Tag: inverse of a matrix and linear equations

Questions Related to inverse of a matrix and linear equations

The area of a triangle, whose vertices are $(3, 2), (5, 2)$ and the point of intersection of the lines $x = a$ and $y = 5$, is $3$ square units. What is the value of $a$?

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. $2$

  2. $3$

  3. $4$

  4. $5$

Show answer
Correct Option: B
Explanation:
$A(3,2) B(5,2)$ intersection of $x=a,y=5$   $(a,5)$
$\cfrac { 1 }{ 2 } \left| \triangle  \right| =3$
$\triangle =6$
$=\begin{vmatrix} 1 & 1 & 1 \\ 3 & 5 & a \\ 2 & 2 & 5 \end{vmatrix}$
$=25-2a-15+2a+6-10$
$=6$
For any value of a area $=3sq.units$

If $P=(x _{1}, y _{1}), Q=(x _{2}, y _{2})$ and $R=(x _{3}, y _{3})$ are three points of a triangle in $\mathbb{R}^{2}$. Then, area of a $\triangle PQR$ in terms of determinant of matrix $M=\begin{bmatrix} 1& 1 & 1 \ x _{1} & x _{2} & x _{3} \ y _{1} & y _{2} & y _{3}\end{bmatrix}$ is

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. $-|det(M)|$

  2. $|det(M)|$

  3. $\dfrac{1}{2}|det(M)|$

  4. $2|det(M)|$

Show answer
Correct Option: C
Explanation:
$P=\left( { x } _{ 1 },{ y } _{ 1 } \right) ,Q=\left( { x } _{ 2 },{ y } _{ 2 } \right) ,R\left( { x } _{ 3 },{ y } _{ 3 } \right) $
$\left| M \right| =\triangle =\begin{vmatrix} 1 & 1 & 1 \\ { x } _{ 1 } & { x } _{ 2 } & { x } _{ 3 } \\ { y } _{ 1 } & { y } _{ 2 } & { y } _{ 3 } \end{vmatrix}$
Area of triangle$=\cfrac { 1 }{ 2 } det(M)$
Answer: Option C

If $\triangle _1,\triangle _2$ be the areas of two triangles with vertices $(b,c), (c,a), (a,b)$, and $ (ac-b^2, ab-c^2),(ba-c^2, bc-a^2), (cb-a^2, ca-b^2)$, then $\ \dfrac{\triangle _1}{\triangle _2}=(a+b+c)^2$

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

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Correct Option: A
Explanation:

$T _1 = (b, c)\ (c, a)\ (a, b) = (x _1, y _1)\ (x _2, y _2)\ (x _3, y _3)$

$T _2 = [(ac - b^2), (ab  - c^2)], [(ba  - c^2), (bc - a^2)], [(cb - a^2), (ca - b^2)]$
$T _1 = \dfrac{1}{2} [c(c - a) + a(a - b) + b(b - c)]$
$= \dfrac{1}{2} [c^2 - ac + a^2 - ab + b^2 - bc]$
$T _2 = \dfrac{1}{2} [(ab - c^2) [ba - c^2 - cb + a^2] + (bc - a^2) [cb - a^2 - ac + b^2] + (ca - b^2) [ac - b^2 - ba + c^2)]$
$= \dfrac{1}{2} [[(ab - c^2)(a - c) (a + b + c)] + (bc - a^2) [(b - a)(a + b + c)] + (ca - b^2) [(c - b) (a + b + c)]]$
$= \dfrac{(a + b + c)}{2} [a^2b - abc - ac^2 + c^3 + b^2 c - abc - a^2b + c^2a - abc - b^2 c + b^3]$
$= \dfrac{(a + b + c)}{2} (a^3 + b^3 + c^3 - 3abc)$
$= \left(\dfrac{a + b + c}{2}\right) [a + b + c] [ a^2 + b^2 + c^2 - ab - bc - ca]$
$\therefore \ \dfrac{\Delta _1}{\Delta _2} = \dfrac{(a + b + c)^2 [a^2 + b^2 + c^2 - bc - ba - ca]}{(a^2 + b^2 + c^2 - bc - ba - ca)}$
$= (a + b + c)^2$

If ${ \left( { x } _{ 1 }-{ { x } _{ 2 } } \right)  }^{ 2 }+{ \left( { y } _{ 1 }-{ y } _{ 2 } \right)  }^{ 2 }={ a }^{ 2 }$, ${ \left( x _{ 2 }-{ x } _{ 3 } \right)  }^{ 2 }+{ \left( { y } _{ 2 }-{ y } _{ 3 } \right)  }^{ 2 }={ b }^{ 2 }$, ${ \left( { x } _{ 3 }-{ x } _{ 1 } \right)  }^{ 2 }+{ \left( { y } _{ 3 }-{ y } _{ 1 } \right)  }^{ 2 }={ c }^{ 2 }$ and $k\begin{vmatrix} { x } _{ 1 } & { y } _{ 1 } & 1 \ { x } _{ 2 } & { y } _{ 2 } & 1 \ { x } _{ 3 } & { y } _{ 3 } & 1 \end{vmatrix}=(a+b+c)(b+c-a)(c+a-b)\times (a+b-c)$, then the value of $k$ is

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. $1$

  2. $2$

  3. $4$

  4. none of these

Show answer
Correct Option: C
Explanation:
Given : $k\begin{vmatrix} { x } _{ 1 } & { y } _{ 1 } & 1 \\ { x } _{ 2 } & { y } _{ 2 } & 1 \\ { x } _{ 3 } & { y } _{ 3 } & 1 \end{vmatrix}^{2}=(a+b+c)(b+c-a)(c+a-b)\times (a+b-c)$ ...... $(i)$

Area of any triangle $PQR$ with vertices $(x _1, y _1) , (x _2, y _2) $ and $(x _3, y _3)$ is given by
$\Delta = \dfrac{1}{2} \begin{vmatrix}x _1 & y _1 & 1\\ x _2 & y _2 & 1\\ x _3 & y _3 & 1\end{vmatrix}$ ......... $(ii)$

${ \left( { x } _{ 1 }-{ { x } _{ 2 } } \right)  }^{ 2 }+{ \left( { y } _{ 1 }-{ y } _{ 2 } \right)  }^{ 2 }={ a }^{ 2 }$, ${ \left( x _{ 2 }-{ x } _{ 3 } \right)  }^{ 2 }+{ \left( { y } _{ 2 }-{ y } _{ 3 } \right)  }^{ 2 }={ b }^{ 2 }$, ${ \left( { x } _{ 3 }-{ x } _{ 1 } \right)  }^{ 2 }+{ \left( { y } _{ 3 }-{ y } _{ 1 } \right)  }^{ 2 }={ c }^{ 2 }$

$\implies a,b,c$ are the length of sides of a $\triangle PQR$

Also, area of $\triangle PQR$ with sides $a, b, c$ is

$\Delta = \sqrt{s (s - a) (s - b) (s - c)}$ ......... $\left[\text{Where } s=\dfrac{a+b+c}{2}\right]$

$= \displaystyle \sqrt{\frac{1}{16}(2s)(2s - 2a)(2s - 2b)(2s - 2c)}$

$\Delta = \displaystyle \sqrt{\left [ \frac{(a + b + c)(b + c - a)(c + a - b)(a + b - c)}{16}\right ]}$ ........... $(iii)$

From $(ii)$ and $(iii)$, we get

$\displaystyle \frac{1}{2} \begin{vmatrix} x _1&  y _1& 1\\ x _2 & y _2 & 1\\ x _3 & y _3 & 1\end{vmatrix} = \sqrt{\left [\frac{(a+b+c)(b+c- a)(c + a - b)(a + b - c)}{16} \right ]}$

Squaring both sides, we have

$\Rightarrow 4 \begin{vmatrix}x _1 & y _1 & 1\\ x _2 & y _2 & 1\\ x _3 & y _3 & 1\end{vmatrix}^2 = (a+b+c)(b+c-a)(c+a- b)(a+b-c)$

Comparing above equation with $(i)$, we get $k=4$

$(x _1 - x _2)^2 + (y _1 - y _2)^2 = a^2$;
$(x _2 - x _3)^2 + (y _2 - y _3)^2 = b^2$;
$(x _3 - x _1)^2 + (y _3 - y _1)^2 = c^2$;
then find $4 \begin{vmatrix}x _1 & y _1 & 1\ x _2 & y _2 & 1\ x _3 & y _3 & 1\end{vmatrix}^2 = $

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. $(a+b+c) (b+c - a) (c + a - b) (a + b - c)$

  2. $-(a+b+c) (b+c - a) (c + a - b) (a + b - c)$

  3. $-(a+b+c) (b+c - a) (c + a - b) (a + b - c)/2$

  4. $(a+b+c) (b+c - a) (c + a - b) (a + b - c)/2$

Show answer
Correct Option: A
Explanation:

Area of the triangle PQR with vertices $(x _1, y _1) , (x _2, y _2) $ and $(x _3, y _3)$ is

$\therefore

\Delta = \dfrac{1}{2} \begin{vmatrix}x _1 & y _1 & 1\ x _2 &

y _2 & 1\ x _3 & y _3 & 1\end{vmatrix}$          ....(1)

Now the area of $\Delta$ $PQR$ with sides $a, b, c$ is

$\Delta = \sqrt{s (s - a) (s - b) (s - c)}$

$= \displaystyle \sqrt{\frac{1}{16}(2s)(2s - 2a)(2s - 2b)(2s - 2c)}$

$\Delta = \displaystyle \sqrt{\left [ \frac{(a + b + c)(b + c - a)(c + a - b)(a + b - c)}{16}\right ]}$          ....(2)

From (1) and (2), we get

$\displaystyle

\frac{1}{2} \begin{vmatrix} x _1&  y _1& 1\ x _2 & y _2 &

1\ x _3 & y _3 & 1\end{vmatrix} = \sqrt{\left [

\frac{(a+b+c)(b+c- a)(c + a - b)(a + b - c)}{16} \right ]}$

Squaring both sides, we have

$\Rightarrow

4 \begin{vmatrix}x _1 & y _1 & 1\ x _2 & y _2 & 1\ x _3

& y _3 & 1\end{vmatrix}^2 = (a+b+c)(b+c-a)(c+a- b)(a+b-c)$

The number of value of $x$ in the closed interval $[-4,-1]$, the matrix $\begin{bmatrix} 3 & -1+x & 2 \ 3 & -1 & x+2 \ x+3 & -1 & 2 \end{bmatrix}$ is singular is 

business maths inverse of a matrix and linear equations non singular matrix singular & non-singular matrix determinants

  1. $0$

  2. $1$

  3. $2$

  4. $3$

Show answer
Correct Option: C
Explanation:
For a singular matrix the value of the determinant = 0 
$ [3(-1)(2)+(-1+x)(x+2)(x+3)+(2)(3)(-1)] $
$\times  [-(2)(-1)(x+3)-(-1+x)(3)(2)-(3)(x+2)(x+3)] = 0 $
$-6+x^3+4x^2+x-6-6+2x+6+6-6x-3x^2-15x-18 = 0 $
$ x^3+x^2-18x-24 = 0 $
$ x = -4 $
$x =(3-\sqrt{33})/2=-1.372 $
$x =(3+\sqrt{33})/2= 4.372 $

Value of x in the closed interval $[-4,-1]$ are $-4,   -1.372$

If $\left[ {\begin{array}{*{20}{c}}1&{ - 1}&x\1&x&1\x&{ - 1}&1\end{array}} \right]$ has no inverse, then the real value of $x$ is 

business maths inverse of a matrix and linear equations non singular matrix singular & non-singular matrix determinants

  1. $2$

  2. $3$

  3. $0$

  4. $1$

Show answer
Correct Option: D
Explanation:

$A=\begin{vmatrix} 1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1 \end{vmatrix}has\quad no\quad inverse\quad x$

$if\,\left| A \right| = 0$

$\begin{vmatrix} 1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1 \end{vmatrix}=0$

$1\left( {x + 1} \right) - 1\left( { - 1 + x} \right) + x\left( { - 1 - {x^2}} \right) = 0$

$x + 1 + 1 - x - x - {x^3} = 0$

$ - {x^3} - x + 2 = 0$

${x^3} + x - 2 = 0$

$\left( {x - 1} \right)\left( {{x^2} + x + 2} \right) = 0$

$x = 1\,is\,real\,value\,$

The matrix $\begin{bmatrix} 1 & 0 & 1 \ 2 & 1 & 0 \ 3 & 1 & 1 \end{bmatrix}$ is:

business maths inverse of a matrix and linear equations non singular matrix singular & non-singular matrix determinants

  1. nonsingular

  2. singular

  3. skew symmetric

  4. symmetric

Show answer
Correct Option: B
Explanation:

$\begin{vmatrix} 1 & 0 & 1 \ 2 & 1 & 0 \end{vmatrix}=1\begin{vmatrix} 1 & 0 \ 1 & 1 \end{vmatrix}-0\begin{vmatrix} 2 & 0 \ 3 & 1 \end{vmatrix}+1\begin{vmatrix} 2 & 1 \ 3 & 1 \end{vmatrix}$

$=1(1-0)-0+1(2-3)$
$=1-1$
$=0$
Therefore, it is a singular matrix.

If $\begin{bmatrix} 1 & 2 & x \  4 & -1 & 7  \  2 & 4 & 6  \end{bmatrix}$ is a singular matrix, then $x=$

business maths inverse of a matrix and linear equations non singular matrix singular & non-singular matrix determinants

  1. $0$

  2. $1$

  3. $-3$

  4. $3$

Show answer
Correct Option: C
Explanation:

$\begin{vmatrix} 1 & 2 & x \ 4 & -1 & 7 \ 2 & 4 & -6  \end{vmatrix}=0$

$\Rightarrow 1\begin{vmatrix} -1 & 7 \ 4 & -6 \end{vmatrix}-2\begin{vmatrix} 4 & 7 \ 2 & -6 \end{vmatrix}+x\begin{vmatrix} 4 & -1 \ 2 & 4 \end{vmatrix}=0$

$\Rightarrow (6-28)-2(-24-14)+x(16+2)=0$
$\Rightarrow -22+76+18x=0$
$\Rightarrow 18x=-54$

$\Rightarrow x=-3$

$A$ and $B$ are two non-zero square matrices such that $AB = 0$. Then

business maths inverse of a matrix and linear equations non singular matrix singular & non-singular matrix determinants

  1. Both $A$ and $B$ are singular

  2. Either of them is singular

  3. Neither matrix is singular

  4. None of these

Show answer
Correct Option: B
Explanation:

Given $AB =0$

$\Rightarrow |AB|=0$

$\Rightarrow |A||B|=0$

$\Rightarrow |A|=0 \,or\, |B|=0$