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CBSE Questions for Class 12 Commerce Maths Determinants Quiz 7 - MCQExams.com
CBSE
Class 12 Commerce Maths
Determinants
Quiz 7
If
A
=
|
a
1
b
1
c
1
a
2
b
2
c
2
a
3
b
3
c
3
|
and
B
=
|
c
1
c
2
c
3
a
1
a
2
a
3
b
1
b
2
b
3
|
then.
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A
=
−
B
0%
A
=
B
0%
B
=
0
0%
B
=
A
2
Explanation
Let determinant of matrix,
|
c
1
c
2
c
3
a
1
a
2
a
3
b
1
b
2
b
3
|
be
y
|
c
1
c
2
c
3
a
1
a
2
a
3
b
1
b
2
b
3
|
exchanging row
1
and row
2
, the determinant becomes
−
y
|
a
1
a
2
a
3
c
1
c
2
c
3
b
1
b
2
b
3
|
exchanging row
2
and row
3
the determinant becomes
y
|
a
1
a
2
a
3
b
1
b
2
b
3
c
1
c
2
c
3
|
Now taking transpose the determinant will remain
y
|
a
1
b
1
c
1
a
2
b
2
c
2
a
3
b
3
c
3
|
So,
A
=
B
(b)
If
|
1
s
i
n
θ
1
−
s
i
n
θ
1
s
i
n
θ
−
1
−
s
i
n
θ
1
|
then,
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0%
Δ
=
0
0%
Δ
∈
(
0
,
∞
)
0%
Δ
∈
[
−
1
,
2
]
0%
Δ
∈
[
2
,
4
]
Explanation
Let
Δ
=
|
1
s
i
n
θ
1
−
s
i
n
θ
1
s
i
n
θ
−
1
−
s
i
n
θ
1
|
=
1
(
1
+
s
i
n
2
θ
)
−
s
i
n
θ
(
0
)
+
1
(
s
i
n
2
θ
+
1
)
=
2
(
1
+
s
i
n
2
θ
)
∵
\Rightarrow 1\leq 1+sin^2 \theta\leq 2
\Rightarrow 2 \leq 2 (1+sin^2 \theta)\leq 4
\Rightarrow 2 \leq \Delta \leq 4
\Rightarrow \Delta \in [2,4]
If
A=\begin{bmatrix} x & 1 & -x \\ 0 & 1 & -1 \\ x & 0 & 7 \end{bmatrix}
and
det(A)=\begin{vmatrix} 3 & 0 & 1 \\ 2 & -1 & 0 \\ 0 & 6 & 7 \end{vmatrix}
then the value of
x
is
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-3
0%
3
0%
2
0%
-8
0%
-2
Explanation
Given,
A=\begin{bmatrix} x & 1 & -x \\ 0 & 1 & -1 \\ x & 0 & 7 \end{bmatrix}
. Then
det
(A) = x7-1(0+x)-x(0-x)
=7x-x+x^2=x^2+6x
...(i)
Also
det(A)=\begin{vmatrix} 3 & 0 & 1 \\ 2 & -1 & 0 \\ 0 & 6 & 7 \end{vmatrix}
=3(-7-0)-0+1(12)=-9
...(ii)
From Eqs(i) and (ii), we get
\quad { x }^{ 2 }+6x=-9
\Rightarrow { x }^{ 2 }+6x+9=0\Rightarrow { (x+3) }^{ 2 }=0
x+3=0
x=-3
If
A=\begin{vmatrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{vmatrix}
, then the value of
\left| A \right| \left| adj\left( A \right) \right|
is
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{ a }^{ 3 }
0%
{ a }^{ 6 }
0%
{ a }^{ 9 }
0%
{ a }^{ 27 }
Explanation
A=\left| \begin{matrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{matrix} \right| =\\ \Rightarrow |A|={ a }^{ 3 }\left| \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right| ={ a }^{ 3 }\\ \left| adj(A) \right| ={ \left| A \right| }^{ n-1 }
Here
n=3
\Rightarrow \left| adj(A) \right| ={ \left| A \right| }^{ 3-1 }={ \left| A \right| }^{ 2 }={ a }^{ 6 }\\ \Rightarrow |A|.\left| adj(A) \right| ={ a }^{ 3 }.{ a }^{ 6 }={ a }^{ 9 }
So option
C
is correct
If
A=\begin{bmatrix} 1 & 2 & -1 \\ -1& 1 & 2 \\ 2 & -1 & 1\end{bmatrix}
, then
\text{det} (\text{adj}(\text{adj} A))
is equal to.
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14^4
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14^3
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14^2
0%
14
Explanation
We have,
A=\begin{bmatrix} 1 & 2 & -1\\ -1 & 1 & 2 \\ 2 & -1 & 1\end{bmatrix}
\Rightarrow |A|=1(1+2)-2(-1-4)-1(1-2)
=3+10+1=14
We know that, for a square matrix of order
n
,
\text{adj}(\text{adj} A)=|A|^{n-2}A,
if
|A|\neq 0
\Rightarrow \text{det}(\text{adj} (\text{adj} A))=||A|^{n-2}A|
\Rightarrow \text{det}(\text{adj} (\text{adj} A))=(|A|^{n-2})^n|A|
\Rightarrow \text{det}(\text{adj} (\text{adj} A))=|A|^{n^2-2n+1}
Here,
n=3
and
|A|=14
Therefore,
\text{det}(\text{adj}(\text{adj} A))=(14)^{3^2-2\times 3+1}=14^4
.
If
\begin{vmatrix} x & 2 & 8 \\ 2 & 8 & x \\ 8 & x & 2 \end{vmatrix}=\begin{vmatrix} 3 & x & 7 \\ x & 7 & 3 \\ 7 & 3 & x \end{vmatrix}=\begin{vmatrix} 5 & 5 & x \\ 5 & x & 5 \\ x & 5 & 5 \end{vmatrix}=0
then
x
is equal to
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0%
0
0%
-10
0%
3
0%
None of these
Explanation
On applying
{ R }_{ 1 }\rightarrow { R }_{ 1 }+{ R }_{ 2 }+{ R }_{ 3 }
in each determinant, we can take out
\left( x+10 \right)
common.
Then, we get
x+10=0
\Rightarrow x=-10
If
A = \begin{bmatrix} \dfrac {-1 + i\sqrt {3}}{2i}& \dfrac {-1 - i\sqrt {3}}{2i}\\ \dfrac {1 + i\sqrt {3}}{2i} & \dfrac {1 - i\sqrt {3}}{2i}\end{bmatrix}, i = \sqrt {-i}
and
f(x) = x^{2} + 2
, then
f(A)
is equal to
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\left (\dfrac {5 - i\sqrt {3}}{2}\right ) \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}
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\left (\dfrac {3 - i\sqrt {3}}{2}\right ) \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}
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\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}
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(2 + i\sqrt {3})\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}
Explanation
From the given matrix, we have
A = \begin{bmatrix}\dfrac {\omega}{i} & \dfrac {\omega^{2}}{i}\\ -\dfrac {\omega^{2}}{i} & -\dfrac {\omega}{i}\end{bmatrix} = \dfrac {\omega}{i} \begin{bmatrix}1 & \omega\\ -\omega & -1\end{bmatrix}
Thus
A^{2} = \omega^{2} \begin{bmatrix}1 - \omega^{2} & 0\\ 0 & 1 - \omega^{2}\end{bmatrix}
= -\begin{bmatrix}-\omega^{2} + \omega^{4}& 0\\ 0 & -\omega^{2} + \omega\end{bmatrix}
= \begin{bmatrix}-\omega^{2} + \omega & 0\\ 0 & -\omega^{2} + \omega\end{bmatrix}
....
\text{Since} \ f(x) = x^{2} + 2
Therefore,
f(A) = A^{2} + 2 = \begin{bmatrix}-\omega^{2} + \omega & 0\\ 0 & -\omega^{2} + \omega\end{bmatrix} + \begin{bmatrix}2 & 0\\ 0 & 2\end{bmatrix}
- [\omega^{2} + \omega + 2] \begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}
= (3 + 2\omega) \begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}
= (2 + i\sqrt {3})\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}
If
3x+2y=I
and
2x-y=O
, where
I
and
O
are unit and null matrices of order
3
respectively, then
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x=\dfrac { 1 }{ 7 } , y=\dfrac { 2 }{ 7 }
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x=\dfrac { 2 }{ 7 } , y=\dfrac { 1 }{ 7 }
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x=\left( \dfrac { 1 }{ 7 } \right) I, y=\left( \dfrac { 2 }{ 7 } \right) I
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x=\left( \dfrac { 2 }{ 7 } \right) I, y=\left( \dfrac { 1 }{ 7 } \right) I
Explanation
We have,
3x+2y=I
...(i)
2x-y=O
....(ii)
On multiplying equation (ii) by
2
, we get
4x-2y=2\cdot O=O
...(iii)
On adding equations (i) and (iii), we get
3x+2y=I
4x-2y=O
____________
7x=I
\Rightarrow x=\dfrac { I }{ 7 }
or
\dfrac { 1 }{ 7 } I
From equation (i), we get
2y=I-\dfrac { 3 }{ 7 } I=\dfrac { 4 }{ 7 } I
\Rightarrow y=\dfrac { 2 }{ 7 } I
\begin{vmatrix} { \left( { a }^{ x }+{ a }^{ -x } \right) }^{ 2 } & { \left( { a }^{ x }-{ a }^{ -x } \right) }^{ 2 } & 1 \\ { \left( b^{ x }+{ b }^{ -x } \right) }^{ 2 } & { \left( { b }^{ x }-{ b }^{ -x } \right) }^{ 2 } & 1 \\ { \left( { c }^{ x }+{ c }^{ -x } \right) }^{ 2 } & { \left( { c }^{ x }-{ c }^{ -x } \right) }^{ 2 } & 1 \end{vmatrix}
is equal to
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0
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2abc
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{ a }^{ 2 }{ b }^{ 2 }{ c }^{ 2 }
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None of these
Explanation
Let
\Delta=\begin{vmatrix} { \left( { a }^{ x }+{ a }^{ -x } \right) }^{ 2 } & { \left( { a }^{ x }-{ a }^{ -x } \right) }^{ 2 } & 1 \\ { \left( b^{ x }+{ b }^{ -x } \right) }^{ 2 } & { \left( { b }^{ x }-{ b }^{ -x } \right) }^{ 2 } & 1 \\ { \left( { c }^{ x }+{ c }^{ -x } \right) }^{ 2 } & { \left( { c }^{ x }-{ c }^{ -x } \right) }^{ 2 } & 1 \end{vmatrix}
Put
x=0
in the given determinant, we get
\Delta =\begin{vmatrix} { \left( 1+1 \right) }^{ 2 } & { \left( 1-1 \right) }^{ 2 } & 1 \\ { \left( 1+1 \right) }^{ 2 } & { \left( 1-1 \right) }^{ 2 } & 1 \\ { \left( 1+1 \right) }^{ 2 } & { \left( 1-1 \right) }^{ 2 } & 1 \end{vmatrix}
=\begin{vmatrix} 4 & 0 & 1 \\ 4 & 0 & 1 \\ 4 & 0 & 1 \end{vmatrix}
=0
If the determinant
\Delta =\begin{vmatrix} 3 & -2 & \sin { 3\theta } \\ -7 & 8 & \cos { 2\theta } \\ -11 & 14 & 2 \end{vmatrix}=0
, then the value of
\sin { \theta }
is
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\dfrac { 1 }{ 3 }
or
1
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\dfrac { 1 }{ \sqrt { 2 } }
or
\dfrac { \sqrt { 3 } }{ 2 }
0%
0
or
\dfrac { 1 }{ 2 }
0%
None of these
Explanation
Applying
{ R }_{ 2 }\rightarrow { R }_{ 2 }+4{ R }_{ 1 }
and
{ R }_{ 3 }\rightarrow { R }_{ 3 }+7{ R }_{ 1 }
we get
\begin{vmatrix} 3 & -2 & \sin { 3\theta } \\ 5 & 0 & \cos { 2\theta } +4\sin { 3\theta } \\ 10 & 0 & 2+7\sin { 3\theta } \end{vmatrix}=0
\Rightarrow 2\left[ 5\left( 2+7\sin { 3\theta } \right) -10\left( \cos { 2\theta } +4\sin { 3\theta } \right) \right] =0
\Rightarrow 2+7\sin { 3\theta } -2\cos { 2\theta } -8\sin { 3\theta } =0
\Rightarrow 2-2\cos { 2\theta } -\sin { 3\theta } =0
\Rightarrow \sin { \theta } \left( 4\sin ^{ 2 }{ \theta } +4\sin { \theta } -3 \right) =0
\Rightarrow \sin { \theta } =0
or
\left( 2\sin { \theta } -1 \right) =0
or
\left( 2\sin { \theta } +3 \right) =0
\Rightarrow \sin { \theta } =0
or
\sin { \theta } =\dfrac { 1 }{ 2 }
If
A, B, C
are collinear points such that
A(3, 4), C(11, 10)
and
AB = 2.5
then point
B
is
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\left(5,\dfrac{11}{2}\right)
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\left(\dfrac{5}{2},11\right)
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(5, 5)
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(5, 6)
Explanation
Given that the points
A,B,C
are collinear such that
A(3,4),C(11,10)
and
AB=2.5
,
Consider the line joining
A
and
C
,
Slope of that line is
\tan { \alpha } =\dfrac { 10-4 }{ 11-3 } =\dfrac { 3 }{ 4 }
,
\Longrightarrow \cos { \alpha } =\dfrac { 4 }{ 5 }
and
\sin { \alpha } =\dfrac { 3 }{ 5 }
,
We know that a point in a line at a distance
r
from that point is
y={ y }_{ 1 }+r\sin { \alpha }
and
x={ x }_{ 1 }+r\cos { \alpha }
,
Where
\tan{\alpha}
is the slope of that line
Let the coordinates of the point
B
be
(x,y)
,
\Longrightarrow x=3+2.5(0.2)=5
and
y=3+2.5(0.6)=\dfrac { 11 }{ 2 }
,
\therefore
The coordinates of
B
are
\left(5,\dfrac{11}{2}\right)
If
A = \begin{bmatrix} 2& -3\\ 4 & 1\end{bmatrix}
, then adjoint of matrix
A
is _______.
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\begin{bmatrix} 1& 3\\ -4 & 2\end{bmatrix}
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\begin{bmatrix} 1& -3\\ -4 & 2\end{bmatrix}
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\begin{bmatrix} 1& 3\\ 4 & -2\end{bmatrix}
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\begin{bmatrix} -1& -3\\ -4 & 2\end{bmatrix}
Explanation
Let
A=\begin{bmatrix} 2&-3 \\4&1\end{bmatrix}
Let
C
be the cofactor matrix of matrix
A
.
{C}_{11}={\left(-1\right)}^{1+1}{M}_{11}=1
{C}_{12}={\left(-1\right)}^{1+2}{M}_{12}=-4
{C}_{21}={\left(-1\right)}^{2+1}{M}_{21}=3
{C}_{22}={\left(-1\right)}^{2+2}{M}_{22}=2
\therefore {C}_{ij}=\begin{bmatrix} {C}_{11} & {C}_{12} \\ {C}_{21} & {C}_{22}\end{bmatrix}
=\begin{bmatrix} 1 & -4 \\ 3 & 2\end{bmatrix}
Adj
\left({C}_{ij}\right)={\begin{bmatrix} 1 & -4 \\ 3 & 2\end{bmatrix}}^{T}
=\begin{bmatrix} 1 & 3 \\ -4 & 2\end{bmatrix}
If maximum and minimum values of
D = \begin{vmatrix}1 & -\cos \theta & 1\\ \cos \theta & 1 & -\cos \theta\\ 1 & \cos \theta & 1\end{vmatrix}
are
p
and
q
respectively, then the value of
2p + 3q
is _________.
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0%
16
0%
6
0%
14
0%
8
Explanation
D = \begin{vmatrix}1 & -\cos \theta & 1\\ \cos \theta & 1 & -\cos \theta\\ 1 & \cos \theta & 1\end{vmatrix}
\Rightarrow D = 1(1 + \cos^{2}\theta) + \cos \theta (\cos \theta + \cos \theta) - 1 (\cos^{2}\theta - 1)
= 1 + \cos^{2}\theta + 2\cos^{2}\theta -\cos^{2} \theta + 1
= 2 (1 + \cos^{2} \theta)
Now,
-1\le \cos \theta\le 1\Rightarrow 0\le \cos^{2}\theta\le 1
\therefore p = 4, q = 2
\therefore 2p + 3q = 14
.
The value of the determinant
\begin{vmatrix}b^2-ab & b-c & bc-ac\\ ab-a^2 & a-b & b^2-ab\\ bc-ac & c-a & ab-a^2\end{vmatrix}
=
____________.
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0%
abc
0%
a+b+c
0%
0
0%
ab+bc+ca
If the vectors
\vec {a}, \vec {b}, \vec {c}
are coplanar, then the value of
\begin{vmatrix}\vec {a}& \vec {b} & \vec {c}\\ \vec {a}.\vec {a} & \vec {a}.\vec {b} & \vec {a}.\vec {c}\\ \vec {b}.\vec {a} & \vec {b}.\vec {b} & \vec {b} . \vec {c}\end{vmatrix} =
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1
0%
0
0%
-1
0%
\vec {a} + \vec {b} + \vec {c}
Explanation
\bar { a } \cdot \bar { b } =\bar { a } \bar { b } \cos { \theta }
Co planar
\Rightarrow \theta =0
\Rightarrow \bar { a } \cdot \bar { b } =0
Parallelly
\bar { a } \cdot \bar { c } =\bar { a } \cdot \bar { c } =0
\begin{vmatrix} \bar { a } & \bar { b } & \bar { c } \\ \bar { a } \cdot \bar { a } & \bar { a } \cdot \bar { b } & \bar { a } \cdot \bar { c } \\ \bar { b } \cdot \bar { a } & \bar { b } \cdot \bar { b } & \bar { b } \cdot \bar { c } \end{vmatrix}
=\begin{vmatrix} \bar { a } & \bar { b } & \bar { c } \\ \bar { a } \cdot \bar { a } & 0 & 0 \\ 0 & \bar { b } \cdot \bar { b } & 0 \end{vmatrix}
=\bar { a } \left( 0 \right) -\bar { b } \left( 0 \right) +\bar { c } \left( \bar { a } \cdot \bar { a } -\bar { b } \cdot \bar { b } \right)
=0
Let
A^{-1}\begin{bmatrix} 1 & 2017 & 2\\ 1 & 2017 & 4 \\ 1 & 2018 & 8\end {bmatrix}
. Then
|2A|-|2A^{-1}|
is equal to.
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3
0%
-3
0%
12
0%
-12
Explanation
\left | A^{-1} \right | = \dfrac{1}{\left | A \right |}
given matrix
A^{-1} =
\begin{bmatrix} 1&2017 &2 \\ 1& 2017 &4 \\ 1&2018 &8 \end{bmatrix}
solving the determinant
\left | A^{-1} \right | = \dfrac{1}{\left | A \right |} = -2
\left | kA \right | = k^{n}\left | A \right |
where n is the order of the matrix
-\left | 2A^{-1} \right | + \left | 2A \right |
=-(-2)\times 2^{3}+\dfrac{-1}{2}\times 2^{3} =12
The value of the determinant
\begin{vmatrix} \cos^2 \dfrac{\theta}{2}&\sin^2\dfrac{\theta}{2}\\ \sin^2\dfrac{\theta}{2} &\cos^2\dfrac{\theta}{2} \end{vmatrix}
for all values of
\theta
, is
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1
0%
\cos\theta
0%
\sin\theta
0%
\cos2\theta
Explanation
\begin{vmatrix}\cos ^2\dfrac{\theta}{2}&\sin ^2\dfrac{\theta}{2}\\ \sin ^2\dfrac{\theta}{2} & {\cos }^2 \dfrac{\theta}{2}\end{vmatrix}=\cos ^4\dfrac{\theta}{2}-\sin ^4\dfrac{\theta}{2}
=\left({\cos }^2 \dfrac{\theta}2+{\sin }^2 \dfrac{\theta}2\right)\left({\cos }^2 \dfrac{\theta}2-{\sin }^2 \dfrac{\theta}2\right)
=1\times \cos \theta=\cos \theta
Hence, the answer is
\cos \theta
.
Let
z = \begin{vmatrix} 1& 1 + 2i & -5i\\ 1 - 2i & -3 & 5 + 3i\\ 5i & 5 - 3i & 7\end{vmatrix}
, then
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z
is purely real
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z
is purely imaginary
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(z - \overline {z})i = 0
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(z + \overline {z})i = 0
Explanation
z=\left| \begin{matrix} 1 & 1+2i & -5i \\ 1-2i & -3 & 5+3i \\ 5i & 5-3i & 7 \end{matrix} \right| \\ \quad =1(-21-36)\quad -(1+2i)(7-14i-25i+15)\quad -5i(-13i-1)\\ \quad =-57-(1+2i)(22-39i)-65+5i\\ \quad =-57-(22-39i+44i+78)-65+5i\\ \quad =-57-100-5i-65+5i\\ \quad =-222
\therefore z
is purely real.
The adjoint of the matric
A = \begin{bmatrix}1 & 0 & 2\\ 2 & 1 & 0\\ 0 & 3 & 1\end{bmatrix}
is
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\begin{bmatrix}-1 & 6 & 2\\ -2 & 1 & -4\\ 6 & 3 & 1\end{bmatrix}
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\begin{bmatrix}1 & 6 & -2\\ -2 & 1 & 4\\ 6 & -3 & 1\end{bmatrix}
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\begin{bmatrix}6 & 1 & 2\\ 4 & -1 & 2\\ 6 & 3 & -1\end{bmatrix}
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\begin{bmatrix}-6 & 2 & 1\\ 4 & -2 & 1\\ 3 & 1 & -6\end{bmatrix}
Explanation
adj(A)=cofactor(A)^T=\begin{bmatrix} A_{11} &A_{21}&A_{31}\\A_{12} &A_{22} &A_{32}\\A_{13} &A_{23} &A_{33}\end{bmatrix}=\begin{bmatrix} 1 &6&-2\\-2 &1 &4\\6 &-3 &1\end{bmatrix}
Three distinct points A, B and C are given in the 2-dimensional coordinate plane such that the ratio of the distance of any one of them from the point
(1, 0)
to the distance from the point
(-1, 0)
is equal to
\displaystyle \frac{1}{3}
. Then the circumcentre of the triangle ABC is at the point:
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\displaystyle \left( \frac{5}{2}, 0 \right)
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\displaystyle \left( \frac{5}{3}, 0 \right)
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\displaystyle \left( 0, 0 \right)
0%
\displaystyle \left( \frac{5}{8}, 0 \right)
Explanation
Let (h,k) be the locus of points A, B & C
using given condition
\dfrac{\sqrt{(h-1)^{2}+K^{2}}}{\sqrt{(h+1)^{2}+K^{2}}}=\dfrac{1}{3}
9(h-1)^{2}+9K^{2}=(h+1)^{2}+K^{2}
or,
8h^{2}+8K^{2}-20h+8=0
h^{2}+K^{2}-\dfrac{5h}{4}+1=0
this is a circle and it will be same circle which circumscribe
\Delta ABC
as it contains all three pts on it.
\therefore
center of circumcircle
\equiv \left(\dfrac{5}{8}, 0\right)
If
A + B + C = \pi
, then
\begin{vmatrix} \sin (A + B + C)& \sin B & \cos C\\ -\sin B & 0 & \tan A\\ \cos (A + B) & -\tan A & 0\end{vmatrix}
is equal to
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0
0%
2\sin B \tan A \cos C
0%
1
0%
None of these
Explanation
Given
A+B+C= \pi
=\begin{vmatrix} \sin (A+B+C) & \sin B & \cos C \\ -\sin B & 0 & \tan A \\ \cos A+B & -\tan A & 0 \end{vmatrix}
=\begin{vmatrix} \sin \pi & \sin B & \cos C \\ -\sin B & 0 & \tan A \\ \cos (\pi-C) & -\tan A & 0 \end{vmatrix}
=\sin\pi [\tan^2A]-\sin B[-\tan A \cos (\pi-C)]+\cos C[\sin B\tan A]
=0+\sin B\, \tan A\, \cos(\pi-C)+\cos C \sin B \tan A
=\sin B\, \tan A[\cos\pi \cos C+ \sin\pi \sin C]+\cos C\sin B\tan A
=-\sin B \tan A\cos C +\cos C \sin B \tan A
=0
\left[\therefore \sin \pi=0 ,\cos \pi =-1\right]
If
A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}
satisfies the equation
x^2 - (a + d) x + k = 0
, then ?
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k = bc
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k = ad
0%
k = a^2 + b^2 + c^2 + d^2
0%
ad - bc
Explanation
If a matrix
A
satisfies a polynomial equation, then the product of the roots of that equation is the determinant of
A
.
Here, product of roots of the equation =
k
, also
det(A)=ad-bc
.
Therefore,
k=ad-bc
For a positive numbers
x, y
and
z
the numerical value of the determinant
\begin{bmatrix}1 & \log_{x} y & \log_{x} z \\ \log_{y} x & 1 & \log_{y} z\\ \log_{z} x & \log_{z} y & 1\end{bmatrix}
is:
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0
0%
1
0%
\log_{e} xyz
0%
-\log_{e} xyz
Explanation
D=\left| \begin{matrix} 1 & \log _{ x }{ y } & \log _{ x }{ z } \\ \log _{ y }{ x } & 1 & \log _{ y }{ z } \\ \log _{ z }{ x } & \log _{ z }{ y } & 1 \end{matrix} \right| =(1-\log _{ z }{ y } .\log _{ y }{ z } )-\log _{ x }{ y } \left( \log _{ y }{ x } -\log _{ z }{ x } .\log _{ y }{ z } \right) +\log _{ x }{ z } \left( \log _{ y }{ x } .\log _{ z }{ y } -\log _{ z }{ x } \right) \\ =\left( 1-1 \right) -\log _{ x }{ y } \left( \log _{ y }{ x } -\log _{ y }{ x } \right) +\log _{ x }{ z } \left( \log _{ z }{ x } -\log _{ z }{ x } \right) =0-0+0=0
The value of the determinant
\begin{vmatrix}b^{2}-ab\,\, b-c\,\, bc-ac & & \\ ab-a^{2}\,\, a-b\,\, b^{2}-ab& & \\ bc-ac\,\, c-a\,\, ab-a^{2}& & \end{vmatrix}
=
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abc
0%
a+b+c
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0
0%
ab+bc+ca
Explanation
\begin{vmatrix}b^{2}-ab\,\, b-c\,\, bc-ac & & \\ ab-a^{2}\,\, a-b\,\, b^{2}-ab& & \\ bc-ac\,\, c-a\,\, ab-a^{2}& & \end{vmatrix}=0
a=b=c=1
If the points
A(-2,1),B(a,b)
and
C(4,-1)
are collinear and
a-b=1
, find the values of
a
and
b
.
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a=1,b=5
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a=1,b=0
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a=2,b=0
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None of these
Explanation
Three points
A(-2,1),\ B(a,b)
and
C(4,-1)
are collinear so the area of the triangle is equal to zero
Here the equation
(1)
is
\to a-b=1
Here
x_1=-2\ y_1=1
x_1 =a\ y_1=b
x_3=4\ y_3=-1
Area of the triangle
\triangle ABC=\dfrac {1}{2} [x_1 (y_2 -y_3)+x_2 (y_3 -y_1)+x_3 (y_1 -y_2)]=0
\Rightarrow \ \dfrac {1}{2} [-2(b+1)+a(-1 -1)+4(1-b)]=0
\Rightarrow \ \dfrac {1}{2}[-2b-2-2b+4-4b]=0
\Rightarrow \ \dfrac {1}{2} [-6b+2-2b]=0
\Rightarrow \ -3b+1-a=0
\Rightarrow \ -a-3b=-1
......Equation
2
Equation
(1)\times 1\longrightarrow a-b=1
Equation
(2)\times 1\to \dfrac {-a-2b=-1}{-4b=0\\ \Rightarrow b=0}
Putting the value of
b=0
in Equation
1:-
a-b=1
\Rightarrow \ a=1
So the value of
a=1, b=0
State true or false.
The determinants
\left | \begin{array}{111} 1 & a & bc \\ 1 & b & cd \\ 1 & c & ab \\ \end {array} \right |
and
\left | \begin{array}{111} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \\ \end {array} \right |
are not identically equal.
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True
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False
If
\triangle
=
\left | \begin{array}{111} x_1+y_1\omega & x_1\omega^2+y_1 & x_1+y_1\omega+z_1\omega^2 \\ x_2+y_2\omega & x_2\omega^2+y_2 & x_2+y_2\omega+z_2\omega^2 \\ x_3+y_3\omega & x_3\omega^2+y_3 & x_3+y_3\omega+z_3\omega^2 \\ \end {array} \right |
where
1,\omega,\omega^2
are cube roots of unity then
\triangle
is equal to
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0
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1
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-1
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None of these
\left| \begin{matrix} 1+a & 1 & 1 & 1 \\ 1 & 1+b & 1 & 1 \\ 1 & 1 & 1+c & 1 \\ 1 & 1 & 1 & 1+d \end{matrix} \right|
=abcd\left( 1+\dfrac { 1 }{ a } +\dfrac { 1 }{ b } +\dfrac { 1 }{ c } +\dfrac { 1 }{ d } \right)\\ =s-r
if
a,b,c,d
are the roots of
{ x }^{ 4 }+p{ x }^{ 3 }+q{ x }^{ 2 }+rx+s=0.
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True
0%
False
If
\left |\begin{array}{111}6i & -3i & 1 \\4 & 3i & -1 \\20 & 3 & i \\\end {array}\right | =x+iy
, then
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x=3, y=1
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x = 1, y=3
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x=0, y=3
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x=0, y=0
Explanation
\begin{vmatrix} 6i & -3i & 1 \\ 4 & 3 i & -1 \\ 20 & 3 & i \end{vmatrix}
\Rightarrow 6i((3i. i) - (-3)) + 3i (4 . i + 20) + 1 (12 - 60 i)
\Rightarrow 6i (-3 + 3) + 12 (-1) + 60 i + 12 - 60 i
\Rightarrow 0
[\because i^2 = -1]
x + i y = 0 + i(0)
x = 0 , y = 0
The value of determinant
\begin{vmatrix} x+1 & x+2 & x+4\\ x+3 & x+5 & x+8\\ x+7 & x+10 & x+14\end{vmatrix}
is?
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-2
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x^2+2
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2
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None of these
Explanation
Operating
C_3-C_2
and
C_2-C_1
, we get
\Delta =\begin{vmatrix} x+1 & 1 & 2\\ x+3 & 2 & 3\\ x+7 & 3 & 4\end{vmatrix}
Apply
R_3-R_2, R_2-R_1
\Delta =\begin{vmatrix} x+1 & 1 & 2\\ 2 & 1 & 1\\ 4 & 1 & 1\end{vmatrix}
apply
C_3-C_2
\begin{vmatrix} x+1 & 1 & 1\\ 2 & 1 & 0 \\ 4 & 1 & 0\end{vmatrix}=1.(2-4)=-2
If
\Delta_1=\begin{vmatrix} x & b & b\\ a & x & b\\ a & a & x\end{vmatrix}
and
\Delta_2=\begin{vmatrix} x & b\\ a & x\end{vmatrix}
are the given determinants, then.
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\Delta_1=3(\Delta_2)^2
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(d/dx)\Delta_1=3\Delta_2
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(d/dx)\Delta_1=3(\Delta_2)^2
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\Delta_1=3\Delta_2^{3/2}
If A =
\begin{bmatrix} a & 0 & 0 \\[0.3em] 0 & a & 0 \\[0.3em] 0 & 0 & a \end{bmatrix}
, then the value of |A| |Adj. A|
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a^3
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a^6
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a^9
0%
a^{27}
Explanation
As
|A|=(a)^3
|A|(Adj. A) = |A|
\begin{bmatrix} a & 0 & 0 \\[0.3em] 0 & a & 0 \\[0.3em] 0 & 0 & a \end{bmatrix}
=
\begin{bmatrix} |A| & 0 & 0 \\[0.3em] 0 & |A| & 0 \\[0.3em] 0 & 0 & |A| \end{bmatrix}
where
|A|
=
a^3
Take determinant of both sides
|A|. |Adj. A| =
\begin{vmatrix} |A| & 0 & 0 \\[0.3em] 0 & |A| & 0 \\[0.3em] 0 & 0 & |A| \end{vmatrix} |A|^3 = (a^3)^3 = a^9
If
f(x) =
\left | \begin{array}{111} 1 & x & x+1 \\ 2x & x(x-1) & (x+1)x \\ 3x(x-1) & x(x-1)(x-2) & (x+1)x(x-1) \\ \end {array} \right |
then f(100) is equal to
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0
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1
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100
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-100
Explanation
f(x) = \begin{vmatrix} 1 & x & x + 1 \\ 2x & x(x - 1) & (x + 1)x \\ 3x(x -1) & x(x - 1) (x - 2) & (x + 1) x(x - 1) \end{vmatrix}
x \times x \times (x - 1) \, \begin{vmatrix} 1 & x & x + 1 \\ 2 & x - 1 & x + 1 \\ 3 & x - 2 & x + 1 \end{vmatrix}
x^2 (x - 1)(x + 1) \, \begin{vmatrix} 1 & x & 1 \\ 2 & x - 1 & 1 \\ 3 & x - 2 & 1 \end{vmatrix}
operating :
R_1 \rightarrow R_2 - R_1
&
R_2 \rightarrow R_3 - R_2
x^2 (x^2 - 1) \, \begin{vmatrix} 1 & -1 & 0 \\ 1 & -1 & 0 \\ 3 & x - 2 & 1 \end{vmatrix}
As two rows of the determinant is zero
\therefore
Determinant
= 0
Hence,
\boxed {f(x) = 0}
\therefore
for any value of
x ;\ f(x) = 0
Hence,
\boxed {f(100) = 0}
If A =
\begin{bmatrix} \alpha & 2 \\ 2 & \alpha\end{bmatrix}
and | A
^3
| = 125 then
\alpha
is
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\pm1
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=2
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\pm3
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\pm5
Explanation
A = \begin{bmatrix} \alpha & 2 \\ 2 & \alpha \end{bmatrix}
.
|A^3| = 125
|A| = \alpha^2 - 4
we know by properties of determinants that
|A^3| = |A|^3
\Rightarrow \ (\alpha^2 - 4)^3 = 125
\Rightarrow \ \alpha^2 - 4 = 5
\Rightarrow \ a^2 = 9
\Rightarrow \ \boxed {\alpha = \pm 3}
Hence, option
(C)
is correct
If a, b, c are three non-zero distinct numbers in A.P., then
\triangle =
\left | \begin{array}{111} (b-c)(c-a) & (a-b)(c-a) & (a-b)(b-c) \\ (c-a)(a-b) & (b-c)(a-b) & (b-c)(c-a) \\ (a-b)(b-c) & (c-a)(b-c) & (c-a)(a-b) \\ \end {array} \right |
is always +ve.
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True
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False
If the points
(-2, -5), (2, -2), (8, a)
are collinear, then the value of
a
is ________.
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\dfrac 52
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\dfrac 32
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\dfrac 72
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None of these
Explanation
If three points are collinear, then the slope of the line joining the first two points is equal to that of the last two.
Apply the slope formula
\dfrac{(y2-y1)}{(x2-x1).}
If collinear then
\dfrac { -2+5 }{ 2+2 } =\dfrac { a+2 }{ 6 } \\ \dfrac { 9 }{ 2 } =a+2\\ a=\dfrac { 5 }{ 2 }
State true or false
Following points are collinear.
(-2, 1) , (0, 5) , (-1, 2)
.
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True
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False
The points
(-a, -b), (0, 0), (a, b)
(a^2,ab)
are
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collinear
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vertices of rectangle
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vertices of parallelgram
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none of these
2x - 3y + z = 0
x + 2y - 3z =0
4x - y - 2z = 0
The system of equations have a non trivial solution
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True
0%
False
If the lines
p_{ 1 }x+q_{ 1 }y=1, p_{ 2 }x+q_{ 2 }y=1
and
p_{3}x+q_{3}y=1
be concurrent, show that the points
(p_{1},q_{1}), (p_{2}, q_{2})
and
(p_{3}, q_{3})
are collinear.
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vertices of right angle triangle
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vertices of an equilateral triangle
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vertices of an isosceles triangle
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Collinear
Explanation
Given:
p_{ 1 }x+q_{ 1 }y=1, p_{ 2 }x+q_{ 2 }y=1
and
p_{3}x+q_{3}y=1
The lines will be concurrent if
\begin{vmatrix} { p }_{ 1 } & { q }_{ 1 } & -1 \\ { p }_{ 2 } & { q }_{ 2 } & -1 \\ { p }_{ 3 } & { q }_{ 3 } & -1 \end{vmatrix}=0
Or
\dfrac { 1 }{ 2 } \begin{vmatrix} { p }_{ 1 } & { q }_{ 1 } & 1 \\ { p }_{ 2 } & { q }_{ 2 } & 1 \\ { p }_{ 3 } & { q }_{ 3 } & 1 \end{vmatrix}=0
Or
\triangle=0
i.e., area of a triangle formed by the points
({p}_{1}, {q}_{1})
,
({p}_{2}, {q}_{2})
, and
({p}_{3}, {q}_{3})
is zero and as such the points are collinear.
If A =
\begin{bmatrix} a & 0 & 0 \\[0.3em] 0 & a & 0 \\[0.3em] 0 & 0 & a \end{bmatrix}
, then the value of |Adj. A| is equal to
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a^3
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a^6
0%
a^9
0%
a^{27}
Explanation
A(Adj. A) = |A|
I_3
= |A|
\begin{bmatrix} 1 & 0 & 0 \\[0.3em] 0 & 1 & 0 \\[0.3em] 0 & 0 & 1 \end{bmatrix}
=
\begin{bmatrix} |A| & 0 & 0 \\[0.3em] 0 & |A| & 0 \\[0.3em] 0 & 0 & |A| \end{bmatrix}
where |A| =
a^3
Take determinant of both sides
|A|. |Adj. A| =
\begin{vmatrix} |A| & 0 & 0 \\[0.3em] 0 & |A| & 0 \\[0.3em] 0 & 0 & |A| \end{vmatrix} |A|^3 =|Adj. A| = |A|^{n-1} = |A|^2 = a^6
If
\begin{vmatrix}a & a & x \\ m & m & m \\b & x & b\end{vmatrix}=0
then
x
is:
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a
0%
b
0%
a
or
b
0%
0
Explanation
Given
\begin{vmatrix} a & a & x \\ m & m & m \\ b & x & b \end{vmatrix}=0
Expanding the determinant we get
a(mb-mx)-a(mb-mb)+x(mx-mb)=0\\ \Rightarrow a(mb-mx)-x(mb-mx)=0\\ \Rightarrow (a-x)(mb-mx)=0\\ \Rightarrow (a-x)(b-x)=0
The solution for this quadratic equation is either
x=a
or
x=b
\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
0%
1
0%
2
0%
3
Explanation
\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}
C_{1}\rightarrow C_{1}+C_{2}
=\begin{vmatrix} 2^{ 3 } + 3.2^{ 3 }+3.2+1 & 3^{ 3 }& 3.2^{2}+3.2+1 \\ 3^{ 3 } +3.3^{2}+3.3+1& 4^{ 3 } & 3.3^{ 2 }+3.3+1 \\ 4^{ 3 }+3.4^{2}+3.4+1 & 5^{ 3 } & 3.4^{ 2 }+3.4+1 \end{vmatrix}
=\begin{vmatrix} 3^{ 3 } & 3^{ 3 } & 3.2^{ 2 }+3.2+1 \\ 4^{ 3 } & 4^{ 3 } & 3.3^{ 2 }+3.3+1 \\ 5^{ 3 } & 5^{ 3 } & 3.4^{ 2 }+3.4+1 \end{vmatrix}
= 0
Find
\begin{vmatrix}\log e & \log e^{2} & \log e^{3} \\ \log e^{2} & \log e^{3} & \log e^{4} \\ \log e^{3} & \log e^{4} & \log e^{5}\end{vmatrix}
.
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0
0%
1
0%
4 \log e
0%
5 \log e
Explanation
As we know that,
loge^n=nloge
So, We can write given Determinant as
\begin{vmatrix}log e & 2log e & 3log e \\ 2log e & 3log e & 4log e \\ 3log e &\ 4log e & 5log e \end{vmatrix}=(loge)^3\begin{vmatrix}1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix}
.... [Taking
loge
common from each row]
Expanding the Given Determinant , we get
=(loge)^3|(15-16)-2(10-12)+3(8-9)|=(loge)^3\times 0=0
If
\left| {\begin{array}{*{20}{c}}1&3&2\\1&{x - 1}&{2x + 2}\\2&5&9\end{array}} \right| = 0
, then
x
is equal to :-
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2
0%
1
0%
4
0%
0
Explanation
Given,
\left| {\begin{array}{*{20}{c}}1&3&2\\1&{x - 1}&{2x + 2}\\2&5&9\end{array}} \right| = 0
\begin{array}{l}\left| {\begin{array}{*{20}{c}}1&3&2\\1&{x - 1}&{2x + 2}\\2&5&9\end{array}} \right|\\{R_2} = {R_2} - {R_1}\\{R_3} = {R_3} - 2{R_1}\\\left| {\begin{array}{*{20}{c}}1&3&4\\0&{x - 4}&{2x - 2}\\0&{ - 1}&1\end{array}} \right|\\\left| {\begin{array}{*{20}{c}}{x - 4}&{2x - 2}\\{ - 1}&1\end{array}} \right| = 0\\x - 4 + 2x - 2 = 0\\3x - 6 = 0\\x = 2\end{array}
If
\Delta = \begin{vmatrix} x-3 & 2x+1 & 2 \\ 3x+2 & x+2 & 1 \\ 5x+1 & 5x+4 & 5 \end{vmatrix}
, then
\Delta
is
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multiple of
x^2
0%
15
0%
a multiple of
x
0%
-15
Explanation
\left| \begin{matrix} x-3 & 2x+1 & 2 \\ 3x+2 & x+2 & 1 \\ 5x+1 & 5x+4 & 5 \end{matrix} \right|
=\cfrac { 1 }{ 2 } \left| \begin{matrix} 2(x-3) & 2(2x+1) & 2(2) \\ 3x+2 & x+2 & 1 \\ 5x+1 & 5x+4 & 5 \end{matrix} \right|
=\cfrac { 1 }{ 2 } \left| \begin{matrix} 2x-6 & 4x+2 & 4 \\ 3x+2 & x+2 & 1 \\ 5x+1 & 5x+4 & 5 \end{matrix} \right|
Apply
R_{1}\rightarrow R_{1}+R_{2}-R_{3}
=\cfrac { 1 }{ 2 } \left| \begin{matrix} -5 & 0 & 0 \\ 3x+2 & x+2 & 1 \\ 5x+1 & 5x+4 & 5 \end{matrix} \right|
=\cfrac{1}{2}\times (-5)\left \{ 5(x+2)-1(5x+4)\right \}
=\cfrac{-5}{2}\times 6
=-15
Find the determinant of given matrix
\left[ \begin{matrix} a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{matrix} \right]
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0%
2(a+b+c)^{3}
0%
(a-b-c)^{3}
0%
2(a-b-c)^{3}
0%
(a+b+c)^{3}
Explanation
\left| \begin{matrix} a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{matrix} \right|
=\left| \begin{matrix} a+b+c & a+b+c & a+b+c \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{matrix} \right| { R }_{ 1 }\rightarrow { R }_{ 1 }+{ R }_{ 2 }+{ R }_{ 3 }
Taking out common
a+b+c
from
{R_1}
=\left( a+b+c \right) \left| \begin{matrix} 1 & 1 & 1 \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{matrix} \right|
=\left( a+b+c \right) \left| \begin{matrix} 1 & 0 & 0 \\ 2b & -\left( a+b+c \right) & 0 \\ 2c & 0 & -\left( a+b+c \right) \end{matrix} \right| { R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 },{ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 1 }
Expanding along
{R_1}
={ \left( a+b+c \right) }^{ 3 }
If
\triangle =\begin{bmatrix} { a }_{ 1 } & { b }_{ 1 } & { c }_{ 1 } \\ { a }_{ 2 } & { b }_{ 2 } & { c }_{ 2 } \\ { a }_{ 3 } & { b }_{ 3 } & { c }_{ 3 } \end{bmatrix}
and
{A}_{2},{B}_{2},{C}_{2}
are respectively cofactors of
{a}_{2},{b}_{2},{c}_{2}
then
{a}_{1}{A}_{2}+{b}_{1}{B}_{2}+{c}_{1}{C}_{2}
is equal to ?
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-\triangle
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0
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\triangle
0%
none\ of\ these
Explanation
Co-factor of
\displaystyle a_2 = (-1)^{i+j}\left|\begin{matrix} b_1 & c_1 \\ b_3 & c_3 \end{matrix}\right| = - [b_1 \, c_3 - c_1 \, b_3] = A_2
i=2 j=1
of
\displaystyle b_2 = (-1)^{2+2} \left|\begin{matrix} a_1 & c_1 \\ a_3 & c_3 \end{matrix}\right| = [a_1 \, c_3 - c_1 \, a_3] = B_2
of
\displaystyle c_2 = (-1)^{2+3} \left|\begin{matrix} a_1 & b_1 \\ a_3 & b_3 \end{matrix}\right| = -[a_1 \, b_3 \, - b_1 \, a_3 ] = C_2
a_1 \, A_2 + b_1 \, B_2 +c_1 \, C_2
\displaystyle -a_1 \, b_1 \, c_3 + a_1 \, c_1 \, b_3 + b_1 \, a_1 \, c_3 - b_1 c_1 \, a_3 -c_1 \, a_1 \, b_3 + c_1 \, b_1 \, a_3
\displaystyle = 0
the below matrix relation is
\begin{vmatrix} 1 & { a }^{ 2 } & { a }^{ 3 } \\ 1 & { b }^{ 2 } & { b }^{ 3 } \\ 1 & { c }^{ 2 } & { c }^{ 3 } \end{vmatrix}=\begin{vmatrix} 0 & { a }^{ 2 }-{ c }^{ 2 } & { a }^{ 3 }-{ c }^{ 3 } \\ 0 & { b }^{ 2 }-{ c }^{ 2 } & { b }^{ 3 }-{ c }^{ 3 } \\ 1 & { c }^{ 2 } & { c }^{ 3 } \end{vmatrix}
=\left( a-b \right) \left( b-c \right) \begin{vmatrix} 0 & { a }-{ c } & { a }^{ 2 }+ac{ +c }^{ 2 } \\ 0 & { b }-{ c } & { b }^{ 2 }+bc+{ c }^{ 2 } \\ 1 & { c }^{ 2 } & { c }^{ 3 } \end{vmatrix}
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True
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False
Let the matrix A and B be defined as
A = \left( {\matrix{ 3 & 2 \cr 2 & 1 \cr } } \right)
and
B = \left( {\matrix{ 3 & 1 \cr 7 & 3 \cr } } \right)
then the value of
Det.\left( {2{A^9}{b^{ - 1}}} \right),
is
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0%
2
0%
1
0%
-1
0%
-2
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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