Determinants Test

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Determinants Test - 56

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Weekly Quiz Competition

Question 1
1 / -0

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

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

B
$$abc({ a }+{ b }+{ c })$$

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

D
$$abc\left( 1+\frac { 1 }{ a } +\dfrac { 1 }{ b } +\frac { 1 }{ c } \right) $$

Solution
Question 2
1 / -0

To solve $$x + y = 3 : 3 x - 2 y - 4 = 0$$ by determinant method find $$D.$$

A
$$5$$

B
$$1$$

C
$$-5$$

D
$$-1$$

Solution

$${ a }_{ 1 }x+{ b }_{ 1 }y-{ c }_{ 1 }=0\quad \Rightarrow { a }_{ 1 }x+{ b }_{ 1 }y={ c }_{ 1 }$$
$${ a }_{ 2 }x+{ b }_{ 2 }y-{ c }_{ 2 }=0\quad \Rightarrow { a }_{ 2 }x+{ b }_{ 2 }y={ c }_{ 2 }$$
then the solution of $$x$$ and $$y$$ can be obtained by evaluating the following integral :
$$x=\frac { \left| \underset { { c }_{ 2 } }{ { c }_{ 1 } } \quad \underset { { b }_{ 2 } }{ { b }_{ 1 } } \right| }{ \left| \underset { { a }_{ 2 } }{ { a }_{ 1 } } \quad \underset { { b }_{ 2 } }{ { b }_{ 1 } } \right| } $$ and $$y=\dfrac { \left| \underset { { a }_{ 2 } }{ { a }_{ 1 } } \quad \underset { { c }_{ 2 } }{ { c }_{ 1 } } \right| }{ \left| \underset { { a }_{ 2 } }{ { a }_{ 1 } } \quad \underset { { b }_{ 2 } }{ { b }_{ 1 } } \right| } $$
$$\therefore$$ $$x+y=3$$
$$3x-2y=4$$
can be solved using the above method
$$x=\dfrac { \left| \underset { 4 }{ 3 } \quad \underset { -2 }{ 1 } \right| }{ \left| \underset { 3 }{ 1 } \quad \underset { -2 }{ 1 } \right| } \quad ;\quad y=\dfrac { \left| \underset { 3 }{ 1 } \quad \underset { 4 }{ 3 } \right| }{ \left| \underset { 3 }{ 1 } \quad \underset { -2 }{ 1 } \right| } $$
$$x=\dfrac { -6-4 }{ -2-3 } \quad ;\quad y=\dfrac { 4-9 }{ -5 } $$
$$x=\dfrac { 10 }{ -5 } \quad ;\quad y=\dfrac { -5 }{ -5 } $$
$$x=2\quad ;\quad y=1$$
now the quantity $$\left| \underset { 3 }{ 1 } \quad \underset { -2 }{ 1 } \right| =D$$ (determinant)
$$D=\left| \underset { 3 }{ 1 } \quad \underset { -2 }{ 1 } \right| =-5$$
So, answer is option C.
Question 3
1 / -0

If $$(k,2-2k),(-k+1,2k),(-4-k,6-2k)$$ are collinear, then $$k=$$

A
$$+1$$

B
$$-1$$

C
$$-2$$

D
$$2$$

Solution

$$\begin{array}{l} where\, \, more\, than\, \, three\, \, po{ { int } }s\, are\, collinear\, slope\, of\, the\, line\, are\, same \\ So, \\ slope:\, \, \left[ { \left( { k,2-2k } \right) ,\left( { -k+1,2k } \right) } \right] =\left[ { \left( { -k+1,\, 2k } \right) ,\left( { -4-k,6-2k } \right) } \right] \\ \Rightarrow \dfrac { { 2-2k-2k } }{ { k+k-1 } } =\dfrac { { 2k-6+2k } }{ { -k+1+4+k } } \\ \Rightarrow \dfrac { { 2-4k } }{ { 2k-1 } } =\dfrac { { 4x-6 } }{ 5 } \\ \Rightarrow \left( { 1-2k } \right) 5=\left( { 2k-3 } \right) \left( { 2k-1 } \right) \end{array}$$
$$5-10k=4k^2-2k-6k+3$$
$$4k^2+2k-2=0$$
$$2k^2+k-1=0$$
$$2k^2+2k-k-1=0$$
$$2k(k+1)-1(k+1)=0$$
$$(2k-1)(k+1)=0$$
$$k=-1\, or\, \dfrac{1}{2}$$

Hence,option $$B$$ is the correct answer.
Question 4
1 / -0

If $$A = \left[ \begin{array} { l l } { 1 } & { 2 } \\ { 2 } & { 1 } \end{array} \right]$$ then $$adj (A) =?$$

A
$$\left[ \begin{array} { c c } { 1 } & { - 2 } \\ { - 2 } & { 1 } \end{array} \right]$$

B
$$\left[ \begin{array} { l l } { 2 } & { 1 } \\ { 1 } & { 1 } \end{array} \right]$$

C
$$\left[ \begin{array} { c c } { 1 } & { - 2 } \\ { - 2 } & { - 1 } \end{array} \right]$$

D
$$\left[ \begin{array} { c c } { - 1 } & { 2 } \\ { 2 } & { - 1 } \end{array} \right]$$

Solution

Given, $$A = \left[ \begin{array} { l l } { 1 } & { 2 } \\ { 2 } & { 1 } \end{array} \right]$$

The adjoin of a square matrix of order 2 is obtained by interchanging the diagonal elements & changing signs of off diagonal elements.
adj (A) = $$A = \left[ \begin{array} { l l } { 1 } & { -2 } \\ { -2 } & { 1 } \end{array} \right]$$
Question 5
1 / -0

If a,b,c are distinct and $$\left| \begin{matrix} a & { a }^{ 2 } & { a }^{ 3 }-1 \\ b & { b }^{ 2 } & { b }^{ 3 }-1 \\ c & { c }^{ 2 } & { c }^{ 3 }-1 \end{matrix} \right| =0$$ then

A
$$a+b+c=1$$

B
$$ab+bc+ca=0$$

C
$$a+b+c=0$$

D
$$abc=1$$

Solution

$$\begin{vmatrix}a & a^2 & a^3-1\\ b & b^2 & b^3-1\\ c & c^2 & c^3-1\end{vmatrix} =0$$

$$\Rightarrow a(b^2c^3-b^2-c^2b^3+c^2)-a^2(bc^3-b-cb^3+c)+(a^3-1)(bc^2-cb^2)=0$$

$$\Rightarrow a (b^2c^2(c-b)+c^2-b^2)-a^2(bc(c^2-b^2)+c-b)+(a^3-1)bc(c-b)=0$$

$$\Rightarrow a(b^2c^2(c-b)+(c-b)(c+b))-a^2(bc(c-b)(c+b)+(c-b))+bc(a^3-1)(c-b)=0$$

$$\Rightarrow a(c-b)(b^2c^2+c+b)-a^2(c-b)(bc(c+b)+1)+bc(a^3-1)(c-b)=0$$

$$\Rightarrow (c-b)\left \{ a(b^2c^2+c+b)-a^2(bc(c+b)+1)+bc(a^3-1) \right \}=0$$

$$\Rightarrow(c-b)\left \{ ab^2c^2+ac+ab-a^2bc^2-a^2b^2c-a^2+a^3bc-bc \right \}=0$$

$$\Rightarrow (c-b)\left \{ abc (bc-ac-ab+b^2)-(bc-ac-ab+a^2) \right \}=0$$

$$\Rightarrow(c-b)(bc-ac-ab+a^2)(abc-1)=0$$

$$\Rightarrow (c-b)(b(c-a)-a(c-a))(abc-1)=0$$

$$\Rightarrow (c-b)(c-a)(b-a)(abc-1)=0$$

$$\Rightarrow c-b=0 $$ or $$c-a =0$$ or $$ b-a =0$$

$$\Rightarrow c=b$$ $$ c=a$$ $$b=a$$

which is not possible since $$a, b, c$$ are distinct.

$$\therefore abc -1=0$$

$$\Rightarrow abc=1$$
Question 6
1 / -0

If f(x), g(x), h(x) are polynomials in x of degree 2 and F(x)=$$\left| \begin{matrix} f & g & h \\ { f' } & g' & h' \\ f" & g" & h" \end{matrix} \right| ,$$ , then F(x) is equal to

A
1

B
0

C
-1

D
$$f(x).g(x).h(x)$$

Solution

$$\begin{array}{l} Let, \\ f\left( x \right) =a{ x^{ 2 } }+bx+c \\ f'\left( x \right) =2ax+b \\ f''\left( x \right) =2a \\ g\left( x \right) ={ a_{ 1 } }{ x^{ 2 } }+{ b_{ 1 } }x+{ c_{ 1 } } \\ g'\left( x \right) =2{ a_{ 1 } }x+{ b_{ 1 } } \\ g''\left( x \right) =2{ a_{ 1 } } \\ and, \\ h\left( x \right) ={ a_{ 2 } }{ x^{ 2 } }+{ b_{ 2 } }x+{ c_{ 2 } } \\ h'\left( x \right) =2{ a_{ 2 } }x+{ b_{ 2 } } \\ f\left( x \right) =\left| { \begin{array} { *{ 20 }{ c } }{ a{ x^{ 2 } }+bx+c } & { { a_{ 1 } }{ x^{ 2 } }+{ b_{ 1 } }x+{ c_{ 1 } } } & { { a_{ 2 } }{ x^{ 2 } }+{ b_{ 2 } }x+{ c_{ 2 } } } \\ { 2ax+b } & { 2{ a_{ 1 } }x+{ b_{ 1 } } } & { 2{ a_{ 2 } }x+{ b_{ 2 } } } \\ { 2a } & { 2{ a_{ 1 } } } & { 2{ a_{ 2 } } } \end{array} } \right| \\ f'\left( x \right) =\left| { \begin{array} { *{ 20 }{ c } }{ 2ax+b } & { 2{ a_{ 1 } }x+{ b_{ 1 } } } & { 2{ a_{ 2 } }x+{ b_{ 2 } } } \\ { 2ax+b } & { 2{ a_{ 1 } }x+{ b_{ 1 } } } & { 2{ a_{ 2 } }x+{ b_{ 2 } } } \\ { 2a } & { 2{ a_{ 1 } } } & { 2{ a_{ 2 } } } \end{array} } \right| +0+0 \\ =0 \end{array}$$
Hence, the option $$B$$ is the correct answer.
Question 7
1 / -0

If the point $$\left(\lambda+1,1\right),\left(2\lambda+1,3\right)$$ and $$\left(2\lambda+2,2\lambda\right)$$ are collinear then the possible value of $$\lambda$$ is

A
$$2$$

B
$$1/2$$

C
$$3$$

D
$$1/3$$

Solution
Question 8
1 / -0

Solve $$\begin{vmatrix} 1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y \end{vmatrix}$$

A
$$1$$

B
$$0$$

C
$$x$$

D
$$xy$$

Solution

Let
$$\begin{array}{l}A= \left| { \begin{array} { *{ 20 }{ c } }1 & 1 & 1 \\ 1 & { 1+x } & 1 \\ 1 & 1 & { 1+y } \end{array} } \right| \\ { R_{ 1 } }\to { R_{ 1 } }-{ R_{ 3 } }\, \, \, \, \, \, \, \, \, \, \, { R_{ 2 } }\to { R_{ 2 } }-{ R_{ 3 } } \\ A=\left| { \begin{array} { *{ 20 }{ c } }0 & 0 & { -y } \\ 0 & x & { -y } \\ 1 & 1 & { 1+y } \end{array} } \right| \\ =-y\left( { 0-x } \right) \\ =xy \end{array}$$
Question 9
1 / -0

If $${ D }_{ P }=\left| \begin{matrix} P & 15 & 8 \\ { P }^{ 2 } & 35 & 9 \\ { P }^{ 3 } & 25 & 10 \end{matrix} \right| ,$$ then $${ D }_{ 1 }+{ D }_{ 2 }+{ D }_{ 3 }+{ D }_{ 4 }+{ D }_{ 5 }$$ is equal to -

A
$$-29000$$

B
$$-25000$$

C
$$25000$$

D
none of these

Solution

Given
$$D_p=\begin{vmatrix} p & 15 & 8\\ p^2 & 35 & 9\\ p^3 & 25 & 10\end{vmatrix}$$
$$=p(350-225)-15(10p^2-9p^3)+8(25p^2-35p^3)$$
$$D_1=1(350-225)-15(10(1)^2-9(1)^3)+8(25(1)^2-35(1)^3)$$
$$=125-15.(1)+8(-10)$$
$$=125-80-15$$
$$=125-95$$
$$=30$$
$$D_2=2(350-225)-15(10(2)^2-9(2)^3)+8(25(2)^2-35(2)^3)$$
$$=2.(125)-15(40-72)+8(100-280)$$
$$=250+15(32)-1440$$
$$=730-1440$$
$$=-710$$
$$D_3=3(350-225)-15(10(3)^2-9(3)^3)+8(25(3)^2-35(3)^3)$$
$$=3.(125)-15(90-243)+8(225-945)$$
$$=375+15(153)-8(720)$$
$$=375+2,295-5760$$
$$=-3090$$.
$$D_4=4(350-225)-15(10(4)^2-9(4)^3)+8(25(4)^2-35(4)^3)$$
$$=4.(125)-15(160-576)+8(400-2240)$$
$$=500+6240-8.(1840)$$
$$=5740-14720$$
$$=-8980$$
$$D_5=5(350-225)-15(10(5)^2-9(5)^3)+8(25(5)^2-35(5)^3)$$
$$=5(125)-15(250-1125)+8(625-4375)$$
$$=625+13125+(-30,000)$$
$$=-16250$$.
$$\therefore D_1+D_2+D_3+D_4+D_5=30-710-3090-8980-16250$$
$$=-29000$$.
Question 10
1 / -0

The value of determinant $$\left| \begin{matrix} { bc-a }^{ 2 } & { ac-b }^{ 2 } & ab-c^{ 2 } \\ { ac-b }^{ 2 } & { ab-c }^{ 2 } & { bc-a }^{ 2 } \\ { ab-c }^{ 2 } & { bc-a }^{ 2 } & ac-b^{ 2 } \end{matrix} \right| $$ is

A
always non-negative

B
always non-positive

C
always zero

D
can't say anything

Solution

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Re-Attempt Weekly Quiz Competition

Determinants Test - 56

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Determinants Test - 56

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Question 1

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

$$abc({ a }+{ b }+{ c })$$

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

$$abc\left( 1+\frac { 1 }{ a } +\dfrac { 1 }{ b } +\frac { 1 }{ c } \right) $$

Solution

Question 2

$$5$$

$$1$$

$$-5$$

$$-1$$

Solution

Question 3

$$+1$$

$$-1$$

$$-2$$

$$2$$

Solution

Question 4

$$\left[ \begin{array} { c c } { 1 } & { - 2 } \\ { - 2 } & { 1 } \end{array} \right]$$

$$\left[ \begin{array} { l l } { 2 } & { 1 } \\ { 1 } & { 1 } \end{array} \right]$$

$$\left[ \begin{array} { c c } { 1 } & { - 2 } \\ { - 2 } & { - 1 } \end{array} \right]$$

$$\left[ \begin{array} { c c } { - 1 } & { 2 } \\ { 2 } & { - 1 } \end{array} \right]$$

Solution

Question 5

$$a+b+c=1$$

$$ab+bc+ca=0$$

$$a+b+c=0$$

$$abc=1$$

Solution

Question 6

1

0

-1

$$f(x).g(x).h(x)$$

Solution

Question 7

$$2$$

$$1/2$$

$$3$$

$$1/3$$

Solution

Question 8

$$1$$

$$0$$

$$x$$

$$xy$$

Solution

Question 9

$$-29000$$

$$-25000$$

$$25000$$

none of these

Solution

Question 10

always non-negative

always non-positive

always zero

can't say anything

Solution

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