Determinants Test

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Determinants Test - 48

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Question 1
1 / -0

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

A
$$\left (\dfrac {5 - i\sqrt {3}}{2}\right ) \begin{bmatrix}

1 & 0\\

0 & 1

\end{bmatrix}$$

B
$$\left (\dfrac {3 - i\sqrt {3}}{2}\right ) \begin{bmatrix}

1 & 0\\

0 & 1

\end{bmatrix}$$

C
$$\begin{bmatrix}

1 & 0\\

0 & 1

\end{bmatrix}$$

D
$$(2 + i\sqrt {3})\begin{bmatrix}

1 & 0\\

0 & 1

\end{bmatrix}$$

Solution

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, $$ f(A) = A^{2} + 2 = \begin{bmatrix}-\omega^{2} + \omega & 0\\ 0 & -\omega^{2} + \omega\end{bmatrix} + \begin{bmatrix}2 & 0\\ 0 & 2\end{bmatrix}$$
$$- [\omega^{2} + \omega + 2] \begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}$$
$$= (3 + 2\omega) \begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}$$
$$ = (2 + i\sqrt {3})\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}$$
Question 2
1 / -0

$$\begin{vmatrix} { \left( { a }^{ x }+{ a }^{ -x } \right) }^{ 2 } & { \left( { a }^{ x }-{ a }^{ -x } \right) }^{ 2 } & 1 \\ { \left( b^{ x }+{ b }^{ -x } \right) }^{ 2 } & { \left( { b }^{ x }-{ b }^{ -x } \right) }^{ 2 } & 1 \\ { \left( { c }^{ x }+{ c }^{ -x } \right) }^{ 2 } & { \left( { c }^{ x }-{ c }^{ -x } \right) }^{ 2 } & 1 \end{vmatrix}$$ is equal to

A
$$0$$

B
$$2abc$$

C
$${ a }^{ 2 }{ b }^{ 2 }{ c }^{ 2 }$$

D
None of these

Solution

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

$$\Delta =\begin{vmatrix} { \left( 1+1 \right) }^{ 2 } & { \left( 1-1 \right) }^{ 2 } & 1 \\ { \left( 1+1 \right) }^{ 2 } & { \left( 1-1 \right) }^{ 2 } & 1 \\ { \left( 1+1 \right) }^{ 2 } & { \left( 1-1 \right) }^{ 2 } & 1 \end{vmatrix}$$

$$=\begin{vmatrix} 4 & 0 & 1 \\ 4 & 0 & 1 \\ 4 & 0 & 1 \end{vmatrix}$$

$$=0$$
Question 3
1 / -0

If $$x=cy+bz,y=az+cx,z=bx+ay$$, where $$x,y,z$$ are not all zero, then the value of $${ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+2abc$$

A
$$0$$

B
$$1$$

C
$$-1$$

D
None of these

Solution

Given equations are $$x=cy+bz, y=az+cx, z=bx+ay$$
Given system has non-trivial solution
$$\quad \begin{vmatrix} 1 & -c & -b \\ c & -1 & a \\ b & a & -1 \end{vmatrix}=0$$
$$\Rightarrow 1-abc-abc-{ b }^{ 2 }-{ a }^{ 2 }-{ c }^{ 2 }=0$$
$$\Rightarrow { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+2abc=1$$
Question 4
1 / -0

If the vectors $$\vec {a}, \vec {b}, \vec {c}$$ are coplanar, then the value of $$\begin{vmatrix}\vec {a}& \vec {b} & \vec {c}\\ \vec {a}.\vec {a} & \vec {a}.\vec {b} & \vec {a}.\vec {c}\\ \vec {b}.\vec {a} & \vec {b}.\vec {b} & \vec {b} . \vec {c}\end{vmatrix} =$$

A
$$1$$

B
$$0$$

C
$$-1$$

D
$$\vec {a} + \vec {b} + \vec {c}$$

Solution

$$\bar { a } \cdot \bar { b } =\bar { a } \bar { b } \cos { \theta } $$
Co planar
$$\Rightarrow \theta =0$$
$$\Rightarrow \bar { a } \cdot \bar { b } =0$$
Parallelly $$\bar { a } \cdot \bar { c } =\bar { a } \cdot \bar { c } =0$$
$$\begin{vmatrix} \bar { a } & \bar { b } & \bar { c } \\ \bar { a } \cdot \bar { a } & \bar { a } \cdot \bar { b } & \bar { a } \cdot \bar { c } \\ \bar { b } \cdot \bar { a } & \bar { b } \cdot \bar { b } & \bar { b } \cdot \bar { c } \end{vmatrix}$$
$$=\begin{vmatrix} \bar { a } & \bar { b } & \bar { c } \\ \bar { a } \cdot \bar { a } & 0 & 0 \\ 0 & \bar { b } \cdot \bar { b } & 0 \end{vmatrix}$$
$$=\bar { a } \left( 0 \right) -\bar { b } \left( 0 \right) +\bar { c } \left( \bar { a } \cdot \bar { a } -\bar { b } \cdot \bar { b } \right) $$
$$=0$$
Question 5
1 / -0

Let $$A^{-1}\begin{bmatrix} 1 & 2017 & 2\\ 1 & 2017 & 4 \\ 1 & 2018 & 8\end {bmatrix}$$. Then $$|2A|-|2A^{-1}|$$ is equal to.

A
$$3$$

B
$$-3$$

C
$$12$$

D
$$-12$$

Solution

$$\left | A^{-1} \right | = \dfrac{1}{\left | A \right |}$$

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

$$-\left | 2A^{-1} \right | + \left | 2A \right |$$

$$=-(-2)\times 2^{3}+\dfrac{-1}{2}\times 2^{3} =12 $$
Question 6
1 / -0

If $$A, B, C$$ are collinear points such that $$A(3, 4), C(11, 10)$$ and $$AB = 2.5$$ then point $$B$$ is

A
$$\left(5,\dfrac{11}{2}\right)$$

B
$$\left(\dfrac{5}{2},11\right)$$

C
$$(5, 5)$$

D
$$(5, 6)$$

Solution

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 $$\left(5,\dfrac{11}{2}\right)$$
Question 7
1 / -0

Let $$z = \begin{vmatrix} 1& 1 + 2i & -5i\\ 1 - 2i & -3 & 5 + 3i\\ 5i & 5 - 3i & 7\end{vmatrix}$$, then

A
$$z$$ is purely real

B
$$z$$ is purely imaginary

C
$$(z - \overline {z})i = 0$$

D
$$(z + \overline {z})i = 0$$

Solution

$$z=\left| \begin{matrix} 1 & 1+2i & -5i \\ 1-2i & -3 & 5+3i \\ 5i & 5-3i & 7 \end{matrix} \right| \\ \quad =1(-21-36)\quad -(1+2i)(7-14i-25i+15)\quad -5i(-13i-1)\\ \quad =-57-(1+2i)(22-39i)-65+5i\\ \quad =-57-(22-39i+44i+78)-65+5i\\ \quad =-57-100-5i-65+5i\\ \quad =-222$$
$$ \therefore z$$ is purely real.
Question 8
1 / -0

If $$A = \begin{bmatrix} 2& -3\\ 4 & 1\end{bmatrix}$$, then adjoint of matrix $$A$$ is _______.

A
$$ \begin{bmatrix} 1& 3\\ -4 & 2\end{bmatrix}$$

B
$$ \begin{bmatrix} 1& -3\\ -4 & 2\end{bmatrix}$$

C
$$ \begin{bmatrix} 1& 3\\ 4 & -2\end{bmatrix}$$

D
$$ \begin{bmatrix} -1& -3\\ -4 & 2\end{bmatrix}$$

Solution

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$$\left({C}_{ij}\right)={\begin{bmatrix} 1 & -4 \\ 3 & 2\end{bmatrix}}^{T}$$

$$=\begin{bmatrix} 1 & 3 \\ -4 & 2\end{bmatrix}$$
Question 9
1 / -0

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:

A
$$\displaystyle \left( \frac{5}{2}, 0 \right)$$

B
$$\displaystyle \left( \frac{5}{3}, 0 \right)$$

C
$$\displaystyle \left( 0, 0 \right)$$

D
$$\displaystyle \left( \frac{5}{8}, 0 \right)$$

Solution

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 $$\equiv \left(\dfrac{5}{8}, 0\right)$$
Question 10
1 / -0

The number of values of k for which the system of linear equations, $$(2k+1)x+5ky=k+2$$ and $$kx+(k+2)y=2$$ has no solution, is:

A
$$Infinitely\ many$$

B
$$3$$

C
$$1$$

D
$$2$$

Solution

A non-homogeneous system of linear equations has a unique non-trivial solution if and only if its determinant is non-zero. If this determinant is zero, then the system has either no nontrivial solutions or an infinite number of solutions.

$$ \begin{bmatrix} (k+2)& 10\\ k & (k+3) \end{bmatrix}$$ $$ \begin{bmatrix} \ x\\ y\end{bmatrix}$$ $$ = \begin{bmatrix} k \\ k-1 \end{bmatrix}$$

Now it is of the Form $$Ax=B$$

Now to for the system to have no Solution , determinant of $$A$$ must be $$0$$,as follows

$$\Rightarrow |A|=(k+2)(k+3)-k\times 10=0\Rightarrow k^2-5k+6 =(k-2)(k-3)=0$$

Therefore for $$k=2,3$$ system will have no solution.

For $$k=2$$, we get infinitely many solutions, after substituting the value of $$k=2$$ in the equations.

Thus $$k=3$$

Thus, the number of solutions is $$1$$.

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Determinants Test - 48

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Determinants Test - 48

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Question 1

$$\left (\dfrac {5 - i\sqrt {3}}{2}\right ) \begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}$$

$$\left (\dfrac {3 - i\sqrt {3}}{2}\right ) \begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}$$

$$\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}$$

$$(2 + i\sqrt {3})\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}$$

Solution

Question 2

$$0$$

$$2abc$$

$${ a }^{ 2 }{ b }^{ 2 }{ c }^{ 2 }$$

None of these

Solution

Question 3

$$0$$

$$1$$

$$-1$$

None of these

Solution

Question 4

$$1$$

$$0$$

$$-1$$

$$\vec {a} + \vec {b} + \vec {c}$$

Solution

Question 5

$$3$$

$$-3$$

$$12$$

$$-12$$

Solution

Question 6

$$\left(5,\dfrac{11}{2}\right)$$

$$\left(\dfrac{5}{2},11\right)$$

$$(5, 5)$$

$$(5, 6)$$

Solution

Question 7

$$z$$ is purely real

$$z$$ is purely imaginary

$$(z - \overline {z})i = 0$$

$$(z + \overline {z})i = 0$$

Solution

Question 8

$$ \begin{bmatrix} 1& 3\\ -4 & 2\end{bmatrix}$$

$$ \begin{bmatrix} 1& -3\\ -4 & 2\end{bmatrix}$$

$$ \begin{bmatrix} 1& 3\\ 4 & -2\end{bmatrix}$$

$$ \begin{bmatrix} -1& -3\\ -4 & 2\end{bmatrix}$$

Solution

Question 9

$$\displaystyle \left( \frac{5}{2}, 0 \right)$$

$$\displaystyle \left( \frac{5}{3}, 0 \right)$$

$$\displaystyle \left( 0, 0 \right)$$

$$\displaystyle \left( \frac{5}{8}, 0 \right)$$

Solution

Question 10

$$Infinitely\ many$$

$$3$$

$$1$$

$$2$$

Solution

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$$\left (\dfrac {5 - i\sqrt {3}}{2}\right ) \begin{bmatrix}

1 & 0\\

0 & 1

\end{bmatrix}$$

$$\left (\dfrac {3 - i\sqrt {3}}{2}\right ) \begin{bmatrix}

1 & 0\\

0 & 1

\end{bmatrix}$$

$$\begin{bmatrix}

1 & 0\\

0 & 1

\end{bmatrix}$$

$$(2 + i\sqrt {3})\begin{bmatrix}

1 & 0\\

0 & 1

\end{bmatrix}$$