Tag: applications of matrices and determinants

Questions Related to applications of matrices and determinants

If $-9$ is a root of the equation $\begin{vmatrix} x & 3 & 7 \ 2 & x & 2 \ 7 & 6 & x \end{vmatrix}=0$, then the other two roots are

business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

  1. $2,7$

  2. $-2,7$

  3. $2,-7$

  4. $-2,-7$

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Correct Option: A
Explanation:
Given $\begin{vmatrix} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \end{vmatrix}=0$
Simplifying matrix we get,
$x(x^2-12)-3(2x-14)+7(12-7x)=x^3-12x-6x+42+84-49x=x^3-67x+126$
The equation can be simplified by, $x^3-67x+126(x+9)(x^2-9x+14)$
$(x^2-9x+14)=(x^2-7x-2x+14)=(x-2)(x-7),x=2,7$
Hence the roots are $2,7,-9$.

For what value of $K$, the equation $kx-9y=66$ and $2x-3y=8$ will have no solutions?

business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

  1. $-6$

  2. $6$

  3. $\dfrac{33}{4}$

  4. none of these

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Correct Option: B
Explanation:

$kx-ay=66-----(1)$

$2x-3y=8-----(2)$
$\begin{bmatrix} k & -9 \ 2 & -3 \end{bmatrix}=0$
$-3k+18=0\Rightarrow 3k=18$
$\ { k=6 } $

If $a,\ b,\ c$ are non zeros, then the system of equations $\left( \alpha +a \right) x+\alpha y+\alpha z=0,\ \alpha x+\left( \alpha +b \right) y+\alpha z=0,\ \alpha x+\alpha y+\left( \alpha +c \right) z=0$ has a non trivial solution if

business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

  1. ${ \alpha }^{ -1 }=-\left( { a }^{ -1 }+{ b }^{ -1 }+{ c }^{ -1 } \right) $

  2. ${ \alpha }^{ -1 }=a+b+c$

  3. $\alpha +a+b+c=1$

  4. $\alpha =a+b+c$

Show answer
Correct Option: A
Explanation:

Homogeneous steamy has non- trivial solution it's mean determinant is zero

So, $\begin{vmatrix} (\alpha +a) & \alpha  & \alpha  \ \alpha  & (\alpha +b) & \alpha  \ \alpha  & \alpha  & \alpha +c \end{vmatrix}=0$
$\Rightarrow (\alpha+a)[(\alpha +b)(\alpha +c)(\alpha +c)-\alpha ^2]-[\alpha ^2+ \alpha C-\alpha^2]+\alpha [\alpha ^2-\alpha^2-2b]=0$
$\Rightarrow (\alpha +a)[\alpha ^2+\alpha c \alpha b+b^c-\alpha^2]-\alpha^2c-\alpha ^2b=0$
$\Rightarrow \alpha 2/c+\alpha2/b+\alpha ac+\alpha ab+abc-\alpha ^2c-\alpha ^2b=0$
$\Rightarrow \alpha(ab+bc+ac)=-abc $
$\Rightarrow \alpha =\dfrac{-abc}{ab+bc+ac}$
$\Rightarrow \dfrac{1}{\alpha}=-\dfrac{-(ab+bc+ac)}{abc}$
$\Rightarrow \dfrac{1}{\alpha}=-\left(\dfrac{1}{c}+\dfrac{1}{b}+\dfrac{1}{a}\right)$
$\Rightarrow \boxed{\alpha ^{-1}=-(a^{-1}+b^{-1}+c^{-1})}$

The system of equation $5x+2y=4$,$7x+3y=5$ are inconsistent.

business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

  1. True

  2. False

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Correct Option: B
Explanation:
$5x + 2y - 4 = 0$, $7x + 3y - 5 = 0$
$\dfrac{{{a _1}}}{{{a _2}}} = \dfrac{5}{7}$
$\dfrac{{{b _1}}}{{{b _2}}} = \dfrac{2}{3}$
$\dfrac{{{c _1}}}{{{c _2}}} = \dfrac{{ - 4}}{{ - 5}} = \dfrac{4}{5}$
Since $\dfrac{{{a _1}}}{{{a _2}}} \ne \dfrac{{{b _1}}}{{{b _2}}}$
The system of equations has unique solution
Therefore it is consistent
Thus it is false

If $3x-4y+2z=-1$, $2x+3y+5z=7$, $x+z=2$, then $x=?$

business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

  1. $3$

  2. $2$

  3. $1$

  4. $-1$

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Correct Option: A

The number of values of $k$ for which the system of equations 
$(k+1)x+8y = 4 $
$kx+(k+3)y = 3k-1$
has infinitely many solutions is

business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

  1. $0$

  2. $1$

  3. $2$

  4. $infinite$

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Correct Option: B
Explanation:

For infinitely many solutions,
$\displaystyle \frac { k+1 }{ k } =\frac { 8 }{ k+3 } =\frac { 4 }{ 3k-1 } $

$\Rightarrow k^{2}-4k+3=0 $ and $24k-8=4k+12$
$\Rightarrow k=3,k=1$ and $k=1$

Hence, $k=1$

If $f(x),g(x)$ and $h(x)$ are three polynomials of degree $2$ and $\Delta(x) \left| \begin{matrix} f\left( x \right)  \ f'\left( x \right)  \ f"\left( x \right)  \end{matrix}\begin{matrix} g\left( x \right)  \ g'\left( x \right)  \ g"\left( x \right)  \end{matrix}\begin{matrix} h\left( x \right)  \ h'\left( x \right)  \ g"\left( x \right)  \end{matrix} \right|$ then polynomial of degree (whenever defined)

business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

  1. $2$

  2. $3$

  3. $0$

  4. $1$

Show answer
Correct Option: C
Explanation:
$\triangle \left( x \right) =\begin{vmatrix} f\left( x \right)  & g\left( x \right)  & h\left( x \right)  \\ f^{ ' }\left( x \right)  & g^{ ' }\left( x \right)  & h^{ ' }\left( x \right)  \\ f^{ '' }\left( x \right)  & g^{ '' }\left( x \right)  & h^{ '' }\left( x \right)  \end{vmatrix}$
$\Rightarrow f\left( x \right) \left[ g^{ ' }\left( x \right) .h^{ '' }\left( x \right) -h^{ ' }\left( x \right) g^{ '' }\left( x \right)  \right] -g\left( x \right) \left[ f^{ ' }\left( x \right) h^{ '' }\left( x \right) -h^{ ' }\left( x \right) f^{ '' }\left( x \right)  \right] +h\left( x \right) \left[ f^{ ' }\left( x \right) g^{ '' }\left( x \right) -g^{ ' }\left( x \right) f^{ '' }\left( x \right)  \right] $
$\because f\left( x \right) ,g\left( x \right) ,h\left( x \right) $ are polynomial of degree $2$
$\therefore f^{ ' }\left( x \right) ,g^{ ' }\left( x \right) ,h^{ ' }\left( x \right) $ are of degree $1$
$\therefore f^{ '' }\left( x \right) ,g^{ '' }\left( x \right) ,h^{ '' }\left( x \right) $ are of degree $0$
$\therefore \triangle \left( x \right) $ is polynomial of degree $3$

The system of linear equations$X-Y+Z=1$$X+Y-Z=3$$X-4Y+4Z=\alpha $ has:

business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

  1. A unique solution when $\alpha =2$

  2. A unique solution when $\alpha \neq 2$

  3. An infinite number of solutions, when $\alpha =2$

  4. An infinite number of solution, when $\alpha =-2$

Show answer
Correct Option: A

Solve the equation for $x$.
$\left|\begin{matrix} a^2 & a & 1 \ \sin(n+1)x & \sin{nx} & \sin(n-1)x \ \cos(n+1)x & \cos{nx} & \cos(n-1)x\end{matrix}\right| = 0$. Given that $ a>0$

business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

  1. $x = n\pi$

  2. $x = (n-1)\pi$

  3. $x = (n+1)\pi$

  4. None of these

Show answer
Correct Option: A
Explanation:

$\left|\begin{matrix} a^2 & a & 1 \ \sin(n+1)x & \sin{nx} & \sin(n-1)x \ \cos(n+1)x & \cos{nx} & \cos(n-1)x\end{matrix}\right| = 0$
 
$ \Rightarrow \displaystyle { a }^{ 2 }\left[ \sin { nx } .\cos { \left( n-1 \right) x } -\cos { nx } .\sin { \left( n-1 \right) x }  \right] +a\left[ \sin  (n-1)x.\cos  (n+1)x-\cos  (n-1)x.\sin  (n+1)x \right] \ +1\left[ \sin  (n+1)x.\cos { nx } -\cos  (n+1)x.\sin { nx }  \right] =0$

$ \displaystyle \Rightarrow a^{ 2 }\sin  x-a\sin  2x+\sin  x=0\ \displaystyle \Rightarrow \sin  x\left( { a }^{ 2 }+1-2a\cos { x }  \right) =0\ \displaystyle \Rightarrow \sin  x=0\; { or }\; \cos { x } =\frac { { a }^{ 2 }+1 }{ 2a } \ \displaystyle \Rightarrow \sin  x=0\; { or }\; \cos { x } =1\quad \left( \because { a }^{ 2 }+1\ge 2a,\; a>0\Rightarrow \frac { { a }^{ 2 }+1 }{ 2a } \ge 1 \right) \ \displaystyle \Rightarrow x=n\pi \; or\; x=2n\pi, \; n \in I \ \displaystyle\Rightarrow x=n\pi $

If the system of linear equations 
$x+ay+z=3$
$x+2y+2z=6$
$x+5y+3z=b$
Has infinitely many solutions, then 

business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

  1. $a=1, b\neq 9$

  2. $a \neq-1, b=9$

  3. $a=-1, b=9$

  4. $a=-1, b \neq 9$

Show answer
Correct Option: A