Tag: non-homogeneous linear equations

Quizzes (0) Questions (57)

Questions Related to non-homogeneous linear equations

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

Which of the given values of $x$ and $y$ make the following pair of matrices equal.
$\displaystyle \begin{bmatrix} 3x+7 & 5 \ y+1 & 2-3x \end{bmatrix}=\begin{bmatrix} 0 & y-2 \ 8 & 4 \end{bmatrix}$

  1. $\displaystyle x=\frac { -1 }{ 3 } ,y=7$

  2. Not possible to find
  3. $\displaystyle y=7,x=\frac { -2 }{ 3 } $
  4. $\displaystyle x=\frac { -1 }{ 3 } ,y=\frac { -2 }{ 3 } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\displaystyle \begin{bmatrix} 3x+7 & 5

\ y+1 & 2-3x \end{bmatrix}=\begin{bmatrix} 0 & y-2 \ 8 & 4

\end{bmatrix}$
Equating the corresponding elements, we get,
$\displaystyle 3x+7=0\Rightarrow x=-\frac { 7 }{ 3 } $
$\displaystyle 5=y-2\Rightarrow y=7$
$\displaystyle y+1=8\Rightarrow y=7$
$\displaystyle 2-3x=4\Rightarrow x=-\frac { 2 }{ 3 } $
We find that on comparing the corresponding elements of the two matrices, we get two different values of $x$, which is not possible.
Hence, it is not possible to find the values of $x$ and $y$ for which the given matrices are equal.

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


Let $A$ be the set of all $3 \times  3$ symmetric matrices all of whose entries are either $0$ or $1$. Five of these entries are $1$ and four of them are $0$.

The number of matrices $\mathrm{A}$ in $d$ for which the system of linear equations $\mathrm{A}\begin{bmatrix}\mathrm{X}\\mathrm{Y}\\mathrm{Z}\end{bmatrix}=\begin{bmatrix}1\0\0\end{bmatrix}$ has a unique solution, is

  1. less than 4

  2. at least 4 but less than 7

  3. atleast 7 but less than 10

  4. at least 10
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\begin{bmatrix}0 & \mathrm{a} & \mathrm{b}\\mathrm{a} & 0 & \mathrm{c}\\mathrm{b} & \mathrm{c} & 1\end{bmatrix}$
either $\mathrm{b}=0$ or $\mathrm{c}=0\Rightarrow |\mathrm{A}|\neq 0$
$ \Rightarrow  2$ matrices
$\begin{bmatrix}0 & \mathrm{a} & \mathrm{b}\\mathrm{a} & 1 & \mathrm{c}\\mathrm{b} & \mathrm{c} & 0\end{bmatrix}$
either $\mathrm{a}=0$ or $\mathrm{c}=0\Rightarrow |\mathrm{A}|\neq 0 $
$\Rightarrow  2$ matrices
$\begin{bmatrix}1 & \mathrm{a} & \mathrm{b}\\mathrm{a} & \mathrm{o} & \mathrm{c}\\mathrm{b} & \mathrm{c} & \mathrm{o}\end{bmatrix}$
either $\mathrm{a}=0$ or $\mathrm{b}=0\Rightarrow |\mathrm{A}|\neq 0 $
$\Rightarrow  2$ matrices.
$\begin{bmatrix}1 & \mathrm{a} & \mathrm{b}\\mathrm{a} & 1 & \mathrm{c}\\mathrm{b} & \mathrm{c} & 1\end{bmatrix}$
If $\mathrm{a}=\mathrm{b}=0\Rightarrow |\mathrm{A}|=0$
If $\mathrm{a}=\mathrm{c}=0\Rightarrow |\mathrm{A}|=0$
If $\mathrm{b}=\mathrm{c}=0\Rightarrow |\mathrm{A}|=0$
$\Rightarrow $ there will be only 6 matrices.

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

Number of real values of $'a'$ for which the system of equations $2ax-2y+3z=0, x+ay+2z=0$ and $2x+az=0$ has a non-trivial solution, is equal to

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

For a homogeneous system of linear equations to have a non-trivial solution, the determinant of its coefficient matrix must be zero. Setting the determinant of the 3x3 matrix to zero gives a cubic equation in a, yielding 1 real value for a.

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

Let $X=\begin{bmatrix} { x } _{ 1 } \ { x } _{ 2 } \ { x } _{ 3 } \end{bmatrix};A=\begin{bmatrix} 1 & -1 & 2 \ 2 & 0 & 1 \ 3 & 2 & 1 \end{bmatrix}$ and $B=\begin{bmatrix} 3 \ 1 \ 4 \end{bmatrix}$. If $AX=B$, then $X$ is equal to

  1. $\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$
  2. $\begin{bmatrix} -1 \\ -2 \\ -3 \end{bmatrix}$
  3. $\begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix}$
  4. $\begin{bmatrix} 0 \\ 2 \\ 1 \end{bmatrix}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$AX=B$

$X=A^{-1}B$

$A^{-1}=\dfrac{1}{|A|}adj(A)$

Given,

$X=\begin{bmatrix}x _1\\ x _2\\ x _3\end{bmatrix}$

$A=\begin{bmatrix}1 &-1  &2 \\  2& 0 &1 \\  3& 2 &1 \end{bmatrix}$

$B=\begin{bmatrix}3\\ 1\\ 4\end{bmatrix}$

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

$=1(0-2)+1(2-3)+2(4-0)=-2-1+8=5$

$C _A=\begin{bmatrix}-2 &1  &4 \\  5& -5 &-5 \\  -1& 3 &2 \end{bmatrix}$

$adj(A)=C _A^T=\begin{bmatrix}-2 &5  &-1 \\  1& -5 &3 \\  4& -5 &2 \end{bmatrix}$

$A^{-1}=\dfrac{1}{5}\begin{bmatrix}-2 &5  &-1 \\  1& -5 &3 \\  4& -5 &2 \end{bmatrix}$

$X=A^{-1}B$

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

$=\dfrac{1}{5}\begin{bmatrix}-6+5-4\\ 3-5+12\\ 12-5+8\end{bmatrix}$

$=\dfrac{1}{5}\begin{bmatrix}-5\\ 10\\ 15\end{bmatrix}$

$=\begin{bmatrix} \left(\dfrac{-5}{5}\right)\\ \left(\dfrac{10}{5}\right)\\ \left(\dfrac{15}{5}\right)\end{bmatrix}$

$\begin{bmatrix}x _1\\ x _2\\ x _3\end{bmatrix}=\begin{bmatrix}-1\\ 2\\ 3\end{bmatrix}$
Multiple choice business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

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

  1. $2,7$
  2. $-2,7$
  3. $2,-7$
  4. $-2,-7$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $2,7,-9$.
Multiple choice business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

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

  1. ${ \alpha }^{ -1 }=-\left( { a }^{ -1 }+{ b }^{ -1 }+{ c }^{ -1 } \right) $
  2. ${ \alpha }^{ -1 }=a+b+c$
  3. $\alpha +a+b+c=1$
  4. $\alpha =a+b+c$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $
$\Rightarrow \alpha =\dfrac{-abc}{ab+bc+ac}$
$\Rightarrow \dfrac{1}{\alpha}=-\dfrac{-(ab+bc+ac)}{abc}$
$\Rightarrow \dfrac{1}{\alpha}=-\left(\dfrac{1}{c}+\dfrac{1}{b}+\dfrac{1}{a}\right)$
$\Rightarrow \boxed{\alpha ^{-1}=-(a^{-1}+b^{-1}+c^{-1})}$

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

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

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
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