Tag: applications of determinants

Questions Related to applications of determinants

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 

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

  1. (a, b)

  2. (b,a)

  3. (-a,b)

  4. (-a, -b)

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Correct Option: B
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 $(b,a)$

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}}$

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

  1. 1

  2. 4

  3. 8

  4. 16

Show answer
Correct Option: C

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

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

  1. 1 and 2 only

  2. 2 and 3 only

  3. 1 and 3 only

  4. 1, 2 and 3

Show answer
Correct Option: A
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.

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

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

  1. $5$

  2. $1$

  3. $-5$

  4. $-1$

Show answer
Correct Option: A
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.

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 $(p _{3},q _{3})$ ,

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

  1. are collinear

  2. form an equilateral triangle

  3. form a scalene triangle

  4. form a right angled triangle

Show answer
Correct Option: A
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.

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

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

  1. $(1, -1)$

  2. $(1, -2)$

  3. $(2, -3)$

  4. $(0, 0)$

Show answer
Correct Option: B
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 $ax+by+c=0$

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
$\displaystyle \frac{p}{p-1}+\frac{q}{q-1}+\frac{r}{r-1}=$

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

  1. $1$

  2. $-1$

  3. $2$

  4. $-2$

Show answer
Correct Option: A
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

If $\begin{vmatrix} x _1 & y _1 & 1 \ x _2 & y _2 & 1 \ x _3 & y _3 & 1\end{vmatrix}=\begin{vmatrix} a _1 & b _1 & 1\ a _2 & b _2 & 1 \ a _3 & b _3 & 1\end{vmatrix}$, then the two triangles with vertices $(x _1, y _1), (x _2, y _2), (x _3, y _3)$ and $(a _1,b _1)$, $(a _2, b _2)$, $(a _3, b _3)$ must be congruent.

mathematics and statistics inverse of a matrix and linear equations application of determinants area of triangle and collinearity of three points applications of determinants

  1. True

  2. False

Show answer
Correct Option: B
Explanation:

The two determinants denote twice the area of $\Delta^s$ whose vertices are $(x _1,y _1), (x _2, y _2)$, $(x _3, y _3)$ and $(a _1, b _1)$, $(a _2, b _2), (a _3, b _3)$. This the equality of two determinants implies that their areas are equal. But equality of the areas of two triangles does not imply that they are congruent.

If the area of the triangle with vertices $(2, 5), (7, k)$ and $(3, 1)$ is $10$, then find the value of $k$.

mathematics and statistics inverse of a matrix and linear equations application of determinants area of triangle and collinearity of three points applications of determinants

  1. $-5$ or $35$

  2. $5$ or $-35$

  3. $15$ or $-5$

  4. $-5$ or $-25$

Show answer
Correct Option: B
Explanation:
If $(x _1,y _1), (x _2, y _2)$ ans $(x _3, y _3)$ are the vertices of a triangle, then its area is given by $\pm \dfrac {1}{2}[x _1(y _2-y _3)+x _2(y _3-y _1)+x _3(y _1-y _2)]$ 
Given vertices are $(2,5), (7,k), (3,1)$ and area is $10$.
Therefore, $\pm 10 = \dfrac {1}{2}[2(k-1)+7(1-5)+3(5-k)]$
$\Rightarrow \pm 20=2k-2-28+15-3k$
$\Rightarrow \pm 20=-k-15$
$\Rightarrow k = 5$ or $-35$

If $\displaystyle \left | \begin{matrix}x _{1} &y _{1}  &1 \ x _{2} &y _{2}  &1 \ x _{3} &y _{3}  &1 \end{matrix} \right |=\left | \begin{matrix}1 &1  &1 \ b _{1} &b _{2}  &b _{3} \ a _{1} &a _{2}  &a _{3}\end{matrix} \right |$ then the two triangles whose vertices are $\displaystyle \left ( x _{1},y _{1} \right ), \left ( x _{2},y _{2} \right ), ( \left ( x _{3},y _{3} \right ) $ and $\displaystyle\left ( a _{1},b _{1} \right ), \left ( a _{2},b _{2} \right ), \left ( a _{13},b _{3} \right ),$ are

mathematics and statistics inverse of a matrix and linear equations application of determinants area of triangle and collinearity of three points applications of determinants

  1. congruent

  2. similar

  3. equal in area

  4. none of these

Show answer
Correct Option: C
Explanation:

If $\left( x _{ 1 },y _{ 1 } \right) ,\left( x _{ 2 },y _{ 2 } \right) ,(\left( x _{ 3 },y _{ 3 } \right) $ are the vertices of triangle , then its area is 

$A _{1}=\dfrac { 1 }{ 2 } \left| \begin{matrix} x _{ 1 } & y _{ 1 } & 1 \ x _{ 2 } & y _{ 2 } & 1 \ x _{ 3 } & y _{ 3 } & 1 \end{matrix} \right| $

If $\left( a _{ 1 },b _{ 1 } \right) ,\left( a _{ 2 },b _{ 2 } \right) ,\left( a _{ 3 },b _{ 3 } \right) $ are the vertices of triangle , then its area is

$A _{2}=\dfrac { 1 }{ 2 } \left| \begin{matrix} a _{ 1 } & b _{ 1 } & 1 \ a _{ 2 } & b _{ 2 } & 1 \ a _{ 3 } & b _{ 3 } & 1 \end{matrix} \right| $

$A _{2}=\dfrac { 1 }{ 2 } \left| \begin{matrix} a _{ 1 } & a _{ 2 } & a _{ 3 } \ b _{ 1 } & b _{ 2 } & b _{ 3 } \ 1 & 1 & 1 \end{matrix} \right|    (\because |A|=|A^{T}|)$


$A _{2}=-\dfrac{1}{2}\left| \begin{matrix} 1 & 1 & 1 \ b _{ 1 } & b _{ 2 } & b _{ 3 } \ a _{ 1 } & a _{ 2 } & a _{ 3 } \end{matrix} \right| $

Since, area is positive,

$A _{2}=\dfrac{1}{2}\left| \begin{matrix} 1 & 1 & 1 \ b _{ 1 } & b _{ 2 } & b _{ 3 } \ a _{ 1 } & a _{ 2 } & a _{ 3 } \end{matrix} \right| $

Given, $\left| \begin{matrix} x _{ 1 } & y _{ 1 } & 1 \ x _{ 2 } & y _{ 2 } & 1 \ x _{ 3 } & y _{ 3 } & 1 \end{matrix} \right| =\left| \begin{matrix} 1 & 1 & 1 \ b _{ 1 } & b _{ 2 } & b _{ 3 } \ a _{ 1 } & a _{ 2 } & a _{ 3 } \end{matrix} \right| $

$\Rightarrow A _{1}=A _{2}$