Tag: maths

Quizzes (518) Questions (9324)

Questions Related to maths

Multiple choice lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

The points $(k, 2-2k)$, $(-k+1, 2k)$ and $(-4-k, 6-2k)$ are collinear for

  1. all values of k
  2. $k=-1$
  3. $k=1/2$

  4. no value of k.
Reveal answer Fill a bubble to check yourself
B,C Correct answer
Explanation

The given points are collinear if 


$\displaystyle \begin{vmatrix} k  & 2-2k  & 1 \ -k+1  & 2k  & 1 \ -4-k & 6-2k  & 1  \end{vmatrix}=0$

$\Rightarrow  
\begin{vmatrix} k  & 2-2k  & 1 \ -2k+1  & 4k-2  & 0 \ -4-2k & 4  & 0  \end{vmatrix}=0$       $ \left [ R _{2}\rightarrow R _{2}-R _{1}, R _{3}\rightarrow R _{3}-R _{1} \right ]$


$\Rightarrow 4(-2k+1)-(-4-2k)(4k-2)=0$

$\Rightarrow (1-2k)(4-8-4k)=0$

$\Rightarrow (1-2k)(k+1)=0$

$\Rightarrow k=-1 \ or \ k=1/2$

Multiple choice lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

The straight lines $\mathrm{x}+2\mathrm{y}-9=0,3\mathrm{x}+5\mathrm{y}-5=0$ and $\mathrm{a}\mathrm{x}+\mathrm{b}\mathrm{y}-1=0$ are concurrent if the straight line $22\mathrm{x}-35\mathrm{y}-1=0$ passes through the point 

  1. (a, b)

  2. (b,a)

  3. (-a,b)

  4. (-a, -b)
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$x+2y-9=0$ ---(1)


$3x+5y-5=0$ ---(2)


$ax+by-1=0$ ---(3)

Solving (1) and (2) simultaneously we get 

$y=22, x=-35$

Now, equation (1), (2) and (3) will be concurrent, that is they will pass through one point if $y=22, x=-35$ satisfy the third equation $ax+by-1=0$.

Substituting the values of 'x' and 'y' in this equation we get $-35a+22b-1=0$ ---(4)


And another equation given is $22x-35y-1=0$ ---(5)

Equation (5) will be of the form of equation (4), if we substitute

$x=b$ & $y=a$

That is $22x-35y-1=0$ passes through $(b,a)$

Multiple choice lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

If $\mathrm{a}\neq b\neq \mathrm{c}$ and if $ax+by+\mathrm{c}=0\  bx+cy+\mathrm{a}=0$ and $cx+ay+b=0$ are concurrent, 

then find the value of 
$ 2^{\mathrm{a}^{2}b^{-1}\mathrm{c}^{-1}}2^{b^{2}\mathrm{c}^{-1}\mathrm{a}^{-1}}2^{\mathrm{c}^{2}\mathrm{a}^{-1}b^{-1}}$

  1. 1

  2. 4

  3. 8

  4. 16
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For the lines to be concurrent, the determinant of the coefficients must be zero: |a, b, c; b, c, a; c, a, b| = 0. This expands to -(a^3 + b^3 + c^3 - 3abc) = 0, implying a+b+c=0 or a=b=c. Given a!=b!=c, we must have a+b+c=0. The exponent is (a^2/bc + b^2/ca + c^2/ab) = (a^3+b^3+c^3)/abc. Since a^3+b^3+c^3 = 3abc when a+b+c=0, the exponent is 3. Thus, 2^3 = 8.

Multiple choice lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

Which of the following are correct in respect of the system of equations $x + y + z = 8, x - y + 2z = 6$ and $3x - y + 5z = k$?
1. They have no solution, if $k = 15$.
2. They have infinitely many solutions, if $k = 20$.
3. They have unique solution, if $k = 25$.
Select the correct answer using the code given below

  1. 1 and 2 only

  2. 2 and 3 only

  3. 1 and 3 only

  4. 1, 2 and 3
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

First we need to know if the above equations are linearly dependent or no,

In order to figure that out lets find the determinant $D$ which is;
$D=\left| \begin{matrix} 1 & 1 & 1 \ 1 & -1 & 2 \ 3 & -1 & 5 \end{matrix} \right| =0$
as $D=0$ the equations are linearly dependent, which means the system of equations can either be inconsistent or have infinitely many solutions. That shall depend on the value of k.
For there to be inifinitely many solutions , values of ${ D } _{ 1 }, { D } _{ 2 }, { D } _{ 3 }$ should be $0$ , defined as;
${ D } _{ 1 }=\left| \begin{matrix} 8 & 1 & 1 \ 6 & -1 & 2 \ k & -1 & 5 \end{matrix} \right| \ { D } _{ 2 }=\left| \begin{matrix} 1 & 8 & 1 \ 1 & 6 & 2 \ 3 & k & 5 \end{matrix} \right| \ { D } _{ 3 }=\left| \begin{matrix} 1 & 1 & 8 \ 1 & -1 & 6 \ 3 & -1 & k \end{matrix} \right| $
which on solving you get;
${ D } _{ 1 }=\left| \begin{matrix} 8 & 1 & 1 \ 6 & -1 & 2 \ k & -1 & 5 \end{matrix} \right| =3k-60=3(k-20)\ { D } _{ 2 }=\left| \begin{matrix} 1 & 8 & 1 \ 1 & 6 & 2 \ 3 & k & 5 \end{matrix} \right| =k-20\ { D } _{ 3 }=\left| \begin{matrix} 1 & 1 & 8 \ 1 & -1 & 6 \ 3 & -1 & k \end{matrix} \right| =40-2k=-2(k-20)$
Now for all the Determinants to be zero , its evident that $k=20$
Hence statement 2 is correct.
Subsequently for there to be no solution $k\neq 20$, hence statement 1 is correct as well,
As $D$ is zero the system of equations CANNOT have a unique solution hence, statement 3 is WRONG.

Multiple choice lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

To solve  $x + y = 3 : 3 x - 2 y - 4 = 0$  by determinant method find  $D.$

  1. $5$
  2. $1$
  3. $-5$
  4. $-1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
${ a } _{ 1 }x+{ b } _{ 1 }y-{ c } _{ 1 }=0\quad \Rightarrow { a } _{ 1 }x+{ b } _{ 1 }y={ c } _{ 1 }$
${ a } _{ 2 }x+{ b } _{ 2 }y-{ c } _{ 2 }=0\quad \Rightarrow { a } _{ 2 }x+{ b } _{ 2 }y={ c } _{ 2 }$
then the solution of $x$ and $y$ can be obtained by evaluating the following integral :
$x=\frac { \left| \underset { { c } _{ 2 } }{ { c } _{ 1 } } \quad \underset { { b } _{ 2 } }{ { b } _{ 1 } }  \right|  }{ \left| \underset { { a } _{ 2 } }{ { a } _{ 1 } } \quad \underset { { b } _{ 2 } }{ { b } _{ 1 } }  \right|  } $  and  $y=\dfrac { \left| \underset { { a } _{ 2 } }{ { a } _{ 1 } } \quad \underset { { c } _{ 2 } }{ { c } _{ 1 } }  \right|  }{ \left| \underset { { a } _{ 2 } }{ { a } _{ 1 } } \quad \underset { { b } _{ 2 } }{ { b } _{ 1 } }  \right|  } $
$\therefore$    $x+y=3$
  $3x-2y=4$
can be solved using the above method
$x=\dfrac { \left| \underset { 4 }{ 3 } \quad \underset { -2 }{ 1 }  \right|  }{ \left| \underset { 3 }{ 1 } \quad \underset { -2 }{ 1 }  \right|  } \quad ;\quad y=\dfrac { \left| \underset { 3 }{ 1 } \quad \underset { 4 }{ 3 }  \right|  }{ \left| \underset { 3 }{ 1 } \quad \underset { -2 }{ 1 }  \right|  } $
$x=\dfrac { -6-4 }{ -2-3 } \quad ;\quad y=\dfrac { 4-9 }{ -5 } $
$x=\dfrac { 10 }{ -5 } \quad ;\quad y=\dfrac { -5 }{ -5 } $
$x=2\quad ;\quad y=1$
now the quantity $\left| \underset { 3 }{ 1 } \quad \underset { -2 }{ 1 }  \right| =D$ (determinant)
$D=\left| \underset { 3 }{ 1 } \quad \underset { -2 }{ 1 }  \right| =-5$
So, answer is option C.
Multiple choice lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

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, then the points $(p _{1},q _{1}),(p _{2},q _{2})$ and $(p _{3},q _{3})$ ,

  1. are collinear

  2. form an equilateral triangle

  3. form a scalene triangle

  4. form a right angled triangle
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$p _{1}x+q _{1}y=1,\ p _{2}x+q _{2}y =1\ p _{3}x+q _{3}y=1$
Given lines are concurrent
$\Rightarrow \begin{vmatrix} p _{1} & q _{1} & 1 \ p _{2} & q _{2}& 1 \ p _{3} & q _{3} & 1 \end{vmatrix}=0$

$\Rightarrow p _{1}(q _{2}-q _{3})-q _{1}(p _{2}-p _{3})+(p _{2}q _{3}-p _{3}q _{2})=0$

$\Rightarrow (p _{1}q _{2}-p _{2}q _{1})+(p _{2}q _{3}-p _{3}q _{2})+(p _{3}q _{1}-p _{1}q _{3})=0$
The left hand side of the above equation is also equal to twice the area of a triangle with coordinates $(p _1, q _1),\; (p _2, q _2),\; (p _3,q _3)$
Since it is equal to zero, $(p _{1},q _{1}),(p _{2},q _{2}),(p _{3},q _{3})$ are collinear.

Multiple choice lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

If $\Delta =\begin{vmatrix}
x+1 & x+2 & x+a\
x+2 & x+3 & x+b\
x+3 & x+4 & x+c
\end{vmatrix}=0$, then
the family of lines $ax+by+c=0$ passes through

  1. $(1, -1)$
  2. $(1, -2)$
  3. $(2, -3)$
  4. $(0, 0)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\Delta =\begin{vmatrix} x+1 & x+2 & x+a \ x+2 & x+3 & x+b \ x+3 & x+4 & x+c \end{vmatrix}=0$


Applying ${ C } _{ 2 }\rightarrow { C } _{ 2 }-{ C } _{ 1 },{ C } _{ 3 }\rightarrow { C } _{ 3 }-{ C } _{ 1 }$

$\Delta =\begin{vmatrix} x+1 & 1 & a-1 \ x+2 & 1 & b-2 \ x+3 & 1 & c-3 \end{vmatrix}=0$

Applying ${ R } _{ 2 }\rightarrow { R } _{ 2 }-{ R } _{ 1 },{ R } _{ 3 }\rightarrow { R } _{ 3 }-{ R } _{ 1 }$

$\Delta =\begin{vmatrix} x+1 & 1 & a-1 \ 1 & 0 & b-a-1 \ 2 & 0 & c-a-2 \end{vmatrix}=0$

Expanding along ${ C } _{ 2 }$
$1\left( c-a-2 \right) -2\left( b-a-1 \right) =0\ \Rightarrow a-2b+c=0$
From options for point $(1,-2)$ lies on the line $ax+by+c=0$

Multiple choice lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

If the lines $\mathrm{x}+\mathrm{p}\mathrm{y}+\mathrm{p}=0,\ \mathrm{q}\mathrm{x}+\mathrm{y}+\mathrm{q}=0$ and $\mathrm{r}\mathrm{x}+\mathrm{r}\mathrm{y}+1 =0 (\mathrm{p},\mathrm{q}, \mathrm{r}$ being distinct and $ \neq$ 1) are concurrent, then the value of
$\displaystyle \frac{p}{p-1}+\frac{q}{q-1}+\frac{r}{r-1}=$

  1. $1$
  2. $-1$
  3. $2$
  4. $-2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$x+py+p=0$
$qx+y+q=0$
$rx+ry+1=0$
$\begin{vmatrix} 1 & p & p \\ q & 1 & q \\ r & r & 1 \end{vmatrix}$
$\left( 1-qr \right) -p\left( q-pqr \right) +p\left( qr-r \right) =0$
$1-qr-pq+pqr+pqr-pr=0$
$qr(p-1)+pr(q-1)-pq(r-1)-pqr=0$
$\Rightarrow \cfrac { p }{ p-1 } +\cfrac { q }{ q-1 } +\cfrac { r }{ r-1 } =1$
Option A