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CBSE Questions for Class 12 Commerce Maths Determinants Quiz 2 - MCQExams.com
CBSE
Class 12 Commerce Maths
Determinants
Quiz 2
If two rows of a determinant are identical, then what is the value of the determinant ?
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0
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1
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-1
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Can be any real value.
Explanation
Let determinant of this matrix is $$x$$, if we interchange the two identical rows of the matrix then by property the determinant of the new matrix is $$-x$$, but overall the matrix will be same as we have interchanged only the two identical rows.
So, $$x=-x$$, we have $$x=0$$ .
Hence, the determinant is zero.
The value of k for which $$kx+3y-k+3=0$$ and $$12x+ky=k$$, have infinite solutions, is?
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$$0$$
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$$-6$$
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$$6$$
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$$1$$
Explanation
For infinite many solution $$\displaystyle\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$$
$$\Rightarrow \displaystyle\dfrac{k}{12}=\dfrac{3}{k}=\dfrac{-(k-3)}{-k}, \dfrac{3}{k}=\dfrac{(k-3)}{k}$$
$$k^2=36$$ $$k-6=0$$
$$k=\pm 6$$ $$k=6$$.
The number of line segments possible with three collinear points is ________.
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$$1$$
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$$2$$
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$$3$$
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Infinite
Explanation
Let three collinear points be $$ A , B , C$$
They can represent three line segments namely, $$AB, BC, AC$$.
Thus namely $$3$$ line segments are possible with three collinear points.
For positive numbers $$x, y$$ and $$z$$ the numerical value of the determinant $$\begin{vmatrix} 1 & \log_x y & \log_x z \\ \log_y x & 1 & \log_y z \\ \log_z x & \log_z y & 1 \end{vmatrix}$$ is
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$$0$$
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$$1$$
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$$\log_e xyz$$
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$$-\log_{e} xyz$$
Explanation
Let $$D=\begin{vmatrix} 1 & \log_x y & \log_x z \\ \log_y x & 1 & \log_y z \\ \log_z x & \log_z y & 1 \end{vmatrix}$$
$$=1(1\times 1-\log_yz\times \log_zy)-\log_xy(\log_yx\times 1-\log_zx\times \log_yz)+\log_xz(\log_yx\times \log_zy-1\times \log_zx)$$
Since, $$\log_ab=\dfrac{\log a}{\log b}$$
So simplifying above further we get
$$\log_yz\times \log_zy=\dfrac{\log z}{\log y}\times \dfrac{\log y}{\log z}=1$$
$$\Rightarrow D=1(1-1)-\log_xy(\log_yx-\log_yx)+\log_xz(\log_zx-\log_zx)$$
$$\Rightarrow D=0$$
If any two adjacent rows or columns of a determinant are interchanged in position, the value of the determinant :
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Becomes zero
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Remains the same
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Changes its sign
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Is doubled
Explanation
If two rows or columns are interchanged
then the value of determinant changes its sign
$$A=\begin{vmatrix} a & b \\ c & d \end{vmatrix}$$
Interchanging column
$$B=\begin{vmatrix} b & a \\ d & c \end{vmatrix}\\ A=-B$$
Similarly, interchanging rows will give
$$C=\begin{vmatrix} c & d \\ a & b \end{vmatrix}\\ A=-C$$
Hence, C is correct.
If $$a, b, c$$ are non-zero and different from $$1$$, then the value of $$\begin{vmatrix}\log_a 1 & \log_a b & \log_ac\\ \log_a \left( \dfrac{1}{b} \right ) & \log_b 1 &\log_a \left( \dfrac{1}{c} \right ) \\ \log_a \left( \dfrac{1}{c}\right ) & \log_a c & \log_c 1\end{vmatrix}$$ is
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$$0$$
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$$1 + \log_a (a +b + c)$$
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$$\log_a (ab + bc + ca)$$
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$$1$$
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$$\log_a(a + b + c)$$
Explanation
$$\Delta =\left| \begin{matrix} \log _{ a }{ 1 } & \log _{ a }{ b } & \log _{ a }{ c } \\ \log _{ a }{ \dfrac { 1 }{ b } } & \log _{ b }{ 1 } & \log _{ a }{ \dfrac { 1 }{ c } } \\ \log _{ a }{ \dfrac { 1 }{ c } } & \log _{ a }{ c } & \log _{ c }{ 1 } \end{matrix} \right| \\ \Rightarrow \Delta =\left| \begin{matrix} 0 & \log _{ a }{ b } & \log _{ a }{ c } \\ -\log _{ a }{ b } & 0 & -\log _{ a }{ c } \\ -\log _{ a }{ c } & \log _{ a }{ c } & 0 \end{matrix} \right| \\ \Rightarrow \Delta =0-\log _{ a }{ b } \{ 0-{ (\log _{ a }{ c) } }^{ 2 }\} +\log _{ a }{ c } \{ -\log _{ a }{ b } \log _{ a }{ c } -0\} \\ \Rightarrow \Delta =\log _{ a }{ b } { (\log _{ a }{ c) } }^{ 2 }-\log _{ a }{ b } { (\log _{ a }{ c) } }^{ 2 }=0$$
So, option A is correct.
The points (2, -3), (4,3) and (5, k/2) are on the same straight line. The value(s) of k is (are):
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$$12$$
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$$-12$$
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$$\pm 12$$
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$$12$$ or $$6$$
Explanation
Given (2, -3), (4, 3) and (5, $$\cfrac{k}{2}$$) are on same straight line.
$$\therefore \cfrac { 3-(-3) }{ 4-2 } =\cfrac { \cfrac { k }{ 2 } -3 }{ 5-4 } \Longrightarrow \cfrac { 6 }{ 2 } =\cfrac { \cfrac { k }{ 2 } -3 }{ 1 } \Longrightarrow 3=\cfrac { k }{ 2 } -3\\ \therefore k=12$$
Let a be the square matrix of order 2 such that $$A^2 - 4A + 4I =0$$ where I is an identify matrix of order .If$$ B = A ^5 - 4A^4 + 6 A^3 + 4A^2 + A $$ then Det (B) is equal to
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162
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$$(162)^2$$
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256
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$$(256)^2$$
Explanation
$$A^2 - 4A + 4I = O \Rightarrow ( A- 2 I)^2 = O \Rightarrow A = 2 I$$
$$ B = A (A^4 \, + \, 4A^3\, + \, 6A^2 \, + \, 4A +I) = A(A \, + \, I)^4 = 2I (3I)^4=162 I$$
det$$(B) = (162)^2$$
If $$A$$ is a skew symmetric matrix, then $$\left| A \right| $$ is
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$$1$$
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$$-1$$
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$$0$$
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none
Explanation
SINCE THE SKEW SYMMETRIC MATRIX CONSIST OF ELEMENTS OF OPPOSITE SIGN AT OPPOSITE SIDE OF MATRIX DIAGONAL WITH ALL THE DIAGONAL ELEMENTS AS ZERO THEREFORE THE DETERMINANT OF SKEW SYMMETRIC MATRIX IS ZERO.
The point $$(-a,-b),(0,0),(a,b)$$ and $$(a^{2},ab)$$ are-
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collinear
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concyclic
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vertices of a rectangle
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vertices of a parallelogram
Explanation
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.
Let $$\omega\neq{1}$$ be a cube root of unity and $$S$$ be the set of all non-singular matrices of the form $$ \begin{bmatrix} 1 & a & b \\ \omega & 1 & c \\ { \omega }^{ 2 } & \omega & 1 \end{bmatrix}$$Where each of $$a,\ b$$ and $$c$$ is either $$\omega$$ or $${\omega}^{2}$$. Then the number of distinct matrices in the set $$S$$ is
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$$2$$
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$$6$$
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$$4$$
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$$8$$
Explanation
For the given matrix to be non-singular
$$1-(a+c)w+acw^2 \ne 0$$
$$(1-aw)(1-cw) \ne 0$$
$$a \ne w^2$$ and $$c \ne w^2$$
where $$w$$ is complex cube root of unity
And $$a,b,c$$ are complex cube root of unity
therefore, $$a$$ and $$c$$ can take two values $$w$$ and $$w^2$$
Hence, total numbers of distinct matrices
$$=1 \times 1\times 2=2$$
$$A=\begin{bmatrix} 5 & 5a & a \\ 0 & a & 5a \\ 0 & 0 & 5 \end{bmatrix}$$ If $$\left| A^{ 2 } \right| =25$$ then $$|a|=$$
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$$5$$
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$$5^{2}$$
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$$1$$
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$$\dfrac{1}{5}$$
Explanation
$$A=\left[ { \begin{array} { *{ 20 }{ c } }5 & { 5a } & a \\ 0 & a & { 5a } \\ 0 & 0 & 5 \end{array} } \right] $$
$$\left| A \right| =\left[ { \begin{array} { *{ 20 }{ c } }5 & { 5a } & a \\ 0 & a & { 5a } \\ 0 & 0 & 5 \end{array} } \right] =25a$$
$$|A^2|=|A|^2=25$$
$$(25a)^2=25$$
$$a=\pm \frac{1}{5}$$
$$|a|=\frac{1}{5}$$
If $$a\neq6,b,c$$ satisfy $$\begin{vmatrix} a&2b&2c \\ 3&b&c \\ 4&a&b \end{vmatrix}=0,$$ then $$abc =$$
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$$a+b+c$$
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$$0$$
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$$b^{3}$$
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$$ab +b -c$$
Explanation
Expanding the Determinant we get,
$$a(b^2-ac)-2b(3b-4c)+2c(3a-4b)=0\Rightarrow ab^2-a^2c-6b^2+8bc+6ca-8bc=0\Rightarrow ab^2-6b^2=a^2c-6ca\\\Rightarrow (a-6)b^2=(a-6)ca\Rightarrow b^2=ca$$
Multilpying $$b$$ both sides we get
$$b^3=abc$$
The value of (adj $$A$$) is equal to
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$$2A$$
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$$4A$$
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$$8A$$
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$$16A$$
Explanation
The value of (adj A) is equal to $$2A$$.
Option $$A$$ is correct answer.
Two points $$(a, 0)$$ and $$(0, b)$$ are joined by a straight line. Another point on this line is
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$$(3a, -2b)$$
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$$({a}^{2}, ab)$$
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$$(-3a, 2b)$$
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$$(a, b)$$
Explanation
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
$$\begin{vmatrix} x & y & z \\ x^2 & y^2 & z^2 \\ x^3 & y^3 & z^3 \end{vmatrix} =xyz(x-y)(y-z)(z-x)$$
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True
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False
Explanation
Let $$A=\begin{vmatrix} x & y & z \\ x^2 & y^2 & z^2 \\ x^3 & y^3 & z^3 \end{vmatrix}$$
Taking $$x,y,z$$ common from $$C_{1},C_{2},C_{3}$$ respectively, we get,
$$A=xyz\begin{vmatrix} 1 & 1 & 1 \\ x & y & z \\ x^2 & y^2 & z^2 \end{vmatrix}$$
Applying $$C_{2}\rightarrow C_{2}-C_{1}$$ and $$C_{3}\rightarrow C_{3}-C_{1}$$, we get,
$$A=xyz\begin{vmatrix} 1 & 0 & 0 \\ x & y-x & z-x \\ x^2 & y^2-x^2 & z^2-x^2 \end{vmatrix}$$
Taking $$(y-x)$$ and $$(z-x)$$ common from $$C_{2}$$ and from $$C_{3}$$
$$A=xyz(y-x)(z-x)\begin{vmatrix} 1 & 0 & 0 \\ x & 1 & 1 \\ x^2 & y+x & z+x \end{vmatrix}$$
Expanding along $$R_{1}$$,
$$A=xyz(y-x)(z-x)(z+x-y-x)$$
$$A=xyz(x-y)(y-z)(z-x)$$
$$\begin{vmatrix} 2^3 & 3^3 & 3.2^2+3.2+1\\ 3^3 & 4^3 & 3.3^2+3.3+1\\ 4^3 & 5^3 & 3.4^2+3.4+1\end{vmatrix}$$ is equal to?
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$$0$$
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$$1$$
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$$92$$
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None of these
Explanation
$$\begin{array}{l} \left| { \begin{array} { *{ 20 }{ c } }{ { 2^{ 3 } } } & { { 3^{ 3 } } } & { 3\cdot { 2^{ 2 } }+3\cdot 2+1 } \\ { { 3^{ 3 } } } & { { 4^{ 3 } } } & { 3\cdot { 3^{ 2 } }+3\cdot 3+1 } \\ { { 4^{ 3 } } } & { { 5^{ 3 } } } & { 3\cdot { 4^{ 2 } }+3\cdot 4+1 } \end{array} } \right| \\ \left| { \begin{array} { *{ 20 }{ c } }{ { 2^{ 3 } } } & { { 3^{ 3 } } } & { { 3^{ 3 } }-{ 2^{ 3 } } } \\ { { 3^{ 3 } } } & { { 4^{ 3 } } } & { { 4^{ 3 } }-{ 3^{ 3 } } } \\ { { 4^{ 3 } } } & { { 5^{ 3 } } } & { { 5^{ 3 } }-{ 4^{ 3 } } } \end{array} } \right| \\ { C_{ 3 } }\to { C_{ 3 } }+{ C_{ 1 } } \\ \left| { \begin{array} { *{ 20 }{ c } }{ { 2^{ 3 } } } & { { 3^{ 3 } } } & { { 3^{ 3 } } } \\ { { 3^{ 3 } } } & { { 4^{ 3 } } } & { { 4^{ 3 } } } \\ { { 4^{ 3 } } } & { { 5^{ 3 } } } & { { 5^{ 3 } } } \end{array} } \right| =0 \end{array}$$
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-
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$$a, b, c$$ are in AP
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$$a, b, c$$ are in GP
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$$\alpha $$ is a root of the equation $$a{x^2} + bx + c = 0$$
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$$\left( {x - \alpha } \right)$$ is a factor of $$a{x^2} + 2bx + c$$
Explanation
$$\begin{vmatrix} a & b & a\alpha +b \\ b & c & b\alpha +c \\ a\alpha +b & b\alpha +c & 0 \end{vmatrix}=0$$
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 $$ax^2+2bx+c$$
$$A = \left[ \begin{array}{l}1\,\,\,\,\,\,\,\,1\\3\,\,\,\,\,\,\,4\end{array} \right]$$ and $$A\left( {adj\,A} \right) = KI$$, then the value of $$'K'$$
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$$1$$
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$$-2$$
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$$10$$
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$$-10$$
$$D=\begin{vmatrix} 18 & 40 & 89 \\ 40 & 89 & 198 \\ 89 & 198 & 440 \end{vmatrix}=$$
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$$1$$
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$$-1$$
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zero
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$$2$$
Explanation
We have,
$$D=\left| \begin{matrix} 18 & 40 & 89 \\ 40 & 89 & 198 \\ 89 & 198 & 440 \end{matrix} \right| $$
$$ D = 18\times (89\times 440 - 198^{2})- 40\times (40\times 440-198\times 89) + 89 \times(40\times 198 - 89^{2}) $$
$$ D = 18\times (39160 - 39204)- 40\times (17600-17622) + 89 \times(7920 - 7921) $$
$$ D = 18\times (-44)- 40\times (-22) + 89 \times(-1) $$
$$ D = -792+880 -89$$
$$D =-1$$
Therefore, the value of the determinant is $$ -1$$.
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
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$$1$$
0%
$$2$$
0%
$$-1$$
0%
$$-2$$
Explanation
$$|A|=a_{11}A_{11}+a_{12}A_{21}+a_{13}A_{31}$$
$$|A|=1(-2)+3(-1)+2(3\alpha)$$
$$|A|=-5+6\alpha$$
$$adj \ A =\begin{bmatrix} -2 & 4 & \alpha \\ -1 & 2 & 2 \\ 3\alpha & -5 & -2 \end{bmatrix}$$
$$|adj \ A|=-2(-4+5)-4(2-3\alpha)+\alpha(5-6\alpha)$$
$$=-2-8+12\alpha +5\alpha - 6\alpha^2$$
$$|A|^2=-6\alpha^2 +17\alpha-10$$
$$(6\alpha-5)^2=-6\alpha^2 +17\alpha-10$$
$$36\alpha^2 +25-60\alpha = -6\alpha^3 +17\alpha -10$$
$$42 \alpha^2 -77\alpha +35=0$$
$$6\alpha^2-11\alpha+5=0$$
$$(6\alpha-5)(\alpha-1)=0$$
$$\alpha=\dfrac{5}{6}$$ or $$\alpha=1$$
$$|A|=-5+6\alpha$$
When, $$\alpha=\dfrac{5}{6}, |A|=0$$
When, $$\alpha=1, |A|=1$$
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?
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$$4$$
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$$11$$
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$$5$$
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$$0$$
The value of determinant $$ \begin{vmatrix} 19 & 6 & 7 \\ 21 & 3 & 15 \\ 28 & 11 & 6 \end{vmatrix} $$ is :
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$$ 150 $$
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$$-110$$
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$$0$$
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None of these
Explanation
$$ \begin{vmatrix} 19 & 6 & 7 \\ 21 & 3 & 15 \\ 28 & 11 & 6 \end{vmatrix} $$
$$ = 19(3\times 6-11\times 15 ) $$
$$ -6(21\times 6-15\times 28) $$
$$ +7(21\times 11-3\times 28) $$
$$ = 19(-147)-6(-294) $$
$$ +7(147) $$
$$ = 0 $$
$$\left| \begin{matrix} 1+i & 1-i & i \\ 1+i & i & 1+i \\ i & 1+i & 1-i \end{matrix} \right| $$ (where $$i=\sqrt {-1}$$) equals.
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$$5i-2$$
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$$7 4i$$
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$$4 7i$$
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$$48i$$
Explanation
$${ \begin{vmatrix} 1+i & 1-i & i \\ 1+i & i & 1+i \\ i & 1+i & 1-i \end{vmatrix}}$$ = $$(1+i)[i(1-i) - (1+i)^{2}] - (1+i)[(1-i)^{2} - i(1+i)] + i[(1-i)(1+i) - i^{2}]$$
=$$(1+i)[i-i^{2}-1-i^{2}-2i] - (1+i)[1+i^{2}-2i-i-i^{2}] + i[1-i^{2}-i^{2}]$$
=$$(1+i)(1-i) - (1+i)(1-3i) + 3i$$
=$$1-i^{2}-1(1-3i)-i(1-3i)+3i$$
=$$1+1-1+3i-i+3i^{2}+3i$$
=$$1-3+6i$$
=$$5i-2$$
If $$A=\left[ \begin{matrix} 0 & 1 \\ -1 & 0 \end{matrix} \right] $$ then determinant of $$[A]$$ is
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$$1$$
0%
$$-1$$
0%
$$0 $$
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$$2$$
Explanation
Determinants of a $$2\times 2$$ matrix is given by:
If $$A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$$
$$Det(A)=ad-bc$$
Hence using this above property:
If $$A=\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$$, then
$$Det(A)=0\times0-1\times (-1)=0+1=1$$
`If $$(8,1),(k,-4),(2,-5)$$ are collinaer, then $$k=$$
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$$1$$
0%
$$2$$
0%
$$3$$
0%
$$4$$
Explanation
Let the given point be $$A(8,1),B(K, - 4),C(2, - 5)$$
if the above point are collinear,
they will lie on same line
i.e. they will not form triangle,
Area of $$\Delta$$ABC$$=0$$
$$ \Rightarrow \dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right] = 0$$
Here,
$$\begin{array}{l} { x_{ 1 } }=8,{ y_{ 1 } }=1 \\ { x_{ 2 } }=k,{ y_{ 2 } }=-4 \\ { x_{ 3 } }=2,{ y_{ 3 } }=-5 \end{array}$$
putting value,
$$\begin{array}{l} \Rightarrow \dfrac { 1 }{ 2 } \left[ { 8\left( { -4+5 } \right) +k\left( { -5-1 } \right) +2\left( { 1-4 } \right) } \right] =0 \\ \Rightarrow 8\left( { -4+5 } \right) +k\left( { -5-1 } \right) +2\left( { 1+4 } \right) =0\times 2 \\ \Rightarrow -32+40-5k-k+2+8=0 \\ \Rightarrow 8+10-6k=0 \\ \Rightarrow -6k=-18 \\ \Rightarrow 6k=18 \\ \Rightarrow k=\dfrac { { 18 } }{ 6 } \\ \Rightarrow k=3 \end{array}$$
hence,we get
$$k=3$$
so, option $$C$$ is correct.
If $$A=\begin{bmatrix} 5 & 1 \\ 2 & 3 \end{bmatrix}$$, the determinant of matrix $$A$$ is
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0%
$$13$$
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$$12$$
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$$17$$
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$$-13$$
Explanation
Determinants of a $$2\times 2$$ matrix is given by:
If $$A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$$
$$Det(A)=ad-bc$$
Hence using this above property:
If $$A=\begin{bmatrix} 5 & 1 \\ 2 & 3 \end{bmatrix}$$, then
$$Det(A)=5\times3-1\times 2=15-2=13$$
Find the determinant:
$$\begin{vmatrix} 1 & 2 & 1 \\ 2 & 2 & 2 \\ 3 & 1 & 4 \end{vmatrix}$$
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$$-2$$
0%
$$0$$
0%
$$3$$
0%
$$1$$
Explanation
$$\begin{array}{l} Let\, \, the\, \, er\min ant\, \, be\, \Delta \\ Now, \\ \Delta =\left| { \begin{array} { *{ 20 }{ c } }1 & 2 & 1 \\ 2 & 2 & 2 \\ 3 & 1 & 4 \end{array} } \right| \\ \Delta =1\left| { \begin{array} { *{ 20 }{ c } }2 & 2 \\ 1 & 4 \end{array} } \right| -2\left| { \begin{array} { *{ 20 }{ c } }2 & 2 \\ 3 & 4 \end{array} } \right| +1\left| { \begin{array} { *{ 20 }{ c } }2 & 2 \\ 3 & 1 \end{array} } \right| \\ \Delta =1\left( { 8-2 } \right) -2\left( { 8-6 } \right) +1\left( { 2-6 } \right) \\ \Delta =6-4-4 \\ \Delta =-2 \\ Hence, \\ option\, \, A\, \, is\, \, correct\, \, answer. \end{array}$$
$$\left| \begin{matrix} 1& a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{matrix} \right| =$$
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$$(a-b)(b-c)(c-a)$$
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$$(a+b)(c-a)$$
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$$(a+b+c)^2$$
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$$2(a+b+c)^2$$
Explanation
$$\begin{array}{l} Let\, \, the\, \, given\, \quad determinant\, be\, \Delta .Then \\ \Delta =\left| { \begin{array} { *{ 20 }{ c } }1 & a & { { a^{ 2 } } } \\ 1 & b & { { b^{ 2 } } } \\ 1 & c & { { c^{ 2 } } } \end{array} } \right| \\ =\left| { \begin{array} { *{ 20 }{ c } }1 & a & { { a^{ 2 } } } \\ 0 & { b-a } & { { b^{ 2 } }-{ a^{ 2 } } } \\ 0 & { c-a } & { { c^{ 2 } }-{ a^{ 2 } } } \end{array} } \right| \, \, \left[ { applying\, \, { R_{ 2 } }\to \left( { { R_{ 2 } }-{ R_{ 1 } } } \right) and\, { R_{ 3 } }\to \left( { { R_{ 3 } }-{ R_{ 1 } } } \right) } \right] \\ =\left( { b-a } \right) \left( { c-a } \right) .\left| { \begin{array} { *{ 20 }{ c } }1 & a & { { a^{ 2 } } } \\ 0 & 1 & { b-a } \\ 0 & 1 & { c+a } \end{array} } \right| \\ \left[ { taking\, \, \left( { b-a } \right) common\, \, from\, \, { R_{ 2 } }\, \, and\, \, \left( { c-a } \right) common\, \, from\, \, { R_{ 3 } } } \right] \\ =\left( { b-a } \right) \left( { c-a } \right) \times 1.\left| { \begin{array} { *{ 20 }{ c } }1 & { b+a } \\ 1 & { c+a } \end{array} } \right| \, \, \left[ { anded\, \, by\, \, { C_{ 1 } } } \right] \\ =\left( { b-a } \right) \left( { c-a } \right) \left\{ { \left( { c+a } \right) -\left( { b+a } \right) } \right\} \\ =\left( { b-a } \right) \left( { c-a } \right) \left( { c-b } \right) =\left( { a-b } \right) \left( { b-c } \right) \left( { c-a } \right) . \\ Hence, \\ \Delta =\left( { a-b } \right) \left( { b-c } \right) \left( { c-a } \right) . \end{array}$$
So, option $$A$$ is correct answer.
If the points $$(a, 1), (2, -1)$$ and $$\left(\dfrac{1}{2}, 2\right)$$ are collinear, then $$a$$ is equal to:
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$$1$$
0%
$$0$$
0%
$$2$$
0%
$$\dfrac{1}{4}$$
Explanation
Slope of $$AB=$$ slope of $$BC$$
$$ \Rightarrow \dfrac{{ - 1 - 1}}{{2 - a}} = \dfrac{{2 - \left( { - 1} \right)}}{{\dfrac{1}{2} - 2}}$$
$$ \Rightarrow \dfrac{{ - 2}}{{2 - a}} = \dfrac{{3 \times 2}}{{ - 3}}$$
$$ \Rightarrow 6 = 12 - 6a$$
$$ \Rightarrow 6a = 6$$
$$a=1$$
Hence, option $$A$$ is correct answer.
If $$\displaystyle A = \begin {vmatrix} \dfrac{1}{2}\left ( e^{\alpha} + e^{\alpha} \right ) & \dfrac{1}{2}\left ( e^{\alpha} e^{\alpha} \right ) \\ \dfrac{1}{2}\left ( e^{\alpha} e^{\alpha} \right ) & \dfrac{1}{2}\left ( e^{\alpha} + e^{\alpha} \right ) \end {vmatrix} $$ then $$A^{1}$$ exists
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0%
For all real $$\alpha$$
0%
For positive real $$\alpha$$ only
0%
For negative real $$\alpha$$ only
0%
None of these
Explanation
$$\displaystyle A = \begin {vmatrix} \dfrac{1}{2}\left ( e^{\alpha} + e^{\alpha} \right ) & \dfrac{1}{2}\left ( e^{\alpha} e^{\alpha} \right ) \\ \dfrac{1}{2}\left ( e^{\alpha} e^{\alpha} \right ) & \dfrac{1}{2}\left ( e^{\alpha} + e^{\alpha} \right ) \end {vmatrix}$$
$$\Rightarrow \displaystyle A =\frac{1}{2} \begin {vmatrix} \left ( e^{\alpha} + e^{\alpha} \right ) & \left ( e^{\alpha} e^{\alpha} \right ) \\ \left ( e^{\alpha} e^{\alpha} \right ) & \left ( e^{\alpha} + e^{\alpha} \right ) \end {vmatrix}$$
$$\Rightarrow A=(e^{\alpha}+e^{\alpha})^2-(e^{\alpha}e^{\alpha})^2$$
$$\Rightarrow A=4e^{2\alpha}-e^{4\alpha}$$
So, $$A$$ is correct option.
What is the value of the determinant
$$\begin{vmatrix} 1!& 2! & 3!\\ 2! & 3! & 4! \\ 3!& 4!& 5!\end{vmatrix}$$$$?$$
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0%
$$0$$
0%
$$12$$
0%
$$24$$
0%
$$36$$
Explanation
Given,
$$\begin{pmatrix}1!&2!&3!\\ 2!&3!&4!\\ 3!&4!&5!\end{pmatrix}$$
$$=1!\cdot \det \begin{pmatrix}3!&4!\\ 4!&5!\end{pmatrix}-2!\cdot \det \begin{pmatrix}2!&4!\\ 3!&5!\end{pmatrix}+3!\cdot \det \begin{pmatrix}2!&3!\\ 3!&4!\end{pmatrix}$$
$$=1!\cdot \:144-2!\cdot \:96+3!\cdot \:12$$
$$=144-192+72$$
$$=24$$
lf $$\mathrm{A}=\left[\begin{array}{lll}
1 & 5 & -6\\
-8 & 0 & 4\\
3 & -7 & 2
\end{array}\right]$$, then the cofactor of -7=......
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0%
44
0%
43
0%
40
0%
39
Explanation
By property of cofactor of matrix,
$$A=\begin{bmatrix}
1 & 5 & -6\\
-8 & 0 & 4\\
3 & -7 & 2
\end{bmatrix}$$
minor of $$-7=\begin{vmatrix}
1 & -6\\
-8 & 4
\end{vmatrix}=4-48=-44$$
So, cofactor of (-7)=minor of $$(-7) \times (-1)^{3+2}=-[minor \;of \; (-7)]$$
$$=-(-44)$$
$$=44$$
The vectorial angle of a point $$P$$ on the line joining the points $$(r_{1}, \theta _{1})$$ and $$(r_{2},\theta _{2})$$ is $$\dfrac{\theta_{1} +\theta _{2}}{2}$$ then the length of radius vector of $$P$$ is
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$$\displaystyle \frac{r_{1} - r_{2}}{r_{1} + r_{2}} \cos (\frac{\theta _{1} - \theta _{2}}{2})$$
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$$\displaystyle \frac{r_{1} + r_{2}}{r_{1} r_{2}} \cos (\frac{\theta _{1} - \theta _{2}}{2})$$
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$$\displaystyle \frac{r_{1} r_{2}}{r_{1} + r_{2}} \cos (\frac{\theta _{1} - \theta _{2}}{2})$$
0%
None of these
Explanation
Given points are
$$A(r_{1},\theta_{1}),B(r_{2},\theta_{2})$$
Eq of line passing through two given points is given by:
$$y-y_{1}=\cfrac{y_{2}-y_{1}}{x_{2}-x_{1}}(x-x_{1})$$
$$\therefore y-\theta_{1}=\cfrac{\theta_{2}-\theta_{1}}{r_{2}-r_{1}}(x-r_{1})$$
Now, angle between point P and line is $$\cfrac{\theta_{1}+\theta_{2}}{2}$$
Let point P meet at point Q on the line.
The radius will be
$$R=PQ\cos\theta$$
$$\therefore R=PQ\cos(\cfrac{\theta_{1}+\theta_{2}}{2})$$
If the entries in a $$3\times 3$$ determinant are either $$0$$ or 1, then the greatest value of their determinats is:
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0%
1
0%
2
0%
3
0%
9
Explanation
Case I: If all the entries in one row are 0 , then the determinant is 0.
Case II: If all the entries in one row are 1 , then the maximum value of determinant is 1.
$$\begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{bmatrix}$$
Case II:If exactly one 1 is in one row, then the maximum value is 1.
$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
Case III:If exactly two 1 is in one row, then the maximum value is 2.
$$\begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{bmatrix}$$
Hence, the largest value of the determinant with entries either 0 or 1 is 2.
If $$A+B+C= \pi$$, then $$ \displaystyle \left| \begin{matrix} \tan { \left( A+B+C \right) } & \tan { B } & \tan { C } \\ \tan { (A+C) } & 0 & \tan { A } \\ \tan { (A+B) } & -\tan { A } & 0 \end{matrix} \right| $$ is equal to
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$$0$$
0%
$$1$$
0%
$$\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{A}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{B}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{C}$$
0%
$$-2$$
Explanation
Let $$P=\begin{vmatrix}
tan(A+B+C) & tan B & tan C\\
tan (A+C) & 0 & tan A\\
tan(A+B) & -tan A & 0
\end{vmatrix}$$
$$=\tan(A+B+C) \tan^2A-\tan B[-\tan A \tan (A+B)]+\tan C[-\tan A \tan(A+C)]$$
$$=\tan (A+B+C) \tan^2 A+\tan A \tan B \tan(A+B)-\tan A \tan C \tan (A+C)$$
Given $$A+B+C=\pi \implies A+B=\pi-C$$ or $$A+C=\pi-B$$
$$\therefore P=\tan (\pi) \tan^2 A + \tan A \tan B(-\tan C) -\tan A \tan C(-\tan B)$$
$$=0\cdot \tan^2 A - \tan A \tan B \tan C + \tan A \tan B \tan C$$
$$=0$$
The value of $$\left|\begin{array}{ll}
2+i & 2-i\\
1+i & 1-i
\end{array}\right|$$ is:
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$$\mathrm{A}$$ complex quantity
0%
real quantity
0%
$$0$$
0%
cannot be determined
Explanation
Let $$A=\begin{vmatrix}
2+i & 2-i\\
1+i & 1-i
\end{vmatrix}$$
$$det A=(2+i)(1-i)-(2-i)(1+i)$$
$$=2-2i+i-i^2-(2+2i-i-i^2)$$
$$=3-i-(3+i)=-2i$$
$$\Rightarrow det A= -2i\Rightarrow A $$ complex quantity.
$$I\mathrm{f}\mathrm{A}=\left[\begin{array}{ll}
1 & 3\\
2 & 1
\end{array}\right]$$, then the determinant $$\mathrm{A}^{2}-2\mathrm{A}$$:
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$$5$$
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$$25$$
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$$-5$$
0%
$$-25$$
Explanation
Given $$A=\begin{pmatrix}
1 & 3\\
2 & 1
\end{pmatrix}$$
$$A^2=\begin{bmatrix}
1 & 3\\
2 & 1
\end{bmatrix}\begin{bmatrix}
1 & 3\\
2 & 1
\end{bmatrix}$$
By operation of matrix (4),
$$A^2=\begin{bmatrix}
7 & 6\\
4 & 7
\end{bmatrix}$$
So,$$A^2-2A=\begin{bmatrix}
7 & 6\\
4 & 7
\end{bmatrix}-2\begin{bmatrix}
1 & 3\\
2 & 1
\end{bmatrix}=\begin{bmatrix}
7 & 6\\
4 & 7
\end{bmatrix}-\begin{bmatrix}
2 & 6\\
4 & 2
\end{bmatrix}$$
$$=\begin{bmatrix}
5 & 0\\
0 & 5
\end{bmatrix}$$
So, $$det A^2-2A=|A^2-2A|=25$$
$$\left|\begin{array}{llll}
1 & & \mathrm{l}\mathrm{o}\mathrm{g}_{b}a\\
\mathrm{l}\mathrm{o}\mathrm{g}_{a}b & 1 &
\end{array}\right|$$ =....
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$$ab
$$
0%
$$\mathrm{b}\mathrm{a}$$
0%
$$\mathrm{a}\mathrm{b}$$
0%
$$0$$
Explanation
Let $$A=\begin{vmatrix}
1 & \log_ba\\
\log_ab & 1
\end{vmatrix}$$
Expanding the determinant,
$$A=1-\log_ab \times \log_ba=1-1=0$$
(because $$\log_ab=\frac{1} {\log_ba}$$)
lf $$\left|\begin{array}{lll}
a+x & a-x & a-x\\
a-x & a+x & a-x\\
a-x & a-x & a+x
\end{array}\right|=0$$ then the non-zero value of x=............
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$$a$$
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$$3a$$
0%
$$2a$$
0%
$$4a$$
Explanation
By operation of matrix (5),
$$A=\begin{vmatrix}
a+x & a-x & a-x\\
a-x & a+x & a-x\\
a-x & a-x & a+x
\end{vmatrix}$$
$$=(a+x)[(a+x)^2-(a-x)^2]-(a-x)[a^2-x^2-(a-x)^2]+(a-x)[(a-x)^2-(a^2-x^2)]$$
$$=(a+x)(4ax)-(a-x)(a^2-x^2)+2(a-x)^3-(a-x)(a^2-x^2)$$
$$=(a+x)(4ax)+2(a-x)[a^2+x^2-2ax-a^2+x^2]$$
$$=(a+x)4ax+2(a-x)(2x^2-2ax)$$
$$A=4a^2x+4ax^2+4ax^2-4a^2x-ax^3+4ax^2$$
$$A=8ax^2+4ax^2-4x^3$$
$$A=12ax^2-4x^3$$
Given,$$ A= 12 ax^2-4x^3=0$$
$$x^2(12 a-x)=0$$
$$x=3a$$
lf $$\left|\begin{array}{lll}
1 & 2 & x\\
4 & -1 & 7\\
2 & 4 & -6
\end{array}\right|$$ is a singular matrix, then $$x$$ is equal to
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0%
$$0$$
0%
$$1$$
0%
$$-3$$
0%
$$3$$
Explanation
Given $$A=\begin{vmatrix}
1 & 2 & x\\
4 & -1 & 7\\
2 & 4 & -6
\end{vmatrix}$$ is a singular matrix
So, $$detA=0$$
$$\Rightarrow 1(6-28)-2(-24-14)+x(16+2)=0$$
$$\Rightarrow -22+76+18x=0$$
$$\Rightarrow 18x=-54$$
$$\therefore x=-3$$
If $$A=\left[\begin{array}{lll}
1^{2} & 2^{2} & 3^{2}\\
2^{2} & 3^{2} & 4^{2}\\
3^{2} & 4^{2} & 5^{2}
\end{array}\right]$$, then the minor of $$\mathrm{a}_{22}$$ is
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0%
$$-56$$
0%
$$51$$
0%
$$-43$$
0%
$$41$$
Explanation
Given, $$A=\begin{bmatrix}
1^2 & 2^2 & 3^2\\
2^2 & 3^2 & 4^2\\
3^2 & 4^2 & 5^2
\end{bmatrix}$$
So, By property of minor (1),
minor of $$a_{22}=M_{22}(-1)^{2+2}=M_{22}$$
$$=\begin{vmatrix}
1^2 & 3^2\\
3^2 & 5^2
\end{vmatrix}$$
$$=25 - 9 \times 9$$
$$=-56$$
$$\begin{vmatrix}
1 &4 &20 \\
1 & -2& 5\\
1 &2x & 5x^{2}
\end{vmatrix}=0$$ find x
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-1, 2
0%
0,1
0%
1, 3
0%
2, 0
Explanation
$$\begin{vmatrix}
1 & 4 & 20\\
1 & -2 & 5\\
1 & 2x & 5x^2
\end{vmatrix}=0$$
$$r_1 \rightarrow r_1-r_2$$ and $$r_3\rightarrow r_3-r_2$$
$$\begin{vmatrix}
0 & 6 & 15\\
1 & -2 & 5\\
0 & 2x+2 & 5x^2-5
\end{vmatrix}$$
$$=-(6(5x^2-5)-15(2x+2))$$
$$=-(30x^2-30-30x-30)$$
$$=30(-x^2+x+2)$$
So, $$x^2-x-2=0$$
$$x=-1,2$$
If a, b, c are all positive and not all equal then the value of the determinant $$\begin{bmatrix}
a & b & c\\
b & c &a \\
c & a & b
\end{bmatrix}$$ is
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0%
0
0%
< 0
0%
> 0
0%
cannot be determined
Explanation
Let $$|A|=\begin{vmatrix} a &b &c \\ b & c& a\\ c & a & b\end{vmatrix}$$
$$C_{1}\rightarrow C_{1}+C_{2}+C_{3}$$
$$|A|=\begin{vmatrix} a+b+c &b &c \\ a+b+c & c& a\\ a+b+c & a & b\end{vmatrix}$$
$$=(a+b+c) \begin{vmatrix}1 &b &c \\ 1& c& a\\ 1& a & b\end{vmatrix}$$
$$R_{1}\rightarrow R_{1}-R_{2} , R_{2}\rightarrow R_{2}-R_{3}$$
$$=(a+b+c)\begin{vmatrix}0 &b-a &c-a \\ 0& c-a& a-b\\ 1&a & b\end{vmatrix}$$
$$=(a+b+c)[(b-a)(a-b)-(c-a)^{2}]$$
$$\therefore |A|=-(a+b+c)[(a-b)^{2}+(c-a)^{2}]$$
Now, $$(a-b)^{2}, (c-a)^{2}>0$$ and $$a+b+c>0$$ as $$a,b,c>0$$
$$\therefore |A|<0$$
Hence,
$$\begin{vmatrix} a &b &c \\ b & c& a\\ c & a & b\end{vmatrix}<0$$
$$\left|\begin{array}{lll}
\mathrm{a}+\mathrm{b} & \mathrm{a} & \mathrm{b}\\
\mathrm{a} & \mathrm{a}+\mathrm{c} & \mathrm{c}\\
\mathrm{b} & \mathrm{c} & \mathrm{b}+\mathrm{c}
\end{array}\right|=$$
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4 abc
0%
abc
0%
$$2\mathrm{a}^{2}\mathrm{b}^{2}\mathrm{c}^{2}$$
0%
$$4\mathrm{a}^{2}\mathrm{b}^{2}\mathrm{c}^{2}$$
Explanation
$$\begin{vmatrix}
a+b & a & b\\
a & a+c & c\\
b & c & b+c
\end{vmatrix}$$
$$=(a+b)[(a+c)(b+c)-c^2]-a[ab+ac-bc]+b[ac-ab-bc]$$
$$=(a+b)[ab+ac+cb]-a^2b-a^2c+abc+abc-ab^2-b^2c$$
$$=a^2b+a^2c+abc+ab^2+abc+cb^2+abc+cb^2-a^2b-a^2c+abc+abc-ab^2-b^2c$$
$$=4abc$$
Adj $$\left ( Adj\begin{bmatrix}
2 &-3 \\
4& 6
\end{bmatrix} \right )=$$
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$$\begin{bmatrix}
2 & -3\\
4& 6
\end{bmatrix}$$
0%
$$\begin{bmatrix}
6& 3\\
-4& 2
\end{bmatrix}$$
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$$\begin{bmatrix}
-6& 3\\
-4& -2
\end{bmatrix}$$
0%
$$\begin{bmatrix}
-6& -3\\
4& -2
\end{bmatrix}$$
Explanation
Adjoint of matrix is given by
$$\begin{bmatrix} M11 & -M12 \\ -M21 & M22 \end{bmatrix}$$
Here $$M11 = 6$$
$$M12 = 4$$
$$M21 = -3$$
$$M22 = 2$$
So adjoint of the matrix is
$$\begin{bmatrix} 6& -4 \\ 3 & 2 \end{bmatrix}$$
Again adjoint of this matrix is
$$\begin{bmatrix} 2&-3\\ 4&6\end{bmatrix}$$
Hence the answer is option A.
A= $$\begin{bmatrix}
3 & 0 & 0\\
0& 3 & 0\\
0& 0 & 3
\end{bmatrix}$$ ,then Adj ( A)
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3A
0%
6A
0%
$$9A^{T}$$
0%
$$2A^{T}$$
Explanation
We have, $$A=\begin{bmatrix}
3 & 0 & 0\\
0& 3 & 0\\
0& 0 & 3
\end{bmatrix}$$
co-factor matrix of $$A$$ is,
$$B = \begin{bmatrix}
9 & 0 & 0\\
0& 9 & 0\\
0& 0 & 9
\end{bmatrix}$$
Thus
$$\therefore $$ Adj(A)$$ =B^T=\begin{bmatrix}
9 & 0 & 0\\
0& 9 & 0\\
0& 0 & 9
\end{bmatrix}=3\begin{bmatrix}
3 & 0 & 0\\
0& 3 & 0\\
0& 0 & 3
\end{bmatrix}=3A$$
$$\left[\begin{array}{llll}
\mathrm{c}\mathrm{o}\mathrm{s}\alpha+\mathrm{i}\mathrm{s}\mathrm{i}\mathrm{n}\alpha & \mathrm{c}\mathrm{o}\mathrm{s}\beta+\mathrm{i}\mathrm{s}\mathrm{i}\mathrm{n}\beta\\
\mathrm{s}\mathrm{i}\mathrm{n}\beta+\mathrm{i}\mathrm{c}\mathrm{o}\mathrm{s}\beta\ & \mathrm{s}\mathrm{i}\mathrm{n}\alpha+\mathrm{i}\mathrm{c}\mathrm{o}\mathrm{s}\alpha &
\end{array}\right]$$ is
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2 $$\cos\alpha$$
0%
2 $$\sin\beta$$
0%
$$0$$
0%
$$1$$
Explanation
Let $$A=\begin{bmatrix}
\cos \alpha+i \sin \alpha & \cos \beta + i sin \beta\\
\sin \beta + i \cos \beta & \sin \alpha + i \cos \alpha
\end{bmatrix}$$
$$|A|=(\cos \alpha + i \sin \alpha)(\sin \alpha+ i \cos \alpha)-(\cos \beta + i \sin \beta)(\sin \beta + i \cos \beta)$$
$$= \cos \alpha \sin \alpha + i - \cos \alpha \sin \alpha - [\cos \beta \sin \beta + i \sin \beta \cos \beta]$$
$$=0$$
If $$\mathrm{A}$$ is an unitary matrix then $$|A|$$ is equal to:
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$$1$$
0%
$$-1$$
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$$\pm 1$$
0%
$$2$$
Explanation
Since, A is a unitary matrix
$$AA^{*}=I$$
$$det A \times det A^{*}=det I$$
$$\Rightarrow (det A)^{2}=1$$ ($$\because |A|=|\overline{A^{T}}|$$)
$$\Rightarrow det A=\pm 1$$
If A =$$\begin{bmatrix}
0 &1 & 2\\
1& 2 & 3\\
3 & 1 & 1
\end{bmatrix}$$ then Adj (A) =
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$$\begin{bmatrix}-1 &+8 & -5\\ 1& -6 & 3\\ -1 & 2 & -1\end{bmatrix}$$
0%
$$\begin{bmatrix}-1 &+1 & -1\\ 8& -6 & 2\\ -5 & 3 & -1\end{bmatrix}$$
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$$\begin{bmatrix}1 &-1 & 1\\ 8& -6 & 2\\ -5 & 3 & -1\end{bmatrix}$$
0%
$$\begin{bmatrix}-1 &-8 & 5\\ -1& 6 & -3\\ 1 & -2 & 1\end{bmatrix}$$
Explanation
Cofactor of the matrix is given by
$$\begin{bmatrix} M11 & -M12 & M13 \\ -M21 & M22 & -M23 \\ M31 & -M32 & M33 \end{bmatrix}$$
where $$M11$$ = $$\begin{bmatrix} 2 & 3 \\ 1 & 1 \end{bmatrix} = 2-3 = -1$$
$$M12$$ =
$$\begin{bmatrix} 1 & 3 \\ 3 & 1 \end{bmatrix} = 1-9 = -8$$
$$M13$$ =
$$\begin{bmatrix} 1 & 2 \\ 3 & 1 \end{bmatrix} = 1-6 = -5$$
$$M21$$ =
$$\begin{bmatrix} 1 & 2 \\ 1 & 1 \end{bmatrix} = 1-2 = -1$$
$$M22$$=
$$\begin{bmatrix} 0 & 2 \\ 3 & 1 \end{bmatrix} = 0-6 = -6$$
$$M23$$ =
$$\begin{bmatrix} 0 & 1 \\ 3 & 1 \end{bmatrix} = 0-3 = -3$$
$$M31$$ =
$$\begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix} = 3-4 = -1$$
$$M32$$ =
$$\begin{bmatrix} 0 & 2\\ 1 & 3 \end{bmatrix} = 0-2 = -2$$
$$M33$$ =
$$\begin{bmatrix} 0 & 1 \\ 1 & 2 \end{bmatrix} = 0-1 = -1$$
substituting all the values in the Cofactor of the matrix formula we get,
=
$$\begin{bmatrix} -1 & 8 & -5 \\ 1& -6 & 3\\ -1 & 2 & -1\end{bmatrix}$$
Hence the adjoint of the matrix is transpose of the cofactor of the given matrix
$$\therefore $$ adjoint of the given matrix is
$$\begin{bmatrix} -1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1 \end{bmatrix}$$
Hence the answer is option B.
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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