Let S be the set of all column matrices $\left[ \begin{matrix}
{{b}_{1}} \\
{{b}_{2}} \\
{{b}_{3}} \\
\end{matrix} \right]$ such that ${{b}_{1}},{{b}_{2}},{{b}_{3}}\in R$ and the system of equations in (real variables)
$\begin{align}
& -x+2y+5z={{b}_{1}} \\
& 2x-4y+3z={{b}_{2}} \\
& x-2y+2z={{b}_{3}} \\
\end{align}$
Has at least one solution. Then which of the following system(s) in real variables has/have at least one solution each $\left[ \begin{matrix}
{{b}_{1}} \\
{{b}_{2}} \\
{{b}_{3}} \\
\end{matrix} \right]\in S$?
(a)$x+2y+3z={{b}_{1}},4y+5z={{b}_{2}},x+2y+6z={{b}_{3}}$
(b)$x+y+3z={{b}_{1}},5x+2y+6z={{b}_{2}},-2x-y-3z={{b}_{3}}$
(c)$-x+2y-5z={{b}_{1}},2x-4y+10z={{b}_{2}},x-2y+5z={{b}_{3}}$
(d)$x+2y+5z={{b}_{1}},2x+3z={{b}_{2}},x+4y-5z={{b}_{3}}$
Answer
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Hint: It is given that the system of linear equations shown above has at least one solution. This means that solutions are consistent. Now, find the determinant of the matrix constructed by the coefficients of x, y and z. If the determinant value is non zero then unique solution is possible and if determinant value is coming as 0 then infinitely many solutions are possible. And we can write linear combinations of any two planes as the third plane. And then check the value of the determinants of the coefficient matrix of all the options and see which determinant value is satisfying the determinant value that we got in the given system of equations.
Complete step-by-step answer:
We have given a system of linear equations as:
$\begin{align}
& -x+2y+5z={{b}_{1}}.........Eq.(1) \\
& 2x-4y+3z={{b}_{2}}..........Eq.(2) \\
& x-2y+2z={{b}_{3}}..............Eq.(3) \\
\end{align}$
Now, it is given that the above system of equations has at least one solution. So, two cases could be possible either the above system of equations has a unique solution or infinitely many solutions which we are going to find out by finding the determinant of the matrix constructed by the coefficients of x, y and z.
The determinant of the matrix constructed from the coefficients of x, y and z is:
$\left| \begin{matrix}
-1 & 2 & 5 \\
2 & -4 & 3 \\
1 & -2 & 2 \\
\end{matrix} \right|$
Solving the above determinant by changing row 1 to the addition of row 1 and row 3 we get,
$\begin{align}
& \left| \begin{matrix}
-1+1 & 2-2 & 5+2 \\
2 & -4 & 3 \\
1 & -2 & 2 \\
\end{matrix} \right| \\
& =\left| \begin{matrix}
0 & 0 & 7 \\
2 & -4 & 3 \\
1 & -2 & 2 \\
\end{matrix} \right| \\
\end{align}$
Now, expanding along the first row of the above determinant we get,
$\begin{align}
& 0-0+7\left( 2\left( -2 \right)-\left( -4 \right) \right) \\
& =7\left( -4+4 \right) \\
& =7\left( 0 \right) \\
& =0 \\
\end{align}$
From the above, we have got the value of determinant as 0.
We know that, when the determinant of coefficient matrix is 0 then two possibilities are either no solution or infinitely many solutions. Now, it is given in the above problem that the system of equations has at least one solution so “no solution” could not be the case and infinitely many solutions are possible.
When the system of linear equations is making infinitely many solutions then we can write a linear combination of two linear equations as the third one. Here, linear equations are in the form of plane equations.
Let us assume eq. (1, 2 and 3) as ${{P}_{1}},{{P}_{2}}\And {{P}_{3}}$ and these plane equations are making infinitely many solutions so we can write them as follows:
$\alpha {{P}_{1}}+\beta {{P}_{2}}={{P}_{3}}$
In the above equation, $\alpha ,\beta $ are constants.
Substituting the value of ${{P}_{1}},{{P}_{2}}\And {{P}_{3}}$ as eq. (1, 2 and 3) respectively,
$\alpha \left( -x+2y+5z \right)+\beta \left( 2x-4y+3z \right)=\left( x-2y+2z \right)$………..Eq. (4)
Solving the above equation we get,
$x\left( -\alpha +2\beta \right)+y\left( 2\alpha -4\beta \right)+z\left( 5\alpha +3\beta \right)=\left( x-2y+2z \right)$
Now, L.H.S will be equal to R.H.S when each of the x, y and z coefficients are equal on both the sides.
$\begin{align}
& -\alpha +2\beta =1, \\
& 2\alpha -4\beta =-2, \\
& 5\alpha +3\beta =2 \\
\end{align}$
Multiplying first equation by 5 and then add to the third equation we get,
$\begin{align}
& 10\beta +3\beta =5+2 \\
& \Rightarrow 13\beta =7 \\
& \Rightarrow \beta =\dfrac{7}{13} \\
\end{align}$
Substituting the above value of $\beta $ in the first equation we get,
$\begin{align}
& -\alpha +2\left( \dfrac{7}{13} \right)=1 \\
& \Rightarrow -\alpha +\left( \dfrac{14}{13} \right)=1 \\
& \Rightarrow -\alpha =1-\dfrac{14}{13} \\
& \Rightarrow -\alpha =\dfrac{13-14}{13} \\
& \Rightarrow \alpha =\dfrac{1}{13} \\
\end{align}$
Substituting the values of $\alpha ,\beta $ in eq. (4) we get,
$\begin{align}
& \dfrac{1}{13}\left( -x+2y+5z \right)+\dfrac{7}{13}\left( 2x-4y+3z \right)=\left( x-2y+2z \right) \\
& \Rightarrow \left( -x+2y+5z \right)+7\left( 2x-4y+3z \right)=13\left( x-2y+2z \right) \\
\end{align}$
Similarly, we can write the relationship between ${{b}_{1}},{{b}_{2}}\And {{b}_{3}}$ as:
${{b}_{1}}+7{{b}_{2}}=13{{b}_{3}}$
Now, checking the determinant of the coefficient matrix of all the options and see whether they are getting at least one solution.
Option (a)$x+2y+3z={{b}_{1}},4y+5z={{b}_{2}},x+2y+6z={{b}_{3}}$
Determinant form of the coefficient matrix of the above equations is:
$\left| \begin{matrix}
1 & 2 & 3 \\
0 & 4 & 5 \\
1 & 2 & 6 \\
\end{matrix} \right|$
Solving the above determinant by changing row 1 as the subtraction of row 1 from row 3 we get,
$\begin{align}
& \left| \begin{matrix}
1-1 & 2-2 & 6-3 \\
0 & 4 & 5 \\
1 & 2 & 6 \\
\end{matrix} \right| \\
& =\left| \begin{matrix}
0 & 0 & 3 \\
0 & 4 & 5 \\
1 & 2 & 6 \\
\end{matrix} \right| \\
\end{align}$
Now, expanding along the first row we get,
$\begin{align}
& 0-0+3\left( 0\left( 2 \right)-4\left( 1 \right) \right) \\
& =3\left( 0-4 \right) \\
& =-12 \\
\end{align}$
Hence, we are getting the non zero value of the determinant so this option is correct because we know that when determinant of the coefficient matrix is non zero then unique solution for all real values of ${{b}_{1}},{{b}_{2}}\And {{b}_{3}}$ and question is asking which options have at least one solution.
Checking option (b) $x+y+3z={{b}_{1}},5x+2y+6z={{b}_{2}},-2x-y-3z={{b}_{3}}$
Determinant form of the coefficient matrix of the above equations is:
$\left| \begin{matrix}
1 & 1 & 3 \\
5 & 2 & 6 \\
-2 & -1 & -3 \\
\end{matrix} \right|$
Solving the above determinant by changing the first row as the addition of first and third row we get,
$\begin{align}
& \left| \begin{matrix}
1-2 & 1-1 & 3-3 \\
5 & 2 & 6 \\
-2 & -1 & -3 \\
\end{matrix} \right| \\
& =\left| \begin{matrix}
-1 & 0 & 0 \\
5 & 2 & 6 \\
-2 & -1 & -3 \\
\end{matrix} \right| \\
\end{align}$
Now, expanding along first row we get,
$\begin{align}
& -1\left( 2\left( -3 \right)-6\left( -1 \right) \right)-0+0 \\
& =-1\left( -6+6 \right) \\
& =-1\left( 0 \right) \\
& =0 \\
\end{align}$
In this problem we are getting the determinant as 0. Now, we have to check the relationship between the linear equations as $1{{P}_{1}}+7{{P}_{2}}=13{{P}_{3}}$.
Substituting the plane equations of this option in the above we get,
$\begin{align}
& 1\left( x+y+3z \right)+7\left( 5x+2y+6z \right)=13\left( -2x-y-3z \right) \\
& \Rightarrow x+y+3z+35x+14y+42z=-26x-13y-39z \\
& \Rightarrow 36x+15y+45z=-26x-13y-39z \\
\end{align}$
As you can see that L.H.S is not equal to R.H.S which should be equal in the above system of linear equations so this option is incorrect.
Checking option (c) $-x+2y-5z={{b}_{1}},2x-4y+10z={{b}_{2}},x-2y+5z={{b}_{3}}$ we get,
Determinant form of the coefficient matrix of the above equations is:
$\left| \begin{matrix}
-1 & 2 & -5 \\
2 & -4 & 10 \\
1 & -2 & 5 \\
\end{matrix} \right|$
Solving the above determinant by changing the first row as the addition of the first row and third row we get,
$\begin{align}
& \left| \begin{matrix}
-1+1 & -2 & -5+5 \\
2 & -4 & 10 \\
1 & -2 & 5 \\
\end{matrix} \right| \\
& =\left| \begin{matrix}
0 & 0 & 0 \\
2 & -4 & 10 \\
1 & -2 & 5 \\
\end{matrix} \right| \\
\end{align}$
We know that, if one row becomes all 0 then the value of the determinant is 0.
Now, we have to check the relationship between the three equations using $1{{P}_{1}}+7{{P}_{2}}=13{{P}_{3}}$ this relation we get,
Substituting the plane equations given in this option in the equation of the planes we get,
$\begin{align}
& 1\left( -x+2y-5z \right)+7\left( 2x-4y+10z \right)=13\left( x-2y+5z \right) \\
& \Rightarrow -x+2y-5z+14x-28y+70z=13x-26y+65z \\
& \Rightarrow 13x-26y+65z=13x-26y+65z \\
\end{align}$
As L.H.S is equal to R.H.S so this option is correct.
Checking option (d) $x+2y+5z={{b}_{1}},2x+3z={{b}_{2}},x+4y-5z={{b}_{3}}$ we get,
Determinant form of the coefficient matrix of the above equations is:
$\left| \begin{matrix}
1 & 2 & 5 \\
2 & 0 & 3 \\
1 & 4 & -5 \\
\end{matrix} \right|$
Changing row 1 to the subtraction of row 1 from row 3 we get,
$\left| \begin{matrix}
1-1 & 4-2 & 5+5 \\
2 & 0 & 3 \\
1 & 4 & -5 \\
\end{matrix} \right|$
$=\left| \begin{matrix}
0 & 2 & 10 \\
2 & 0 & 3 \\
1 & 4 & -5 \\
\end{matrix} \right|$
Expanding the above determinant along row 1 we get,
$\begin{align}
& 0-2\left( 2\left( -5 \right)-3\left( 1 \right) \right)+10\left( 2\left( 4 \right)-0\left( 1 \right) \right) \\
& =0-2\left( -10-3 \right)+10\left( 8 \right) \\
& =2\left( 13 \right)+80 \\
& =106 \\
\end{align}$
As the determinant of the coefficient matrix is non zero which determines the unique solution for all real values of ${{b}_{1}},{{b}_{2}},{{b}_{3}}$ so option (d) is also correct.
So, the correct answer is “Option (a),(c) and (d)”.
Note: The point where the possibility of committing mistake is high like while checking the options you found the determinant of the coefficient matrix as 0 and you jump to the conclusion that infinitely many solutions are possible. But “at least one solution” is for the system of equations given in the above problem but not for the options so in options although you are getting the determinant of the coefficient matrix as 0 but it need not necessarily mean that all the system of the equations of the options in which determinant value is 0 is following the same relationship between the equations as the given system of equations. So, be careful in checking the options.
Complete step-by-step answer:
We have given a system of linear equations as:
$\begin{align}
& -x+2y+5z={{b}_{1}}.........Eq.(1) \\
& 2x-4y+3z={{b}_{2}}..........Eq.(2) \\
& x-2y+2z={{b}_{3}}..............Eq.(3) \\
\end{align}$
Now, it is given that the above system of equations has at least one solution. So, two cases could be possible either the above system of equations has a unique solution or infinitely many solutions which we are going to find out by finding the determinant of the matrix constructed by the coefficients of x, y and z.
The determinant of the matrix constructed from the coefficients of x, y and z is:
$\left| \begin{matrix}
-1 & 2 & 5 \\
2 & -4 & 3 \\
1 & -2 & 2 \\
\end{matrix} \right|$
Solving the above determinant by changing row 1 to the addition of row 1 and row 3 we get,
$\begin{align}
& \left| \begin{matrix}
-1+1 & 2-2 & 5+2 \\
2 & -4 & 3 \\
1 & -2 & 2 \\
\end{matrix} \right| \\
& =\left| \begin{matrix}
0 & 0 & 7 \\
2 & -4 & 3 \\
1 & -2 & 2 \\
\end{matrix} \right| \\
\end{align}$
Now, expanding along the first row of the above determinant we get,
$\begin{align}
& 0-0+7\left( 2\left( -2 \right)-\left( -4 \right) \right) \\
& =7\left( -4+4 \right) \\
& =7\left( 0 \right) \\
& =0 \\
\end{align}$
From the above, we have got the value of determinant as 0.
We know that, when the determinant of coefficient matrix is 0 then two possibilities are either no solution or infinitely many solutions. Now, it is given in the above problem that the system of equations has at least one solution so “no solution” could not be the case and infinitely many solutions are possible.
When the system of linear equations is making infinitely many solutions then we can write a linear combination of two linear equations as the third one. Here, linear equations are in the form of plane equations.
Let us assume eq. (1, 2 and 3) as ${{P}_{1}},{{P}_{2}}\And {{P}_{3}}$ and these plane equations are making infinitely many solutions so we can write them as follows:
$\alpha {{P}_{1}}+\beta {{P}_{2}}={{P}_{3}}$
In the above equation, $\alpha ,\beta $ are constants.
Substituting the value of ${{P}_{1}},{{P}_{2}}\And {{P}_{3}}$ as eq. (1, 2 and 3) respectively,
$\alpha \left( -x+2y+5z \right)+\beta \left( 2x-4y+3z \right)=\left( x-2y+2z \right)$………..Eq. (4)
Solving the above equation we get,
$x\left( -\alpha +2\beta \right)+y\left( 2\alpha -4\beta \right)+z\left( 5\alpha +3\beta \right)=\left( x-2y+2z \right)$
Now, L.H.S will be equal to R.H.S when each of the x, y and z coefficients are equal on both the sides.
$\begin{align}
& -\alpha +2\beta =1, \\
& 2\alpha -4\beta =-2, \\
& 5\alpha +3\beta =2 \\
\end{align}$
Multiplying first equation by 5 and then add to the third equation we get,
$\begin{align}
& 10\beta +3\beta =5+2 \\
& \Rightarrow 13\beta =7 \\
& \Rightarrow \beta =\dfrac{7}{13} \\
\end{align}$
Substituting the above value of $\beta $ in the first equation we get,
$\begin{align}
& -\alpha +2\left( \dfrac{7}{13} \right)=1 \\
& \Rightarrow -\alpha +\left( \dfrac{14}{13} \right)=1 \\
& \Rightarrow -\alpha =1-\dfrac{14}{13} \\
& \Rightarrow -\alpha =\dfrac{13-14}{13} \\
& \Rightarrow \alpha =\dfrac{1}{13} \\
\end{align}$
Substituting the values of $\alpha ,\beta $ in eq. (4) we get,
$\begin{align}
& \dfrac{1}{13}\left( -x+2y+5z \right)+\dfrac{7}{13}\left( 2x-4y+3z \right)=\left( x-2y+2z \right) \\
& \Rightarrow \left( -x+2y+5z \right)+7\left( 2x-4y+3z \right)=13\left( x-2y+2z \right) \\
\end{align}$
Similarly, we can write the relationship between ${{b}_{1}},{{b}_{2}}\And {{b}_{3}}$ as:
${{b}_{1}}+7{{b}_{2}}=13{{b}_{3}}$
Now, checking the determinant of the coefficient matrix of all the options and see whether they are getting at least one solution.
Option (a)$x+2y+3z={{b}_{1}},4y+5z={{b}_{2}},x+2y+6z={{b}_{3}}$
Determinant form of the coefficient matrix of the above equations is:
$\left| \begin{matrix}
1 & 2 & 3 \\
0 & 4 & 5 \\
1 & 2 & 6 \\
\end{matrix} \right|$
Solving the above determinant by changing row 1 as the subtraction of row 1 from row 3 we get,
$\begin{align}
& \left| \begin{matrix}
1-1 & 2-2 & 6-3 \\
0 & 4 & 5 \\
1 & 2 & 6 \\
\end{matrix} \right| \\
& =\left| \begin{matrix}
0 & 0 & 3 \\
0 & 4 & 5 \\
1 & 2 & 6 \\
\end{matrix} \right| \\
\end{align}$
Now, expanding along the first row we get,
$\begin{align}
& 0-0+3\left( 0\left( 2 \right)-4\left( 1 \right) \right) \\
& =3\left( 0-4 \right) \\
& =-12 \\
\end{align}$
Hence, we are getting the non zero value of the determinant so this option is correct because we know that when determinant of the coefficient matrix is non zero then unique solution for all real values of ${{b}_{1}},{{b}_{2}}\And {{b}_{3}}$ and question is asking which options have at least one solution.
Checking option (b) $x+y+3z={{b}_{1}},5x+2y+6z={{b}_{2}},-2x-y-3z={{b}_{3}}$
Determinant form of the coefficient matrix of the above equations is:
$\left| \begin{matrix}
1 & 1 & 3 \\
5 & 2 & 6 \\
-2 & -1 & -3 \\
\end{matrix} \right|$
Solving the above determinant by changing the first row as the addition of first and third row we get,
$\begin{align}
& \left| \begin{matrix}
1-2 & 1-1 & 3-3 \\
5 & 2 & 6 \\
-2 & -1 & -3 \\
\end{matrix} \right| \\
& =\left| \begin{matrix}
-1 & 0 & 0 \\
5 & 2 & 6 \\
-2 & -1 & -3 \\
\end{matrix} \right| \\
\end{align}$
Now, expanding along first row we get,
$\begin{align}
& -1\left( 2\left( -3 \right)-6\left( -1 \right) \right)-0+0 \\
& =-1\left( -6+6 \right) \\
& =-1\left( 0 \right) \\
& =0 \\
\end{align}$
In this problem we are getting the determinant as 0. Now, we have to check the relationship between the linear equations as $1{{P}_{1}}+7{{P}_{2}}=13{{P}_{3}}$.
Substituting the plane equations of this option in the above we get,
$\begin{align}
& 1\left( x+y+3z \right)+7\left( 5x+2y+6z \right)=13\left( -2x-y-3z \right) \\
& \Rightarrow x+y+3z+35x+14y+42z=-26x-13y-39z \\
& \Rightarrow 36x+15y+45z=-26x-13y-39z \\
\end{align}$
As you can see that L.H.S is not equal to R.H.S which should be equal in the above system of linear equations so this option is incorrect.
Checking option (c) $-x+2y-5z={{b}_{1}},2x-4y+10z={{b}_{2}},x-2y+5z={{b}_{3}}$ we get,
Determinant form of the coefficient matrix of the above equations is:
$\left| \begin{matrix}
-1 & 2 & -5 \\
2 & -4 & 10 \\
1 & -2 & 5 \\
\end{matrix} \right|$
Solving the above determinant by changing the first row as the addition of the first row and third row we get,
$\begin{align}
& \left| \begin{matrix}
-1+1 & -2 & -5+5 \\
2 & -4 & 10 \\
1 & -2 & 5 \\
\end{matrix} \right| \\
& =\left| \begin{matrix}
0 & 0 & 0 \\
2 & -4 & 10 \\
1 & -2 & 5 \\
\end{matrix} \right| \\
\end{align}$
We know that, if one row becomes all 0 then the value of the determinant is 0.
Now, we have to check the relationship between the three equations using $1{{P}_{1}}+7{{P}_{2}}=13{{P}_{3}}$ this relation we get,
Substituting the plane equations given in this option in the equation of the planes we get,
$\begin{align}
& 1\left( -x+2y-5z \right)+7\left( 2x-4y+10z \right)=13\left( x-2y+5z \right) \\
& \Rightarrow -x+2y-5z+14x-28y+70z=13x-26y+65z \\
& \Rightarrow 13x-26y+65z=13x-26y+65z \\
\end{align}$
As L.H.S is equal to R.H.S so this option is correct.
Checking option (d) $x+2y+5z={{b}_{1}},2x+3z={{b}_{2}},x+4y-5z={{b}_{3}}$ we get,
Determinant form of the coefficient matrix of the above equations is:
$\left| \begin{matrix}
1 & 2 & 5 \\
2 & 0 & 3 \\
1 & 4 & -5 \\
\end{matrix} \right|$
Changing row 1 to the subtraction of row 1 from row 3 we get,
$\left| \begin{matrix}
1-1 & 4-2 & 5+5 \\
2 & 0 & 3 \\
1 & 4 & -5 \\
\end{matrix} \right|$
$=\left| \begin{matrix}
0 & 2 & 10 \\
2 & 0 & 3 \\
1 & 4 & -5 \\
\end{matrix} \right|$
Expanding the above determinant along row 1 we get,
$\begin{align}
& 0-2\left( 2\left( -5 \right)-3\left( 1 \right) \right)+10\left( 2\left( 4 \right)-0\left( 1 \right) \right) \\
& =0-2\left( -10-3 \right)+10\left( 8 \right) \\
& =2\left( 13 \right)+80 \\
& =106 \\
\end{align}$
As the determinant of the coefficient matrix is non zero which determines the unique solution for all real values of ${{b}_{1}},{{b}_{2}},{{b}_{3}}$ so option (d) is also correct.
So, the correct answer is “Option (a),(c) and (d)”.
Note: The point where the possibility of committing mistake is high like while checking the options you found the determinant of the coefficient matrix as 0 and you jump to the conclusion that infinitely many solutions are possible. But “at least one solution” is for the system of equations given in the above problem but not for the options so in options although you are getting the determinant of the coefficient matrix as 0 but it need not necessarily mean that all the system of the equations of the options in which determinant value is 0 is following the same relationship between the equations as the given system of equations. So, be careful in checking the options.
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