Tag: maths

Questions Related to maths

If you switch the first row with the fourth row, what will the new first row be?
$\begin{bmatrix}3&4&2&11\9&1&0&0\0&1&0&2\0&0&6&1\end{bmatrix}$

maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

  1. $3, 4, 2, 11$

  2. $9, 1, 0, 0$

  3. $0, 1, 0, 2$

  4. $0, 0, 6, 1$

  5. $0, 2, 0, 3$

Show answer
Correct Option: D
Explanation:
Given Matrix$=\begin{bmatrix} 3 & 4 & 2 & 11 \\ 9 & 1 & 0 & 0 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 6 & 1 \end{bmatrix}$
First and fourth row are interchanged
New matrix obtained $=\begin{bmatrix} 0 & 0 & 6 & 1 \\ 9 & 1 & 0 & 0 \\ 0 & 1 & 0 & 2 \\ 3 & 4 & 2 & 1 \end{bmatrix}$
New first row$=\begin{bmatrix} 0 & 0 & 6 & 1 \end{bmatrix}$ Option D

Use a transformation matrix to find the image of $D(-7,6)$ after a rotation of $180^0$ counterclockwise around the origin.

maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

  1. $(7,6)$

  2. $(-7,-6)$

  3. $(7,-6)$

  4. $(-7,6)$

Show answer
Correct Option: C
Explanation:

The transformation matrix for rotation  is $\begin{bmatrix} cos\theta  & -sin\theta  \ sin\theta  & cos\theta  \end{bmatrix}$

For $\theta=180^{0}$ , the transformation matrix will be $\quad \begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix}$
So the image of point $(-7,6)$ is $\quad \begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix}\begin{bmatrix} -7 \ 6 \end{bmatrix}=\begin{bmatrix} 7 \ -6 \end{bmatrix}$
Therefore the correct option is $C$

If a matrix A is such that $3{A^3} + 2{A^2} + 5A + I = 0$ , then $A^{-1}$ is equal to

maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

  1. $ - (3{A^2} + 2A + 5)$

  2. $3{A^2} + 2A + 5$

  3. $3{A^2} - 2A - 5$

  4. None of these

Show answer
Correct Option: A
Explanation:

$3A^3+2A^2+5A+I=0$
$3A^3+2A^2+5A+AA^{-1}=0$
$A^{-1}=-3A^2-2A-5$

If $A=\begin{bmatrix} 3 & -2 \ 5 & 8 \end{bmatrix}$, then $A^{-1}=$

maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

  1. $\frac{1}{30}\begin{bmatrix} 8 & 2 \ -5 & 3 \end{bmatrix}$

  2. $\frac{1}{34}\begin{bmatrix} 8 & 2 \ -5 & 3 \end{bmatrix}$

  3. $-\frac{1}{34}\begin{bmatrix} -8 & -2 \ -5 & 3 \end{bmatrix}$

  4. None of these

Show answer
Correct Option: B
Explanation:

$A=\left[{\begin{array}{cc}3&2\5&8\end{array}}\right]$

$A^{-1}=\cfrac{1}{ad-bc}\left[{\begin{array}{cc}d&-b\-c&a\end{array}}\right]$  (determinant)
$=\cfrac{1}{3\times 8-(-2\times 5)}\left[{\begin{array}{cc}8&2\-5&3\end{array}}\right]$  
$=\cfrac{1}{34}\left[{\begin{array}{cc}8&2\-5&3\end{array}}\right]$  

The inverse of the $\begin{bmatrix}- 1 & 5\ - 3 & 2\end{bmatrix}$ is

maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

  1. $\frac{1}{13} \begin{bmatrix}
    2 & - 5\
    3 & - 1
    \end{bmatrix}$

  2. $\frac{1}{13} \begin{bmatrix}
    - 1 & 5\
    - 3 & 2
    \end{bmatrix}$

  3. $\frac{1}{13} \begin{bmatrix}
    - 1 & - 3\
    5 & 2
    \end{bmatrix}$

  4. $\frac{1}{13} \begin{bmatrix}
    1 & 5\
    3 & - 2
    \end{bmatrix}$

Show answer
Correct Option: A
Explanation:
$A=\left[\begin{matrix} -1 & 5 \\ -3  & 2 \end{matrix} \right]$

$\left|A\right|=-2+15=13\neq 0$

Hence ${A}^{-1}$ exists.

${C} _{ij}={\left(-1\right)}^{i+j}{M} _{ij}$

${C} _{11}=2,\,{C} _{12}=3,\,{C} _{21}=-5$  and ${C} _{22}=-1$

${C} _{ij}=\left[\begin{matrix} 2 & 3 \\ -5  & -1 \end{matrix} \right]$

Adj${A}={{C} _{ij}}^{T}=\left[\begin{matrix} 2 & -5 \\ 3  & -1 \end{matrix} \right]$

${A}^{-1}=\dfrac{Adj{\left(A\right)}}{\left|A\right|}=\dfrac{1}{13}\left[\begin{matrix} 2 & -5 \\ 3  & -1 \end{matrix} \right]$

The inverse of the matrix $\begin{bmatrix} 5 & -2 \ 3 & 1 \end{bmatrix}$ is 

maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

  1. $\dfrac { 1 }{ 11 } \begin{bmatrix} 1 & 2 \ -3 & 5 \end{bmatrix}$

  2. $\begin{bmatrix} 1 & 2 \ -3 & 5 \end{bmatrix}$

  3. $\dfrac { 1 }{ 13 } \begin{bmatrix} -2 & 5 \ 1 & 3 \end{bmatrix}$

  4. $\begin{bmatrix} 1 & 3 \ -2 & 5 \end{bmatrix}$

Show answer
Correct Option: A
Explanation:

$\begin{array}{l} A=\left[ \begin{array}{l} 5\, \, \, \, \, -2 \ 3\, \, \, \, \, \, \, \, \, 1 \end{array} \right]  \ \left| A \right| =5+6=11\ne 0 \ so,\, A\, \, is\, \, \, non-\sin  gular\, ,\, { A^{ -1 } }\, \, is\, \, exist \ so,m\, { A _{ 11 } }=1,\, \, \, \, \, { A _{ 12 } }=-3,\, \, \, \, { A _{ 21 } }=2,\, \, \, \, \, \, { A _{ 22 } }=5 \ A=\left( { \begin{array} { *{ 20 }{ c } }1 & { -3 } \ 2 & 5 \end{array} } \right) \Rightarrow AdjA=\left( { \begin{array} { *{ 20 }{ c } }1 & 2 \ { -3 } & 5 \end{array} } \right)  \ { A^{ -1 } }=\frac { 1 }{ { \left| A \right|  } } adjA\, \, \, \, \Rightarrow \, \, \, \, \frac { 1 }{ { 11 } } \left( { \begin{array} { *{ 20 }{ c } }1 & 2 \ { -3 } & 5 \end{array} } \right)  \end{array}$


Hence, this is the answer.

Let $P(x, y)$ be any given point and $\displaystyle P(x _{1},y _{1})$ be the image of $P(x,y)$ after reflection.
The matrix of reflection of point $P$ through the line $\displaystyle y = x $ is given by

maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

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

  2. $\displaystyle \begin{bmatrix}-1 &0 \0 &-1\end{bmatrix}$

  3. $\displaystyle \begin{bmatrix}-1 &0 \0 &1\end{bmatrix}$

  4. $\displaystyle \begin{bmatrix}0 &1 \1&0\end{bmatrix}$

Show answer
Correct Option: D
Explanation:

Reflection of matrix $P(x,y)$ through the line $y=mx$ making an angle $\theta$ with $x-$axis is

 
$\begin{bmatrix} \cos  2\theta  & \sin { 2\theta  }  \ \sin{ 2\theta  } & -\cos { 2\theta  }  \end{bmatrix}$

Given line is $y=x$ which makes an angle of $45^{\circ}$ with $x-$axis

Hence, the transformation matrix is $\begin{bmatrix} 0&1\1&0\end{bmatrix}$

The line $3x-4y+7=0$ is rotated through an angle $\dfrac {\pi}{4}$ in clockwise direction about the point $\left (1,1\right)$. The equation of the line in its new position is

maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

  1. $7y+x-6=0$

  2. $7y-x-6=0$

  3. $x+7y=8$

  4. $7y-x+6=0$

Show answer
Correct Option: A
Explanation:

Given slope $=\dfrac{3}{4} < 1$
$< 45^{o}$
$\therefore $ after rational clock slope because negative
$\dfrac{m _1-m _{2}}{1+m _{1},m _{2}}=\tan 45$
$m _{1}=\dfrac{3}{4} \,\,\,m _{2}=$ new slope
$\left| \dfrac{\dfrac{3}{4}-m}{1+\dfrac{3}{4} m} \right|=1$
$\dfrac{3}{4}-m= \pm \left( 1+\dfrac{3}{4}m \right)$
$\dfrac{3}{4}-m=1+ \dfrac{3}{4} m$
$\dfrac{7}{4}m=\dfrac{-1}{4}$
$m=\dfrac{-1}{7}$ appeared 
$\therefore$ satisfy slope and pt in option to save time.

Let $\displaystyle A\equiv \left( 2,0 \right) $ and $\displaystyle B\equiv \left( 3,1 \right) $. The line $\displaystyle AB$ turns about $\displaystyle A$ through an angle $\displaystyle \frac { \pi  }{ 12 } $ in the clockwise sense, and the new position of $\displaystyle B$ is $\displaystyle B'$. Then $\displaystyle B'$ has the co-ordinates :-

maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

  1. $\displaystyle \left( \frac { 2\sqrt { 2 } -\sqrt { 3 } }{ \sqrt { 2 } } ,\frac { 1 }{ \sqrt { 2 } } \right) $

  2. $\displaystyle \left( \frac { 2\sqrt { 2 } +\sqrt { 3 } }{ \sqrt { 2 } } ,\frac { 1 }{ \sqrt { 2 } } \right) $

  3. $\displaystyle \left( \frac { \sqrt { 3 } -2\sqrt { 2 } }{ \sqrt { 2 } } ,\frac { 1 }{ \sqrt { 2 } } \right) $

  4. $\displaystyle \left( \frac { \sqrt { 3 } -2\sqrt { 2 } }{ 2 } ,\frac { 1 }{ \sqrt { 2 } } \right) $

Show answer
Correct Option: B
Explanation:

Slope of the line $\displaystyle AB=\frac { 0-1 }{ 2-3 } =1$
$\therefore \angle BAX={ 45 }^{ o }$
Given $\angle B'AB={ 15 }^{ o }\Rightarrow \angle B'AX={ 30 }^{ o }$
Therefore slope of the line $\displaystyle AB'=\tan { { 30 }^{ o } } =\frac { 1 }{ \sqrt { 3 }  } $
Now line $AB'$ makes an angle of ${ 30 }^{ o }$ with positive direction of $x$-axis and 
$AB'=AB=\sqrt { { \left( 3-2 \right)  }^{ 2 }+{ \left( 1-0 \right)  }^{ 2 } } =\sqrt { 2 } $
Therefore coordinates are $\displaystyle \left( 2+\sqrt { 2 } \cos { { 30 }^{ o } } ,0+\sqrt { 2 } \sin { { 30 }^{ o } }  \right) =\left( \frac { 2\sqrt { 2 } +\sqrt { 3 }  }{ \sqrt { 2 }  } ,\frac { 1 }{ \sqrt { 2 }  }  \right) $

Without changing the direction of coordinates axes,  origin is transferred to $(\alpha ,\ \beta)$ so that linear term in the equation $x^{2}+y^{2}+2x-4y+6=0$ are eliminated the point $(\alpha ,\ \beta)$ is   

maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

  1. $(-1,\ 2)$

  2. $(1,\ -2)$

  3. $(1,\ 2)$

  4. $(-1,\ -2)$

Show answer
Correct Option: B