, -2b)$$

B
$$({a}^{2}, ab)$$

C
$$(-3a, 2b)$$

D
$$(a, b)$$

Solution

Given that $$(a,0)$$ and $$(0,b)$$ lie on a straight line.

We know that a straight line is represented by $$y=mx+c$$
Substituting co-ordinates in equation, we get
$$(1) 0 = am + c$$
$$(2) b = a(0) + c = c$$

$$\Rightarrow m = \dfrac{-b}{a} , c = b$$

$$\therefore$$ Equation of line is $$ay = ab - bx$$

Substituting options we see that $$(3a,-2b)$$ lies on this line
Question 5
1 / -0

First row of the matrix $$A$$ is $$\begin{bmatrix}1& 3 & 2\end{bmatrix}$$. If $$adj (A)$$ =\begin{bmatrix}-2 & 4 & a\\ -1& 2 & 1\\ 3a& -5 &-2 \end{bmatrix} then a possible value of $$det(A)$$ is

A
$$1$$

B
$$2$$

C
$$-1$$

D
$$-2$$

Solution

$$|A|=a_{11}A_{11}+a_{12}A_{21}+a_{13}A_{31}$$

$$|A|=1(-2)+3(-1)+2(3\alpha)$$

$$|A|=-5+6\alpha$$

$$adj \ A =\begin{bmatrix} -2 & 4 & \alpha \\ -1 & 2 & 2 \\ 3\alpha & -5 & -2 \end{bmatrix}$$

$$|adj \ A|=-2(-4+5)-4(2-3\alpha)+\alpha(5-6\alpha)$$

$$=-2-8+12\alpha +5\alpha - 6\alpha^2$$

$$|A|^2=-6\alpha^2 +17\alpha-10$$

$$(6\alpha-5)^2=-6\alpha^2 +17\alpha-10$$

$$36\alpha^2 +25-60\alpha = -6\alpha^3 +17\alpha -10$$

$$42 \alpha^2 -77\alpha +35=0$$

$$6\alpha^2-11\alpha+5=0$$

$$(6\alpha-5)(\alpha-1)=0$$

$$\alpha=\dfrac{5}{6}$$ or $$\alpha=1$$

$$|A|=-5+6\alpha$$

When, $$\alpha=\dfrac{5}{6}, |A|=0$$

When, $$\alpha=1, |A|=1$$
Question 6
1 / -0

$$\begin{vmatrix} 2^3 & 3^3 & 3.2^2+3.2+1\\ 3^3 & 4^3 & 3.3^2+3.3+1\\ 4^3 & 5^3 & 3.4^2+3.4+1\end{vmatrix}$$ is equal to?

A
$$0$$

B
$$1$$

C
$$92$$

D
None of these

Solution

$$\begin{array}{l} \left| { \begin{array} { *{ 20 }{ c } }{ { 2^{ 3 } } } & { { 3^{ 3 } } } & { 3\cdot { 2^{ 2 } }+3\cdot 2+1 } \\ { { 3^{ 3 } } } & { { 4^{ 3 } } } & { 3\cdot { 3^{ 2 } }+3\cdot 3+1 } \\ { { 4^{ 3 } } } & { { 5^{ 3 } } } & { 3\cdot { 4^{ 2 } }+3\cdot 4+1 } \end{array} } \right| \\ \left| { \begin{array} { *{ 20 }{ c } }{ { 2^{ 3 } } } & { { 3^{ 3 } } } & { { 3^{ 3 } }-{ 2^{ 3 } } } \\ { { 3^{ 3 } } } & { { 4^{ 3 } } } & { { 4^{ 3 } }-{ 3^{ 3 } } } \\ { { 4^{ 3 } } } & { { 5^{ 3 } } } & { { 5^{ 3 } }-{ 4^{ 3 } } } \end{array} } \right| \\ { C_{ 3 } }\to { C_{ 3 } }+{ C_{ 1 } } \\ \left| { \begin{array} { *{ 20 }{ c } }{ { 2^{ 3 } } } & { { 3^{ 3 } } } & { { 3^{ 3 } } } \\ { { 3^{ 3 } } } & { { 4^{ 3 } } } & { { 4^{ 3 } } } \\ { { 4^{ 3 } } } & { { 5^{ 3 } } } & { { 5^{ 3 } } } \end{array} } \right| =0 \end{array}$$
Question 7
1 / -0

If $$P=\begin{bmatrix} 1 & \alpha & 3\\ 1 & 3 & 3\\ 2 & 4 & 4\end{bmatrix}$$ is a $$3\times 3$$ matrix A and $$|A|=4$$, then $$\alpha$$ is equal to?

A
$$4$$

B
$$11$$

C
$$5$$

D
$$0$$

Solution
Question 8
1 / -0

The determinant $$\left| {\begin{array}{*{20}{c}}a&b&{a\alpha + b}\\b&c&{b\alpha + c}\\{a\alpha + b}&{b\alpha + c}&0\end{array}} \right|$$ is equal to zero, if-

A
$$a, b, c$$ are in AP

B
$$a, b, c$$ are in GP

C
$$\alpha $$ is a root of the equation $$a{x^2} + bx + c = 0$$

D
$$\left( {x - \alpha } \right)$$ is a factor of $$a{x^2} + 2bx + c$$

Solution

$$\begin{vmatrix} a & b & a\alpha +b \\ b & c & b\alpha +c \\ a\alpha +b & b\alpha +c & 0 \end{vmatrix}=0$$
By expanding matrix,
$$a(c(0)-(b\alpha+c)(b\alpha+c))-b(b(0)-(a\alpha+b)(b\alpha+c))+(a\alpha+b)(b(b\alpha+c)-c(a\alpha+b))=0$$
$$(a\alpha+b)(b^2\alpha+bc-ac\alpha-bc)=a(b\alpha+c)^2-b(a\alpha+b)(b\alpha+c)$$
$$(a\alpha+b)(b^2\alpha-ac\alpha)=(a(b\alpha+c)-b(a\alpha+b))(b\alpha+c)$$
$$(a\alpha+b)((b^2-ac)\alpha)=(ab\alpha+ac-ba\alpha+b^2))(b\alpha+c)$$
$$(a\alpha+b)(b^2-ac)\alpha=(ac-b^2)(b\alpha+c)$$
$$(a\alpha+b)(-1)\alpha=(b\alpha+c)$$
$$-a\alpha^2-b\alpha=b\alpha+c$$
$$a\alpha^2+2b\alpha+c=0$$
Hence , $$(x-\alpha) $$ is a factor of $$ax^2+2bx+c$$
Question 9
1 / -0

$$A = \left[ \begin{array}{l}1\,\,\,\,\,\,\,\,1\\3\,\,\,\,\,\,\,4\end{array} \right]$$ and $$A\left( {adj\,A} \right) = KI$$, then the value of $$'K'$$

A
$$1$$

B
$$-2$$

C
$$10$$

D
$$-10$$

Solution
Question 10
1 / -0

The value of determinant $$ \begin{vmatrix} 19 & 6 & 7 \\ 21 & 3 & 15 \\ 28 & 11 & 6 \end{vmatrix} $$ is :

A
$$ 150 $$

B
$$-110$$

C
$$0$$

D
None of these

Solution

$$ \begin{vmatrix} 19 & 6 & 7 \\ 21 & 3 & 15 \\ 28 & 11 & 6 \end{vmatrix} $$
$$ = 19(3\times 6-11\times 15 ) $$
$$ -6(21\times 6-15\times 28) $$
$$ +7(21\times 11-3\times 28) $$
$$ = 19(-147)-6(-294) $$
$$ +7(147) $$
$$ = 0 $$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Determinants Test - 23

Result

Determinants Test - 23

Score

-

Rank

-

SHARING IS CARING

Question 1

collinear

concyclic

vertices of a rectangle

vertices of a parallelogram

Solution

Question 2

$$5$$

$$5^{2}$$

$$1$$

$$\dfrac{1}{5}$$

Solution

Question 3

$$2A$$

$$4A$$

$$8A$$

$$16A$$

Solution

Question 4

$$(3a, -2b)$$

$$({a}^{2}, ab)$$

$$(-3a, 2b)$$

$$(a, b)$$

Solution

Question 5

$$1$$

$$2$$

$$-1$$

$$-2$$

Solution

Question 6

$$0$$

$$1$$

$$92$$

None of these

Solution

Question 7

$$4$$

$$11$$

$$5$$

$$0$$

Solution

Question 8

$$a, b, c$$ are in AP

$$a, b, c$$ are in GP

$$\alpha $$ is a root of the equation $$a{x^2} + bx + c = 0$$

$$\left( {x - \alpha } \right)$$ is a factor of $$a{x^2} + 2bx + c$$

Solution

Question 9

$$1$$

$$-2$$

$$10$$

$$-10$$

Solution

Question 10

$$ 150 $$

$$-110$$

$$0$$

None of these

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.