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CBSE Questions for Class 12 Commerce Maths Determinants Quiz 9 - MCQExams.com
CBSE
Class 12 Commerce Maths
Determinants
Quiz 9
$$\left| \begin{matrix} a & b-c & c+b \\ a+c & b & c-a \\ a-b & b+a & c \end{matrix} \right| $$ =
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$$abc({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 })$$
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$$abc({ a }+{ b }+{ c })$$
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$$a+b+c({ a }^{ 2 }+{ b^{ 2 } }+{ c^{ 2 } })$$
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$$abc\left( 1+\frac { 1 }{ a } +\dfrac { 1 }{ b } +\frac { 1 }{ c } \right) $$
If $$A = \left[ \begin{array} { l l } { 1 } & { 2 } \\ { 2 } & { 1 } \end{array} \right]$$ then $$adj (A) =?$$
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$$\left[ \begin{array} { c c } { 1 } & { - 2 } \\ { - 2 } & { 1 } \end{array} \right]$$
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$$\left[ \begin{array} { l l } { 2 } & { 1 } \\ { 1 } & { 1 } \end{array} \right]$$
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$$\left[ \begin{array} { c c } { 1 } & { - 2 } \\ { - 2 } & { - 1 } \end{array} \right]$$
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$$\left[ \begin{array} { c c } { - 1 } & { 2 } \\ { 2 } & { - 1 } \end{array} \right]$$
Explanation
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) = $$A = \left[ \begin{array} { l l } { 1 } & { -2 } \\ { -2 } & { 1 } \end{array} \right]$$
To solve $$x + y = 3 : 3 x - 2 y - 4 = 0$$ by determinant method find $$D.$$
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$$5$$
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$$1$$
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$$-5$$
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$$-1$$
Explanation
$${ a }_{ 1 }x+{ b }_{ 1 }y-{ c }_{ 1 }=0\quad \Rightarrow { a }_{ 1 }x+{ b }_{ 1 }y={ c }_{ 1 }$$
$${ a }_{ 2 }x+{ b }_{ 2 }y-{ c }_{ 2 }=0\quad \Rightarrow { a }_{ 2 }x+{ b }_{ 2 }y={ c }_{ 2 }$$
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.
The number of distinct real roots of the equation,$$\left| \begin{matrix} cosx & sinx & sinx \\ sinx & cosx & sinx \\ sinx & sinx & cosx \end{matrix} \right| =0$$
In t interval $$\left[ -\dfrac { \pi }{ 4 } \dfrac { \pi }{ 4 } \right]$$ is/are:
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$$3$$
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$$2$$
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$$1$$
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$$4$$
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
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$$2$$
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$$1/2$$
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$$3$$
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$$1/3$$
Solve $$\begin{vmatrix} 1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y \end{vmatrix}$$
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$$1$$
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$$0$$
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$$x$$
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$$xy$$
Explanation
Let
$$\begin{array}{l}A= \left| { \begin{array} { *{ 20 }{ c } }1 & 1 & 1 \\ 1 & { 1+x } & 1 \\ 1 & 1 & { 1+y } \end{array} } \right| \\ { R_{ 1 } }\to { R_{ 1 } }-{ R_{ 3 } }\, \, \, \, \, \, \, \, \, \, \, { R_{ 2 } }\to { R_{ 2 } }-{ R_{ 3 } } \\ A=\left| { \begin{array} { *{ 20 }{ c } }0 & 0 & { -y } \\ 0 & x & { -y } \\ 1 & 1 & { 1+y } \end{array} } \right| \\ =-y\left( { 0-x } \right) \\ =xy \end{array}$$
If $$(k,2-2k),(-k+1,2k),(-4-k,6-2k)$$ are collinear, then $$k=$$
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$$+1$$
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$$-1$$
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$$-2$$
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$$2$$
Explanation
$$\begin{array}{l} where\, \, more\, than\, \, three\, \, po{ { int } }s\, are\, collinear\, slope\, of\, the\, line\, are\, same \\ So, \\ slope:\, \, \left[ { \left( { k,2-2k } \right) ,\left( { -k+1,2k } \right) } \right] =\left[ { \left( { -k+1,\, 2k } \right) ,\left( { -4-k,6-2k } \right) } \right] \\ \Rightarrow \dfrac { { 2-2k-2k } }{ { k+k-1 } } =\dfrac { { 2k-6+2k } }{ { -k+1+4+k } } \\ \Rightarrow \dfrac { { 2-4k } }{ { 2k-1 } } =\dfrac { { 4x-6 } }{ 5 } \\ \Rightarrow \left( { 1-2k } \right) 5=\left( { 2k-3 } \right) \left( { 2k-1 } \right) \end{array}$$
$$5-10k=4k^2-2k-6k+3$$
$$4k^2+2k-2=0$$
$$2k^2+k-1=0$$
$$2k^2+2k-k-1=0$$
$$2k(k+1)-1(k+1)=0$$
$$(2k-1)(k+1)=0$$
$$k=-1\, or\, \dfrac{1}{2}$$
Hence,option $$B$$ is the correct answer.
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 -
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$$-29000$$
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$$-25000$$
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$$25000$$
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none of these
Explanation
Given
$$D_p=\begin{vmatrix} p & 15 & 8\\ p^2 & 35 & 9\\ p^3 & 25 & 10\end{vmatrix}$$
$$=p(350-225)-15(10p^2-9p^3)+8(25p^2-35p^3)$$
$$D_1=1(350-225)-15(10(1)^2-9(1)^3)+8(25(1)^2-35(1)^3)$$
$$=125-15.(1)+8(-10)$$
$$=125-80-15$$
$$=125-95$$
$$=30$$
$$D_2=2(350-225)-15(10(2)^2-9(2)^3)+8(25(2)^2-35(2)^3)$$
$$=2.(125)-15(40-72)+8(100-280)$$
$$=250+15(32)-1440$$
$$=730-1440$$
$$=-710$$
$$D_3=3(350-225)-15(10(3)^2-9(3)^3)+8(25(3)^2-35(3)^3)$$
$$=3.(125)-15(90-243)+8(225-945)$$
$$=375+15(153)-8(720)$$
$$=375+2,295-5760$$
$$=-3090$$.
$$D_4=4(350-225)-15(10(4)^2-9(4)^3)+8(25(4)^2-35(4)^3)$$
$$=4.(125)-15(160-576)+8(400-2240)$$
$$=500+6240-8.(1840)$$
$$=5740-14720$$
$$=-8980$$
$$D_5=5(350-225)-15(10(5)^2-9(5)^3)+8(25(5)^2-35(5)^3)$$
$$=5(125)-15(250-1125)+8(625-4375)$$
$$=625+13125+(-30,000)$$
$$=-16250$$.
$$\therefore D_1+D_2+D_3+D_4+D_5=30-710-3090-8980-16250$$
$$=-29000$$.
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
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always non-negative
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always non-positive
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always zero
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can't say anything
If $$A=\begin{bmatrix} 2 & -3 \\ -4 & 1 \end{bmatrix}$$, then $$\text{adj}$$ $$(3{ A }^{ 2 }+12A)$$ is equal to
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$$\begin{bmatrix} 51 & 63 \\ 84 & 72 \end{bmatrix}$$
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$$\begin{bmatrix} 51 & 84 \\ 63 & 72 \end{bmatrix}$$
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$$\begin{bmatrix} 72 & -63 \\ -84 & 51 \end{bmatrix}$$
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$$\begin{bmatrix} 72 & -84 \\ -63 & 51 \end{bmatrix}$$
Explanation
$$A = \begin{bmatrix} 2 & -3 \\ -4 & 1 \end{bmatrix}$$ then
$$A^2 = \begin{bmatrix} 2 & -3 \\ -4 & 1 \end{bmatrix} \, \begin{bmatrix} 2 & -3\\ -4 &1 \end{bmatrix}$$
$$A^2 = \begin{bmatrix} 16 & -9 \\ -12 & 13 \end{bmatrix}$$
$$3A^2 = \begin{bmatrix} 48 & -27 \\ -36 & 39 \end{bmatrix}$$
$$3A^2 + 12 A = \begin{bmatrix} 48 & -27 \\ -36 & 39 \end{bmatrix} + \begin{bmatrix} 24 & -36 \\ -48 & 12 \end{bmatrix} = \begin{bmatrix} 72 & -63 \\ -84 & 51 \end{bmatrix}$$
$$adj (3A^2 + 12 A) = \begin{bmatrix} 51 & 63 \\ 84 & 72 \end{bmatrix}$$
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
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$$a+b+c=1$$
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$$ab+bc+ca=0$$
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$$a+b+c=0$$
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$$abc=1$$
Explanation
$$\begin{vmatrix}a & a^2 & a^3-1\\ b & b^2 & b^3-1\\ c & c^2 & c^3-1\end{vmatrix} =0$$
$$\Rightarrow a(b^2c^3-b^2-c^2b^3+c^2)-a^2(bc^3-b-cb^3+c)+(a^3-1)(bc^2-cb^2)=0$$
$$\Rightarrow a (b^2c^2(c-b)+c^2-b^2)-a^2(bc(c^2-b^2)+c-b)+(a^3-1)bc(c-b)=0$$
$$\Rightarrow a(b^2c^2(c-b)+(c-b)(c+b))-a^2(bc(c-b)(c+b)+(c-b))+bc(a^3-1)(c-b)=0$$
$$\Rightarrow a(c-b)(b^2c^2+c+b)-a^2(c-b)(bc(c+b)+1)+bc(a^3-1)(c-b)=0$$
$$\Rightarrow (c-b)\left \{ a(b^2c^2+c+b)-a^2(bc(c+b)+1)+bc(a^3-1) \right \}=0$$
$$\Rightarrow(c-b)\left \{ ab^2c^2+ac+ab-a^2bc^2-a^2b^2c-a^2+a^3bc-bc \right \}=0$$
$$\Rightarrow (c-b)\left \{ abc (bc-ac-ab+b^2)-(bc-ac-ab+a^2) \right \}=0$$
$$\Rightarrow(c-b)(bc-ac-ab+a^2)(abc-1)=0$$
$$\Rightarrow (c-b)(b(c-a)-a(c-a))(abc-1)=0$$
$$\Rightarrow (c-b)(c-a)(b-a)(abc-1)=0$$
$$\Rightarrow c-b=0 $$ 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.
$$\therefore abc -1=0$$
$$\Rightarrow abc=1$$
If $$\begin{bmatrix} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{bmatrix}$$ is the adjoint of a $$3\times 3$$ matrix $$A$$ and $$|A|=4$$, then $$\alpha$$ is equal to:
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$$4$$
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$$11$$
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$$5$$
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$$0$$
Explanation
$$\begin{vmatrix} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{vmatrix} = 1 \left( 3 \times 4 - 3 \times 4 \right) - \alpha \left( 1 \times 4 - 3 \times 2 \right) + 3 \left( 1 \times 4 - 3 \times 2 \right) = 2 \alpha - 6$$
$$\begin{vmatrix} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{vmatrix} = \begin{vmatrix} adj. \; A \end{vmatrix}$$
$$\Rightarrow \left( 2 \alpha - 6 \right) = \begin{vmatrix} adj. \; A \end{vmatrix}$$
Now,
$$\begin{vmatrix} adj. \; A \end{vmatrix} = {\left| A \right|}^{3 - 1} = {4}^{2} = 16$$
$$\therefore 2 \alpha - 6 = 16$$
$$\Rightarrow 2 \alpha = 16 + 6$$
$$\Rightarrow \alpha = \cfrac{22}{2} = 11$$
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
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1
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0
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-1
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$$f(x).g(x).h(x)$$
Explanation
$$\begin{array}{l} Let, \\ f\left( x \right) =a{ x^{ 2 } }+bx+c \\ f'\left( x \right) =2ax+b \\ f''\left( x \right) =2a \\ g\left( x \right) ={ a_{ 1 } }{ x^{ 2 } }+{ b_{ 1 } }x+{ c_{ 1 } } \\ g'\left( x \right) =2{ a_{ 1 } }x+{ b_{ 1 } } \\ g''\left( x \right) =2{ a_{ 1 } } \\ and, \\ h\left( x \right) ={ a_{ 2 } }{ x^{ 2 } }+{ b_{ 2 } }x+{ c_{ 2 } } \\ h'\left( x \right) =2{ a_{ 2 } }x+{ b_{ 2 } } \\ f\left( x \right) =\left| { \begin{array} { *{ 20 }{ c } }{ a{ x^{ 2 } }+bx+c } & { { a_{ 1 } }{ x^{ 2 } }+{ b_{ 1 } }x+{ c_{ 1 } } } & { { a_{ 2 } }{ x^{ 2 } }+{ b_{ 2 } }x+{ c_{ 2 } } } \\ { 2ax+b } & { 2{ a_{ 1 } }x+{ b_{ 1 } } } & { 2{ a_{ 2 } }x+{ b_{ 2 } } } \\ { 2a } & { 2{ a_{ 1 } } } & { 2{ a_{ 2 } } } \end{array} } \right| \\ f'\left( x \right) =\left| { \begin{array} { *{ 20 }{ c } }{ 2ax+b } & { 2{ a_{ 1 } }x+{ b_{ 1 } } } & { 2{ a_{ 2 } }x+{ b_{ 2 } } } \\ { 2ax+b } & { 2{ a_{ 1 } }x+{ b_{ 1 } } } & { 2{ a_{ 2 } }x+{ b_{ 2 } } } \\ { 2a } & { 2{ a_{ 1 } } } & { 2{ a_{ 2 } } } \end{array} } \right| +0+0 \\ =0 \end{array}$$
Hence, the option $$B$$ is the correct answer.
If $$\left| \begin{array} { c c } { 1 } & { \sin x } & { \sin ^ { 2 } x } \\ { 1 } & { \cos x } & { \cos ^ { 2 } x } \\ { 1 } & { \tan x } & { \tan ^ { 2 } x } \end{array} \right| = 0 , x \in [ 0,2 \pi ] ,$$ then number of possible values of $$x$$ is.
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$$4$$
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$$5$$
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$$6$$
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None of these
Explanation
$$\left| \overset { 1 }{ \underset { 1 }{ 1 } } \quad \overset { sinx }{ \underset { tanx }{ cosx } } \quad \overset { { \sin }^{ 2 }x }{ \underset { { \tan }^{ 2 }x }{ { \cos }^{ 2 }x } } \right| =0$$
$$\left( \underset { { R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 3 } }{ { R }_{ 1 }\rightarrow { R }_{ 1 }-{ R }_{ 2 } } \right) $$
$$\Rightarrow \left| \overset { 0 }{ \underset { 1 }{ 0 } } \quad \quad \overset { sinx-cosx }{ \underset { tanx }{ cosx-tanx } } \quad \quad \overset { { \sin }^{ 2 }x-{ \cos }^{ 2 }x }{ \underset { { \tan }^{ 2 }x }{ { \cos }^{ 2 }x-{ \tan }^{ 2 }x } } \right| =0$$
$$\Rightarrow \underset { \left( cosx-tanx \right) }{ \left( sinx-cosx \right) } \left| \overset { 0 }{ \underset { 1 }{ 0 } } \quad \quad \overset { 1 }{ \underset { tanx }{ 1 } } \quad \quad \overset { sinx+cosx }{ \underset { { \tan }^{ 2 }x }{ cosx+tanx } } \right| =0$$
Expanding along colomn I we get
$$\Rightarrow \left( cosx-tanx \right) \left( sinx-cosx \right) \left( cosx+tanx-sinx-cosx \right) =0$$
$$\Rightarrow \left( cosx-tanx \right) \left( sinx-cosx \right) \left( tanx-sinx \right) =0$$
$$\therefore$$ $$cosx-tanx=0$$ or $$sinx=cosx$$ or $$tanx=sinx$$
or $$cosx=tanx$$ ; $$sinx=cosx$$ ; $$tanx=sinx$$
for $$x\epsilon \left[ 0,2\pi \right] $$
$$csox=tanx$$
from graph we see there are $$2$$ solutions.
for $$x\epsilon \left[ 0,2\pi \right] $$
$$sinx=cosx$$
from graph we have two solutions
for $$x\epsilon \left[ 0,2\pi \right] $$
from graph we have solutions
So a total of $$2+2+3=7$$ solutions
i.e Answer : D.
If $$\begin{vmatrix} x-4 & 2x & 2x \\ 2x & x-4 & 2x \\ 2x & 2x & x-4 \end{vmatrix}$$ = $$(A+Bx)(x-A)^2$$,
then the ordered pair $$(A , B)$$ is equal to:
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( -4 , -5 )
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( -4 , 3 )
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( -4 , 5 )
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( 4 , 5 )
Explanation
$$\left|\begin{matrix} x-4 & 2x & 2x \\ 2x & x-4 & 2x \\ 2x & 2x & x-4 \end{matrix}\right|=$$ $$(x-4)[(x-4)^2-4x^2]-2x[2x(x-4)-4x^2]+2x[4x^2-2x(x-4)]$$
$$=(x-4)[x^2+16-8x-4x^2]-2x[2x^2-8x-4x^2]+2x[4x^2-2x^2+8x]$$
$$=(x-4)[16-8x-3x^2]-2x[-8x-2x^2]+2x[2x^2+8x]$$
$$=(x-4)(16-8x-3x^2)+4x[2x^2+8x]$$
$$=16x-8x^2-3x^3-64+32x+12x^2+8x^3+32x^2$$
$$=5x^3+36x^2+48x-64$$
$$(A+Bx)(x-A)^2=(A+Bx)(x^2+A^2-2xA)$$
$$=Ax^2+A^3-2xA^2+Bx^3+BA^2x-2ABx^2$$
$$=Bx^3+(A-2AB)x^2+(BA^2-2x)x+A^3$$
$$B=5$$ $$A-2AB=36$$
$$A(1-2B)=36$$
$$A(1-10)=36$$
$$A=-4$$
$$\therefore (A,B)=(-4,5)$$ .
If $$A=\begin{vmatrix} 10 & 2 \\ 30 & 6 \end{vmatrix}\\ $$ then $$\left|A\right|=$$
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0
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10
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12
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60
Explanation
$$A=\begin{vmatrix} 10 & 2 \\ 30 & 6 \end{vmatrix}\\ $$
$$=10\times 6-30\times 2$$
$$=60-60=0$$
If $$\left|( {adj\,A}) \right| = 81,$$ for $$3 \times 3$$ matrix, then det $$A$$ is equal to
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$$1$$
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$$2$$
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$$4$$
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$$9$$
Explanation
Given for a square matrix $$A$$ of order $$3$$.
$$|adjA|=|A|^{n-1}$$, where $$n$$ is the order of matrix.
We have, $$|adj(A)|=|A|^{(3-1)}$$
or, $$|A|^2=81$$
or, $$|A|=9$$. [ Taking the positive value]
For what value of x, will the points (-1,x),(-3,2) and (-4,4) lie on a line?
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-3
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3
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-2
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2
Explanation
$$A(-1, x), B(-3, 2), C(-4, 4)$$ lie on a line
$$\therefore \dfrac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|=0$$
$$\dfrac{1}{2}\left|-1(2-4)+-3(4-x)+-4(x-2)\right|=0$$
$$\dfrac{1}{2}\left|-1(-2)-3(4-x)-4(x-2)\right|=0$$
$$\dfrac{1}{2}\left|2-3\times 4+3x-4x+4\times 2\right|=0$$
$$\dfrac{1}{2}\left|2-12+3x-4x+8\right|=0$$
$$\dfrac{1}{2}\left|-2-x\right|=0$$
$$-2-x=0$$
$$x=-2$$.
$$\left| \begin{matrix} \frac { 1 }{ a } & { a }^{ 2 } & bc \\ \frac { 1 }{ b } & { b }^{ 2 } & ca \\ \frac { 1 }{ c } & { c }^{ 2 } & ab \end{matrix} \right| =$$
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abc
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a+b+c
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0
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4abc
Explanation
Given,
$$\begin{vmatrix}\frac{1}{a}&a^2&bc\\ \frac{1}{b}&b^2&ca\\ \frac{1}{c}&c^2&ab\end{vmatrix}$$
$$=\dfrac{1}{a}\det \begin{pmatrix}b^2&ca\\ c^2&ab\end{pmatrix}-a^2\det \begin{pmatrix}\frac{1}{b}&ca\\ \frac{1}{c}&ab\end{pmatrix}+bc\det \begin{pmatrix}\frac{1}{b}&b^2\\ \frac{1}{c}&c^2\end{pmatrix}$$
$$=\dfrac{1}{a}\left(ab^3-ac^3\right)-a^2\cdot \:0+bc\left(\dfrac{c^2}{b}-\dfrac{b^2}{c}\right)$$
$$=b^3-c^3-0+bc\left(\dfrac{c^2}{b}-\dfrac{b^2}{c}\right)$$
$$=b^3+c^3-b^3-c^3$$
$$=0$$
The point $$(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{1}, y_{2})$$ & $$(x_{2}, y_{1})$$ are always
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Collinear
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Concyclic
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Vertices of a square
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Vertices of a rhombus.
Explanation
Let the coordinates be denoted as $$A(x_1,y_2)$$,
$$B(x_2,y_2)$$,
$$C(x_2,y_1)$$ and
$$D(x_1,y_1)$$
Plot the given points on a graph as above,
It is not necessary that
$$|x_2-x_1|=|y_2-y_1|$$
With $$(x_2,y_1)$$ and $$(x_1,y_2)$$ as ends of diameter $$\angle ABC=90^{\circ}$$ and $$\angle ADC=90^{\circ}$$
$$\therefore ABCD$$ are concyclic.
So, $$\text{B}$$ is the correct option.
Find the value of $$\begin{vmatrix} 5 & 3 \\ -7 & -4 \end{vmatrix}$$
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-1
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-41
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41
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1
Explanation
$$\begin{vmatrix} 5 & 3 \\ -7 & -4 \end{vmatrix}$$ $$= 5 \times (-4) - 3 \times (-7) = -20 + 21 = 1$$
Hence, the correct answer is 1.
If $$a,b,c$$ are $$pth$$,$$qth$$and $$rth$$ terms of a GP, then $$\begin{vmatrix} \log { a } & p & 1 \\ \log { b } & q & 1 \\ \log { c } & r & 1 \end{vmatrix}$$ is equal to
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$$0$$
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$$1$$
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$$\log { abc } $$
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none of these
Explanation
It is given that the given terms are in G.P
$$\therefore {p}^{th}$$ term$$=a,{q}^{th}$$ term$$=b$$ and $${r}^{th}$$ term$$=c$$
$$\Rightarrow a=A{R}^{p-1}\Rightarrow \log{a}=\log{A{\left(R\right)}^{p-1}}=\log{A}+\left(p-1\right)\log{R}$$
$$\Rightarrow b=A{R}^{q-1}\Rightarrow \log{b}=\log{A{\left(R\right)}^{q-1}}=\log{A}+\left(q-1\right)\log{R}$$
$$\Rightarrow c=A{R}^{q-1}\Rightarrow \log{c}=\log{A{\left(R\right)}^{r-1}}=\log{A}+\left(r-1\right)\log{R}$$
Now, $$\Delta=\begin{vmatrix} \log{a} & p & 1 \\ \log{b} & q & 1 \\\log{c} & r & 1 \end{vmatrix}$$
$$=\begin{vmatrix} \log{A}+\left(p-1\right)\log{R} & p & 1 \\ \log{A}+\left(q-1\right)\log{R} & q & 1\\\log{A}+\left(r-1\right)\log{R} & r & 1 \end{vmatrix}$$
$${C}_{2}\rightarrow {C}_{2}-{C}_{3}$$
$$=\begin{vmatrix} \log{A}+\left(p-1\right)\log{R} & p-1 & 1 \\ \log{A}+\left(q-1\right)\log{R} & q-1 & 1\\\log{A}+\left(r-1\right)\log{R} & r-1 & 1 \end{vmatrix}$$
$${C}_{1}\rightarrow {C}_{1}-\left(\log{A}\right){C}_{3}-\left(\log{R}\right){C}_{2}$$
$$=\begin{vmatrix} 0 & p-1 & 1 \\0 & q-1 & 1 \\0 & r-1 & 1\end{vmatrix}$$
$$=0$$
$$|A|=6$$ and $$A$$ is $$3\times 3$$ matrix then $$det(2\ adj(2(A^{-1})^{T}))=$$
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0%
$$162$$
0%
$$36$$
0%
$$216$$
0%
$$512$$
If $$\Delta{(x)} =
\begin{vmatrix}
e^x & \sin 2x & \tan x^2\\
ln(1 + x) & \cos x & \sin x\\
\cos x^2 & e^x - 1 & \sin x^2
\end{vmatrix} = A + Bx + Cx^2 + .....$$ then B is equal to
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0%
0
0%
1
0%
2
0%
4
If $$A=\begin{bmatrix} 5 & 5x & x\\ 0 & x & 5x\\ 0 & 0 & 5\end{bmatrix}$$ and $$|A^2|=25$$, then $$|x|$$ is equal to?
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0%
$$\dfrac{1}{5}$$
0%
$$5$$
0%
$$5^2$$
0%
$$1$$
Explanation
Given $$A=\begin{bmatrix} 5 & 5x & x\\ 0 & x & 5x\\ 0 & 0 & 5\end{bmatrix}$$ and $$|A^2|=25$$
$$|A^2|=|A|^2=25$$
$$|A|=5[5x-0]+0=25x$$
$$\Rightarrow (25x)^2=25$$
$$\Rightarrow x^2 =\dfrac {1}{25}$$
$$\Rightarrow |x|=\dfrac{1}{5}$$.
If a determinant of order $$3\times 3$$ is formed by using the numbers $$1$$ or $$-1$$, then the minimum value of the determinant is?
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0%
$$-2$$
0%
$$-4$$
0%
$$0$$
0%
$$-8$$
If $$\alpha, \beta$$ are the roots of $$x^2+x+1=0$$ then $$\begin{vmatrix} y+1 & \beta & \alpha\\ \beta & y+\alpha & 1\\ \alpha & 1 & y+\beta\end{vmatrix}=?$$
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0%
$$y^2-1$$
0%
$$y(y^2-1)$$
0%
$$y^2-y$$
0%
$$y^3$$
$$\begin{vmatrix} \sin ^{ 2 }{ \theta } & \cos ^{ 2 }{ \theta } \\ -\cos ^{ 2 }{ \theta } & \sin ^{ 2 }{ \theta } \end{vmatrix}=$$
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0%
$$\cos { 2\theta } $$
0%
$$\cfrac { 1 }{ 2 } \left( 1+\cos ^{ 2 }{ 2\theta } \right) $$
0%
$$\cfrac { 1 }{ 2 } \left( 1-\sin ^{ 2 }{ 2\theta } \right) $$
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$$\cfrac { 1 }{ 2 } \sin ^{ 2 }{ 2\theta } $$
Explanation
we know that
$$(sin^2x+cos^2x)^2=sin^4x+cos^4x+2sin^2xcos^2x$$
$$\begin{vmatrix} \sin ^{ 2 }{ \theta } & \cos ^{ 2 }{ \theta } \\ -\cos ^{ 2 }{ \theta } & \sin ^{ 2 }{ \theta } \end{vmatrix}=$$
$$\sin ^{ 4 }{ \theta } +\cos ^{ 4 }{ \theta } =1-2\sin ^{ 2 }{ \theta } \cos ^{ 2 }{ \theta }$$
$$ =1-\cfrac { 1 }{ 2 } \sin ^{ 2 }{ 2\theta } $$
$$=1-\cfrac { 1 }{ 2 } \left( 1-\cos ^{ 2 }{ 2\theta } \right)$$
$$ =\cfrac { 1 }{ 2 } \left( 1+\cos ^{ 2 }{ 2\theta } \right) $$
The sum of the real roots of the equation
$$\begin{vmatrix} x & -6 & -1 \\ 2 & -3x & x-3 \\ -3 & 2x & x+2 \end{vmatrix}=0$$ is equal to
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0%
$$6$$
0%
$$1$$
0%
$$0$$
0%
$$-4$$
Explanation
Given $$\begin{vmatrix} x & -6 & -1 \\ 2 & -3x & x-3 \\ -3 & 2x & x+2 \end{vmatrix}=0$$
By expansion, we get
$$x(-3x^2+6x)-(-6)(2x+4-3x+3)+(-1)(4x+9x)$$
$$\Rightarrow -5{ x }^{ 3 }+30x-30+5x=0\Rightarrow -5{ x }^{ 3 }+35x-30=0\Rightarrow { x }^{ 3 }-7x+6=0$$
$$\Rightarrow ({x-1})({x-2})({x-3})$$
$$\Rightarrow {x}=1,2,-3$$
Therefore, All roots are real
So, sum of roots $$=0$$
$$\begin{vmatrix} a+ib & c+id \\ -c+id & a-ib \end{vmatrix}=$$?
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0%
$$(a^2+b^2-c^2-d^2)$$
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$$(a^2-b^2+c^2-d^2)$$
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$$(a^2+b^2+c^2+d^2)$$
0%
$$none\ of\ these$$
For square matrices $$A$$ and $$B$$ of the same order, we have $$adj (AB) = ?$$
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0%
$$(adj\ A)(adj\ B)$$
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$$(adj\ B)(adj\ A)$$
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$$|AB|$$
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None of these
$$\begin{vmatrix} \cos { { 70 }^{ o } } & \sin { { 20 }^{ o } } \\ \sin { { 70 }^{ o } } & \cos { { 20 }^{ o } } \end{vmatrix}=$$?
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$$1$$
0%
$$0$$
0%
$$\cos 50^o$$
0%
$$\sin 50^o$$
Explanation
Let us consider $$\Delta=\begin{vmatrix} \cos { { 70 }^{ o } } & \sin { { 20 }^{ o } } \\ \sin { { 70 }^{ o } } & \cos { { 20 }^{ o } } \end{vmatrix}$$
$$\Delta =\cos 70^0\cos 20^0-\sin 70^0\sin 20^0$$
$$=\cos(70^0+20^0)$$
$$=\cos 90^0$$
$$\Delta =0$$
Option $$B$$.
Evaluate : $$\begin{vmatrix} \sin { { 23 }^{ o } } & -\sin { { 7 }^{ o } } \\ \cos { { 23 }^{ o } } & \cos { { 7 }^{ o } } \end{vmatrix}$$
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$$\dfrac {\sqrt 3}{2}$$
0%
$$\dfrac {1}{2}$$
0%
$$\sin 16^o$$
0%
$$\cos 16^o$$
If $$A = \begin{bmatrix} a& b\\c & d\end{bmatrix}$$ then $$adj\ A = ?$$
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$$\begin{bmatrix} d& -c\\ -b & a\end{bmatrix}$$
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$$\begin{bmatrix} -d& b\\ c & -a\end{bmatrix}$$
0%
$$\begin{bmatrix} d& -b\\ -c & a\end{bmatrix}$$
0%
$$\begin{bmatrix} -d& -b\\ c & a\end{bmatrix}$$
If $$A$$ is a $$3-rowed$$ square matrix and $$|A| = 5$$ then $$|adj\ A| = ?$$
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0%
$$5$$
0%
$$25$$
0%
$$125$$
0%
None of these
Explanation
$$|adj\ A| = |A|^{(n - 1)} = |A|^{2} = 5^{2} = 25$$.
$$\begin{vmatrix} \cos { { 15 }^{ o } } & \sin { { 15 }^{ o } } \\ \sin { { 15 }^{ o } } & \cos { { 15 }^{ o } } \end{vmatrix}=$$?
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$$1$$
0%
$$\dfrac {1}{2}$$
0%
$$\dfrac {\sqrt 3}{2}$$
0%
$$none\ of\ these$$
Explanation
Let us consider $$\Delta=\begin{vmatrix} \cos { { 15 }^{ o } } & \sin { { 15 }^{ o } } \\ \sin { { 15 }^{ o } } & \cos { { 15 }^{ o } } \end{vmatrix}=$$
$$\Delta=\cos^215^0-\sin^2 15^0$$
$$=\cos 30^0$$.........$$(\because \cos 2\theta=\cos^2 \theta-\sin^2 \theta)$$
$$=\dfrac{\sqrt{3}}{2}$$
Hence
$$\begin{vmatrix} \cos { { 15 }^{ o } } & \sin { { 15 }^{ o } } \\ \sin { { 15 }^{ o } } & \cos { { 15 }^{ o } } \end{vmatrix}=\dfrac{\sqrt{3}}{2}$$
Option $$C$$.
The roots of the equation
$$\begin{vmatrix} 3{ x }^{ 2 } & { x }^{ 2 }+x\cos { \theta } +\cos ^{ 2 }{ \theta } & { x }^{ 2 }+x\sin { \theta } +\sin ^{ 2 }{ \theta } \\ { x }^{ 2 }+x\cos { \theta } +\cos ^{ 2 }{ \theta } & 3\cos ^{ 2 }{ \theta } & 1+\dfrac { \sin { 2\theta } }{ 2 } \\ { x }^{ 2 }+x\sin { \theta } +\sin ^{ 2 }{ \theta } & 1+\dfrac { \cos { 2\theta } }{ 2 } & 3\sin ^{ 2 }{ \theta } \end{vmatrix}=0$$ are
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0%
$$\sin {\theta},\ \cos {\theta}$$
0%
$$\sin^2 {\theta},\ \cos^2 {\theta}$$
0%
$$\sin {\theta},\ \cos^2 {\theta}$$
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$$\sin^2 {\theta},\ \cos {\theta}$$
When the determinant $$\begin{vmatrix} \cos { 2x } & \sin ^{ 2 }{ x } & \cos { 4x } \\ \sin ^{ 2 }{ x } & \cos { 2x } & \cos ^{ 2 }{ x } \\ \cos { 4x } & \cos ^{ 2 }{ x } & \cos { 2x } \end{vmatrix}$$ is expanded in powers of $$\sin x$$, then the constant term in that expression is
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0%
$$1$$
0%
$$0$$
0%
$$-1$$
0%
$$2$$
Explanation
$$f(x)=\begin{vmatrix} \cos { 2x } & \sin ^{ 2 }{ x } & \cos { 4x } \\ \sin ^{ 2 }{ x } & \cos { 2x } & \cos ^{ 2 }{ x } \\ \cos { 4x } & \cos ^{ 2 }{ x } & \cos { 2x } \end{vmatrix}$$
the requried constant term is
$$\quad f(0)=\begin{vmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix}=\quad \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \end{vmatrix}=1(0-1)=-1$$
If the determinant $$\begin{vmatrix} \cos 2x & \sin ^{ 2 } x & \cos 4x \\ \sin ^{ 2 } x & \cos 2x & \cos ^{ 2 } x \\ \cos 4x & \cos ^{ 2 } x & \cos 2x \end{vmatrix}$$ is expanded in powers of $$\sin x$$ then the constant term in the expansion is
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0%
$$1$$
0%
$$2$$
0%
$$-1$$
0%
$$-2$$
Explanation
Let $$\begin{vmatrix}\cos 2x & \sin^2 x &\cos 4x \\ \sin^2 x &\cos 2x &\cos^2x \\\cos 4x &\cos^2x &\cos 2x\end{vmatrix}= a_{1} +a_{2}\sin ^2 x +a_{3}\sin ^3 x + ....$$ . . . . . . . . . . . $$(i)$$
The constant term in the expansion is clearly the term $$a_{1}$$.
Putting $$x=0$$ on both sides of $$(i)$$ we get-
$$\begin{vmatrix}1 & 0 & 1\\ 0 & 1 &1 \\ 1 & 1 &1\end{vmatrix} = a_{1}$$
$$\Rightarrow a_{1}=\begin{vmatrix}1 & 0 & 1\\ 0 & 1 &1 \\ 0 & 1 &0\end{vmatrix}$$ ($$R_{3} \rightarrow R_{3} - R_{1}$$)
$$\Rightarrow a_{1} = 1(0 - 1) = -1$$ (Expanding along $$C_{1}$$)
Thus, the constant term is $$-1$$.
The correct answer is option
C.
If $$|A| = 3$$ and $$A^{-1} = \begin{bmatrix}3 & -1\\ \dfrac {-5}{3} & \dfrac {2}{3}\end{bmatrix}$$ then $$adj\ A = ?$$
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0%
$$\begin{bmatrix}
9 & 3\\
-5 & -2
\end{bmatrix}$$
0%
$$\begin{bmatrix}
9 & -3\\
-5 & 2
\end{bmatrix}$$
0%
$$\begin{bmatrix}
-9 & 3\\
5 & -2
\end{bmatrix}$$
0%
$$\begin{bmatrix}
9 & -3\\
5 & -2
\end{bmatrix}$$
Explanation
$$A^{-1} = \dfrac {1}{|A|}\cdot adj\ A \rightarrow adj\ A = |A| \cdot A^{-1} = 3A^{-1} = \begin{bmatrix} 9& -3\\ -5 & 2\end{bmatrix}$$.
If $$\begin{vmatrix} a & b-c & c+b \\ a+c & b & c-a \\ a-b & a+b & c \end{vmatrix}=0$$, then the line $$ax+by+c=0$$ passes through the fixed point whcih is
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$$(1,2)$$
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$$(1,1)$$
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$$(-2,1)$$
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$$(1,0)$$
Explanation
applying $${C}_{1}\rightarrow a{C}_{!}$$ and then $${C}_{1}\rightarrow {C}_{1}+b{C}_{2}+c{C}_{3}$$ and taking $$\left( { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } \right) $$ common from $${C}_{1}$$ we get
$$\Delta =\cfrac { \left( { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } \right) }{ a } \begin{vmatrix} 1 & b-c & c+b \\ 1 & b & c-a \\ 1 & b+a & c \end{vmatrix}=\cfrac { \left( { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } \right) }{ a } \begin{vmatrix} 1 & b-c & c+b \\ 0 & c & -a-b \\ 0 & a+c & -b \end{vmatrix}$$
$$({R}_{2}\rightarrow {R}_{2}-{R}_{1},{R}_{3}\rightarrow {R}_{3}-{R}_{1})$$
$$=\cfrac { \left( { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } \right) }{ a } (-bc+{ a }^{ 2 }+ab+ac+bc)=\left( { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } \right) (a+b+c)$$
hence $$\Delta =0\Rightarrow a+b+c=0$$
therefore line $$ax+by+c=0$$ passes through the fixed point $$(1,1)$$
If $$A = \begin{bmatrix}2 & 5\\ 1 & 3\end{bmatrix}$$ then $$adj\ A = ?$$
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$$\begin{bmatrix}3 & -5\\ -1 & 2\end{bmatrix}$$
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$$\begin{bmatrix}3 & -1\\ -5 & 2\end{bmatrix}$$
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$$\begin{bmatrix}-1 & 2\\ 3 & -5\end{bmatrix}$$
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None of these
If A is singular matrix , then adj A is
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0%
singular
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non-singular
0%
symmetric
0%
not defined
Explanation
$$ A adj A = |A| I $$
$$ \Rightarrow |A adj A | = |A|^n $$ [ if A is of order $$ n \times n ] $$
$$ \Rightarrow |A| |adj A| = |A|^n $$
$$ \Rightarrow |adj A| = |A|^{n-1} $$
Now, A is singular .
$$ \therefore |A| = 0 $$
$$ \Rightarrow |adj A | = 0 $$
Hence adj A is singular.
If $$A = \left[ \begin{matrix} 1 & -1 & 2 \\ 0 & 3 & 1 \\ 0 & 0 & -1/3 \end{matrix} \right] $$, then
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$$ | A | = -1 $$
0%
$$ adj A = \left[ \begin{matrix} -1 & 1 & -2 \\ 0 & -3 & -1 \\ 0 & 0 & -1/3 \end{matrix} \right] $$
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$$ A = \left[ \begin{matrix} 1 & 1/3 & 7 \\ 0 & 1/3 & 1 \\ 0 & 0 & -3 \end{matrix} \right] $$
0%
$$ A= \left[ \begin{matrix} 1 & -1/3 & -7 \\ 0 & -3 & 0 \\ 0 & 0 & 1 \end{matrix} \right] $$
Explanation
$$ |A^{-1}| = -1 \Rightarrow |A|= -1 $$
Now, use $$ adj A=|A|A^{-1} and A=(A^{-1})^{-1} $$
There are two values of a which makes determinant $$ \Delta =\left| \begin{matrix} 1\quad & -2 & 5 \\ 2 & a & -1 \\ 0 & 4 & 2a \end{matrix} \right| =86 $$ then sum of these number is
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0%
$$ 4 $$
0%
$$ 5 $$
0%
$$ - 4 $$
0%
$$ 9 $$
Explanation
we have $$ \Delta =\left| \begin{matrix} 1\quad & -2 & 5 \\ 2 & a & -1 \\ 0 & 4 & 2a \end{matrix} \right| =86 $$
$$ \Rightarrow 1(2a^2 +4) -2(-4a -20) + 0 = 86 $$ [expanding along first column ]
$$ \Rightarrow 2a^2 +4 +8a +40 = 86 $$
$$ \Rightarrow 2a^2 + 8a +44 -86 = 0 $$
$$ \Rightarrow a^2 +4a -21 = 0 $$
$$ \Rightarrow a^2 +7a -3a-21 =0 $$
$$ \Rightarrow (a +7)(a-3) = 0 $$
$$ a = -7 $$ and $$ 3 $$
$$ \therefore $$ Required sum = $$ -7 +3 = -4 $$
If $$ \begin{vmatrix} 2x & 5 \\ 8 & x \end{vmatrix}=\begin{vmatrix} 6 & -2 \\ 7 & 3 \end{vmatrix}$$ then value of $$ x $$ is
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0%
$$ 3 $$
0%
$$ \pm 3 $$
0%
$$ \pm 6 $$
0%
$$ 6 $$
Explanation
$$ \because \begin{vmatrix} 2x & 5 \\ 8 & x \end{vmatrix}=\begin{vmatrix} 6 & -2 \\ 7 & 3 \end{vmatrix} $$
$$ \Rightarrow 2x^2 -40 = 18 +14 $$
$$ \Rightarrow 2x^2 = 32 +40 $$
$$ \Rightarrow x^2 = \frac {72}{2} = 36 $$
$$ \therefore x = \pm 6 $$
Choose the correct answer from the given alternatives in the following question:
If $$ A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} , {\text{adj}} A = \begin{bmatrix} 4 & a \\ -3 & b \end{bmatrix} $$, then the values of $$a$$ and $$b$$ are
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0%
$$ a = -2 , b = 1 $$
0%
$$ a = 2 , b = 4$$
0%
$$ a = 2 , b = -1 $$
0%
$$ a = 1 , b = -2 $$
Choose the correct answer from the given alternatives in the following question:
If $$ A = \begin{bmatrix} 2 & -4 \\ 3 & 1 \end{bmatrix} $$, then the adjoint of matrix $$A$$ is
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0%
$$ \begin{bmatrix} -1 & 3 \\ -4 & 1 \end{bmatrix} $$
0%
$$ \begin{bmatrix} 1 & 4 \\ -3 & 2 \end{bmatrix} $$
0%
$$ \begin{bmatrix} 1 & 3 \\ 4 & -2 \end{bmatrix} $$
0%
$$ \begin{bmatrix} -1 & -3 \\ -4 & 2 \end{bmatrix} $$
State whether true or false:
If the value of a third order determinant is $$12 $$, then the value of determinant formed by replacing each element by its co-factor will be $$ 144 $$
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0%
True
0%
False
Explanation
Let A is determinant
$$ \therefore |A| = 12 $$
Also we know that if $$ A $$ is a square matrix of order $$ n $$ then $$ | adj A| = |A|^{n-1} $$
For $$ n = 3 , |Adj A| = |A|^{3-1} =|A|^2 $$
$$ =(12)^2 = 144 $$
Choose the correct answer from the given alternatives in the following question:
If $$ A = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} $$ and $$ A ({\text{adj}} A) = KI $$ , then the value of $$k$$ is
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0%
$$1$$
0%
$$-1$$
0%
$$0$$
0%
$$-3$$
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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