Mixing
What is the value of the variables in the equation?
$begingroup$
$$x_1 + y_1 = 3$$
$$y_1 - x_3 = -1$$
$$x_3 + y_3 = 7$$
$$x_1 - y_3 = -3$$
Find the values of $x_1$, $x_3$, $y_1$ and $y_3$.
All I am getting is $x_1 + x_3 = 4$ and $y_1 + y_3 = 6$.
linear-algebra systems-of-equations
edited Feb 1 at 16:16
Michael Rozenberg
110k1896201
asked Feb 1 at 15:22
Sagar SharmaSagar Sharma
143
$endgroup$
add a comment |
$begingroup$
$$x_1 + y_1 = 3$$
$$y_1 - x_3 = -1$$
$$x_3 + y_3 = 7$$
$$x_1 - y_3 = -3$$
Find the values of $x_1$, $x_3$, $y_1$ and $y_3$.
All I am getting is $x_1 + x_3 = 4$ and $y_1 + y_3 = 6$.
linear-algebra systems-of-equations
edited Feb 1 at 16:16
Michael Rozenberg
110k1896201
asked Feb 1 at 15:22
Sagar SharmaSagar Sharma
143
$endgroup$
1
$begingroup$
This system has infinitely many solutions of the form, $x_1=a$, $x_3=4-a$, $y_1 = 3-a$ and $y_2 = a+3$ for any $ain R$.
$endgroup$
– Faiq Irfan
Feb 1 at 15:32
add a comment |
$begingroup$
$$x_1 + y_1 = 3$$
$$y_1 - x_3 = -1$$
$$x_3 + y_3 = 7$$
$$x_1 - y_3 = -3$$
Find the values of $x_1$, $x_3$, $y_1$ and $y_3$.
All I am getting is $x_1 + x_3 = 4$ and $y_1 + y_3 = 6$.
linear-algebra systems-of-equations
edited Feb 1 at 16:16
Michael Rozenberg
110k1896201
asked Feb 1 at 15:22
Sagar SharmaSagar Sharma
143
$endgroup$
$$x_1 + y_1 = 3$$
$$y_1 - x_3 = -1$$
$$x_3 + y_3 = 7$$
$$x_1 - y_3 = -3$$
Find the values of $x_1$, $x_3$, $y_1$ and $y_3$.
All I am getting is $x_1 + x_3 = 4$ and $y_1 + y_3 = 6$.
linear-algebra systems-of-equations
linear-algebra systems-of-equations
edited Feb 1 at 16:16
Michael Rozenberg
110k1896201
asked Feb 1 at 15:22
Sagar SharmaSagar Sharma
143
edited Feb 1 at 16:16
Michael Rozenberg
110k1896201
asked Feb 1 at 15:22
Sagar SharmaSagar Sharma
143
edited Feb 1 at 16:16
Michael Rozenberg
110k1896201
edited Feb 1 at 16:16
Michael Rozenberg
110k1896201
edited Feb 1 at 16:16
Michael Rozenberg
110k1896201
110k1896201
asked Feb 1 at 15:22
Sagar SharmaSagar Sharma
143
asked Feb 1 at 15:22
Sagar SharmaSagar Sharma
143
asked Feb 1 at 15:22
Sagar SharmaSagar Sharma
143
143
1
$begingroup$
This system has infinitely many solutions of the form, $x_1=a$, $x_3=4-a$, $y_1 = 3-a$ and $y_2 = a+3$ for any $ain R$.
$endgroup$
– Faiq Irfan
Feb 1 at 15:32
add a comment |
1
$begingroup$
This system has infinitely many solutions of the form, $x_1=a$, $x_3=4-a$, $y_1 = 3-a$ and $y_2 = a+3$ for any $ain R$.
$endgroup$
– Faiq Irfan
Feb 1 at 15:32
1
1
$begingroup$
This system has infinitely many solutions of the form, $x_1=a$, $x_3=4-a$, $y_1 = 3-a$ and $y_2 = a+3$ for any $ain R$.
$endgroup$
– Faiq Irfan
Feb 1 at 15:32
$begingroup$
This system has infinitely many solutions of the form, $x_1=a$, $x_3=4-a$, $y_1 = 3-a$ and $y_2 = a+3$ for any $ain R$.
$endgroup$
– Faiq Irfan
Feb 1 at 15:32
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $x_1=a$, $y_1=b$, $x_3=c$ and $y_3=d$.
Thus, $b=3-a$ and from the second we obtain $c=4-a$, which from the third gives $d=3+a$ and we see that the fourth holds.
Id est, got infinitely many solutions:
$${(a,3-a,4-a,3+a)}$$
answered Feb 1 at 15:31
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
add a comment |
$begingroup$
With this sort of system one should note that it can be represented as a matrix, where the elements are the coefficients for each variable:
$$M = left[begin{array}{cccccc|c} 1 & 0 & 0 & 1 & 0 & 0 & 3\
0 & 0 & -1 & 1 & 0 & 0 & -1\
0 & 0 & 1 & 0 & 0 & 1 & 7\
1 & 0 & 0 & 0 & 0 & -1 & -3end{array} right]$$
Of course, to find a solution (if one exists), one must change the matrix $M$ into reduced row-echelon form, denoted here as $M_{rrf}$, which is as follows:
$$M_{rrf} = left[begin{array}{cccccc|c} 1 & 0 & 0 & 0 & 0 & 0 & 3 \
0 & 0 & 1 & 0 & 0 & 0 & 1 \
0 & 0 & 0 & 1 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 & 1 & 6end{array} right] \ text{this was carried out by the following $it row$ operations:} \
-R_1 + R_4 to R_4 \ R_2 + R_3 to R_3 \R_4 + R_3 to R_3 \ -R_4 to R_4 \ -R_3 + R_2 to R_2 \ -R_3 + R_1 to R_1 \ -R_2 to R_2 \$$
Thus, since the 3rd row of $M_{rrf}$ is $left[begin{array}{cccccc|c} 0 & 0 & 0 & 1 & 0 & 0 & 0end{array}right]$, we have that $text{dim kerM} = 1$ (recall that the dimension of the kernel is defined as the nullity, $text{null M}$ of the matrix). Since, $null(M) = 1$, the system has infinitely many solutions.
Now, as shown by Michael's answer, we can parametrize our solution to create an expression that generates all of the solutions. Begin by defining the following: $$x_1 = 3 = a \ x_3 = 1 = b \ y_1 = 0 = c \ y_3 = 6 = d$$ which gives us $$ a = a
\ a + c = 3 \ -b + c = -1 \ a - d = -3 \ text{which simplifies to:} \ a = a \ b = c + 1 = 4 - a \ c = 3 - a \ d = a + 3 $$
Therefore, the solution set for the system is $ {a, 4-a, 3-a, 3+a }$
answered Feb 1 at 16:24
Victoria MVictoria M
42618
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
Let $x_1=a$, $y_1=b$, $x_3=c$ and $y_3=d$.
Thus, $b=3-a$ and from the second we obtain $c=4-a$, which from the third gives $d=3+a$ and we see that the fourth holds.
Id est, got infinitely many solutions:
$${(a,3-a,4-a,3+a)}$$
answered Feb 1 at 15:31
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
add a comment |
$begingroup$
Let $x_1=a$, $y_1=b$, $x_3=c$ and $y_3=d$.
Thus, $b=3-a$ and from the second we obtain $c=4-a$, which from the third gives $d=3+a$ and we see that the fourth holds.
Id est, got infinitely many solutions:
$${(a,3-a,4-a,3+a)}$$
answered Feb 1 at 15:31
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
add a comment |
$begingroup$
Let $x_1=a$, $y_1=b$, $x_3=c$ and $y_3=d$.
Thus, $b=3-a$ and from the second we obtain $c=4-a$, which from the third gives $d=3+a$ and we see that the fourth holds.
Id est, got infinitely many solutions:
$${(a,3-a,4-a,3+a)}$$
answered Feb 1 at 15:31
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
Let $x_1=a$, $y_1=b$, $x_3=c$ and $y_3=d$.
Thus, $b=3-a$ and from the second we obtain $c=4-a$, which from the third gives $d=3+a$ and we see that the fourth holds.
Id est, got infinitely many solutions:
$${(a,3-a,4-a,3+a)}$$
answered Feb 1 at 15:31
Michael RozenbergMichael Rozenberg
110k1896201
answered Feb 1 at 15:31
Michael RozenbergMichael Rozenberg
110k1896201
answered Feb 1 at 15:31
Michael RozenbergMichael Rozenberg
110k1896201
answered Feb 1 at 15:31
Michael RozenbergMichael Rozenberg
110k1896201
110k1896201
add a comment |
add a comment |
$begingroup$
With this sort of system one should note that it can be represented as a matrix, where the elements are the coefficients for each variable:
$$M = left[begin{array}{cccccc|c} 1 & 0 & 0 & 1 & 0 & 0 & 3\
0 & 0 & -1 & 1 & 0 & 0 & -1\
0 & 0 & 1 & 0 & 0 & 1 & 7\
1 & 0 & 0 & 0 & 0 & -1 & -3end{array} right]$$
Of course, to find a solution (if one exists), one must change the matrix $M$ into reduced row-echelon form, denoted here as $M_{rrf}$, which is as follows:
$$M_{rrf} = left[begin{array}{cccccc|c} 1 & 0 & 0 & 0 & 0 & 0 & 3 \
0 & 0 & 1 & 0 & 0 & 0 & 1 \
0 & 0 & 0 & 1 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 & 1 & 6end{array} right] \ text{this was carried out by the following $it row$ operations:} \
-R_1 + R_4 to R_4 \ R_2 + R_3 to R_3 \R_4 + R_3 to R_3 \ -R_4 to R_4 \ -R_3 + R_2 to R_2 \ -R_3 + R_1 to R_1 \ -R_2 to R_2 \$$
Thus, since the 3rd row of $M_{rrf}$ is $left[begin{array}{cccccc|c} 0 & 0 & 0 & 1 & 0 & 0 & 0end{array}right]$, we have that $text{dim kerM} = 1$ (recall that the dimension of the kernel is defined as the nullity, $text{null M}$ of the matrix). Since, $null(M) = 1$, the system has infinitely many solutions.
Now, as shown by Michael's answer, we can parametrize our solution to create an expression that generates all of the solutions. Begin by defining the following: $$x_1 = 3 = a \ x_3 = 1 = b \ y_1 = 0 = c \ y_3 = 6 = d$$ which gives us $$ a = a
\ a + c = 3 \ -b + c = -1 \ a - d = -3 \ text{which simplifies to:} \ a = a \ b = c + 1 = 4 - a \ c = 3 - a \ d = a + 3 $$
Therefore, the solution set for the system is $ {a, 4-a, 3-a, 3+a }$
answered Feb 1 at 16:24
Victoria MVictoria M
42618
$endgroup$
add a comment |
$begingroup$
With this sort of system one should note that it can be represented as a matrix, where the elements are the coefficients for each variable:
$$M = left[begin{array}{cccccc|c} 1 & 0 & 0 & 1 & 0 & 0 & 3\
0 & 0 & -1 & 1 & 0 & 0 & -1\
0 & 0 & 1 & 0 & 0 & 1 & 7\
1 & 0 & 0 & 0 & 0 & -1 & -3end{array} right]$$
Of course, to find a solution (if one exists), one must change the matrix $M$ into reduced row-echelon form, denoted here as $M_{rrf}$, which is as follows:
$$M_{rrf} = left[begin{array}{cccccc|c} 1 & 0 & 0 & 0 & 0 & 0 & 3 \
0 & 0 & 1 & 0 & 0 & 0 & 1 \
0 & 0 & 0 & 1 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 & 1 & 6end{array} right] \ text{this was carried out by the following $it row$ operations:} \
-R_1 + R_4 to R_4 \ R_2 + R_3 to R_3 \R_4 + R_3 to R_3 \ -R_4 to R_4 \ -R_3 + R_2 to R_2 \ -R_3 + R_1 to R_1 \ -R_2 to R_2 \$$
Thus, since the 3rd row of $M_{rrf}$ is $left[begin{array}{cccccc|c} 0 & 0 & 0 & 1 & 0 & 0 & 0end{array}right]$, we have that $text{dim kerM} = 1$ (recall that the dimension of the kernel is defined as the nullity, $text{null M}$ of the matrix). Since, $null(M) = 1$, the system has infinitely many solutions.
Now, as shown by Michael's answer, we can parametrize our solution to create an expression that generates all of the solutions. Begin by defining the following: $$x_1 = 3 = a \ x_3 = 1 = b \ y_1 = 0 = c \ y_3 = 6 = d$$ which gives us $$ a = a
\ a + c = 3 \ -b + c = -1 \ a - d = -3 \ text{which simplifies to:} \ a = a \ b = c + 1 = 4 - a \ c = 3 - a \ d = a + 3 $$
Therefore, the solution set for the system is $ {a, 4-a, 3-a, 3+a }$
answered Feb 1 at 16:24
Victoria MVictoria M
42618
$endgroup$
add a comment |
$begingroup$
With this sort of system one should note that it can be represented as a matrix, where the elements are the coefficients for each variable:
$$M = left[begin{array}{cccccc|c} 1 & 0 & 0 & 1 & 0 & 0 & 3\
0 & 0 & -1 & 1 & 0 & 0 & -1\
0 & 0 & 1 & 0 & 0 & 1 & 7\
1 & 0 & 0 & 0 & 0 & -1 & -3end{array} right]$$
Of course, to find a solution (if one exists), one must change the matrix $M$ into reduced row-echelon form, denoted here as $M_{rrf}$, which is as follows:
$$M_{rrf} = left[begin{array}{cccccc|c} 1 & 0 & 0 & 0 & 0 & 0 & 3 \
0 & 0 & 1 & 0 & 0 & 0 & 1 \
0 & 0 & 0 & 1 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 & 1 & 6end{array} right] \ text{this was carried out by the following $it row$ operations:} \
-R_1 + R_4 to R_4 \ R_2 + R_3 to R_3 \R_4 + R_3 to R_3 \ -R_4 to R_4 \ -R_3 + R_2 to R_2 \ -R_3 + R_1 to R_1 \ -R_2 to R_2 \$$
Thus, since the 3rd row of $M_{rrf}$ is $left[begin{array}{cccccc|c} 0 & 0 & 0 & 1 & 0 & 0 & 0end{array}right]$, we have that $text{dim kerM} = 1$ (recall that the dimension of the kernel is defined as the nullity, $text{null M}$ of the matrix). Since, $null(M) = 1$, the system has infinitely many solutions.
Now, as shown by Michael's answer, we can parametrize our solution to create an expression that generates all of the solutions. Begin by defining the following: $$x_1 = 3 = a \ x_3 = 1 = b \ y_1 = 0 = c \ y_3 = 6 = d$$ which gives us $$ a = a
\ a + c = 3 \ -b + c = -1 \ a - d = -3 \ text{which simplifies to:} \ a = a \ b = c + 1 = 4 - a \ c = 3 - a \ d = a + 3 $$
Therefore, the solution set for the system is $ {a, 4-a, 3-a, 3+a }$
answered Feb 1 at 16:24
Victoria MVictoria M
42618
$endgroup$
With this sort of system one should note that it can be represented as a matrix, where the elements are the coefficients for each variable:
$$M = left[begin{array}{cccccc|c} 1 & 0 & 0 & 1 & 0 & 0 & 3\
0 & 0 & -1 & 1 & 0 & 0 & -1\
0 & 0 & 1 & 0 & 0 & 1 & 7\
1 & 0 & 0 & 0 & 0 & -1 & -3end{array} right]$$
Of course, to find a solution (if one exists), one must change the matrix $M$ into reduced row-echelon form, denoted here as $M_{rrf}$, which is as follows:
$$M_{rrf} = left[begin{array}{cccccc|c} 1 & 0 & 0 & 0 & 0 & 0 & 3 \
0 & 0 & 1 & 0 & 0 & 0 & 1 \
0 & 0 & 0 & 1 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 & 1 & 6end{array} right] \ text{this was carried out by the following $it row$ operations:} \
-R_1 + R_4 to R_4 \ R_2 + R_3 to R_3 \R_4 + R_3 to R_3 \ -R_4 to R_4 \ -R_3 + R_2 to R_2 \ -R_3 + R_1 to R_1 \ -R_2 to R_2 \$$
Thus, since the 3rd row of $M_{rrf}$ is $left[begin{array}{cccccc|c} 0 & 0 & 0 & 1 & 0 & 0 & 0end{array}right]$, we have that $text{dim kerM} = 1$ (recall that the dimension of the kernel is defined as the nullity, $text{null M}$ of the matrix). Since, $null(M) = 1$, the system has infinitely many solutions.
Now, as shown by Michael's answer, we can parametrize our solution to create an expression that generates all of the solutions. Begin by defining the following: $$x_1 = 3 = a \ x_3 = 1 = b \ y_1 = 0 = c \ y_3 = 6 = d$$ which gives us $$ a = a
\ a + c = 3 \ -b + c = -1 \ a - d = -3 \ text{which simplifies to:} \ a = a \ b = c + 1 = 4 - a \ c = 3 - a \ d = a + 3 $$
Therefore, the solution set for the system is $ {a, 4-a, 3-a, 3+a }$
answered Feb 1 at 16:24
Victoria MVictoria M
42618
answered Feb 1 at 16:24
Victoria MVictoria M
42618
answered Feb 1 at 16:24
Victoria MVictoria M
42618
answered Feb 1 at 16:24
Victoria MVictoria M
42618
42618
add a comment |
add a comment |
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$begingroup$
This system has infinitely many solutions of the form, $x_1=a$, $x_3=4-a$, $y_1 = 3-a$ and $y_2 = a+3$ for any $ain R$.
$endgroup$
– Faiq Irfan
Feb 1 at 15:32