Give the linear systems: $(i) x_1+2x_2=2,3x_1+7x_2=8\(ii) x_1+2x_2=1,3x_1+7x_2=7$ [closed]
$begingroup$
Give the linear systems:
(i) $x_1+2x_2=2\3x_1+7x_2=8$
(ii) $x_1+2x_2=1\3x_1+7x_2=7$
Solve both systems by incorporating the right-hand sides into a $2times2$ matrix $B$ and computing the reduced row echelon form of
$$[A|B]=begin{bmatrix}1&2&Big|&2&1\3&7&Big|&8&7end{bmatrix}$$
Both two sides of this form is $2times2$, so I have no idea how to reduce the row echelon form of this one.
linear-algebra matrix-equations
edited Jan 5 at 14:41
Shubham Johri
5,525818
asked Jan 5 at 13:01
Shadow ZShadow Z
396
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closed as off-topic by Henrik, KM101, darij grinberg, max_zorn, Ali Caglayan Jan 5 at 19:27
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Henrik, KM101, darij grinberg, max_zorn, Ali Caglayan
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Give the linear systems:
(i) $x_1+2x_2=2\3x_1+7x_2=8$
(ii) $x_1+2x_2=1\3x_1+7x_2=7$
Solve both systems by incorporating the right-hand sides into a $2times2$ matrix $B$ and computing the reduced row echelon form of
$$[A|B]=begin{bmatrix}1&2&Big|&2&1\3&7&Big|&8&7end{bmatrix}$$
Both two sides of this form is $2times2$, so I have no idea how to reduce the row echelon form of this one.
linear-algebra matrix-equations
edited Jan 5 at 14:41
Shubham Johri
5,525818
asked Jan 5 at 13:01
Shadow ZShadow Z
396
$endgroup$
closed as off-topic by Henrik, KM101, darij grinberg, max_zorn, Ali Caglayan Jan 5 at 19:27
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Henrik, KM101, darij grinberg, max_zorn, Ali Caglayan
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Welcome to the website. I have done it for you this time, but in future, please typeset your equations using Mathjax for better presentation and readability.
$endgroup$
– Shubham Johri
Jan 5 at 14:43
add a comment |
$begingroup$
Give the linear systems:
(i) $x_1+2x_2=2\3x_1+7x_2=8$
(ii) $x_1+2x_2=1\3x_1+7x_2=7$
Solve both systems by incorporating the right-hand sides into a $2times2$ matrix $B$ and computing the reduced row echelon form of
$$[A|B]=begin{bmatrix}1&2&Big|&2&1\3&7&Big|&8&7end{bmatrix}$$
Both two sides of this form is $2times2$, so I have no idea how to reduce the row echelon form of this one.
linear-algebra matrix-equations
edited Jan 5 at 14:41
Shubham Johri
5,525818
asked Jan 5 at 13:01
Shadow ZShadow Z
396
$endgroup$
Give the linear systems:
(i) $x_1+2x_2=2\3x_1+7x_2=8$
(ii) $x_1+2x_2=1\3x_1+7x_2=7$
Solve both systems by incorporating the right-hand sides into a $2times2$ matrix $B$ and computing the reduced row echelon form of
$$[A|B]=begin{bmatrix}1&2&Big|&2&1\3&7&Big|&8&7end{bmatrix}$$
Both two sides of this form is $2times2$, so I have no idea how to reduce the row echelon form of this one.
linear-algebra matrix-equations
linear-algebra matrix-equations
edited Jan 5 at 14:41
Shubham Johri
5,525818
asked Jan 5 at 13:01
Shadow ZShadow Z
396
edited Jan 5 at 14:41
Shubham Johri
5,525818
asked Jan 5 at 13:01
Shadow ZShadow Z
396
edited Jan 5 at 14:41
Shubham Johri
5,525818
edited Jan 5 at 14:41
Shubham Johri
5,525818
edited Jan 5 at 14:41
Shubham Johri
5,525818
5,525818
asked Jan 5 at 13:01
Shadow ZShadow Z
396
asked Jan 5 at 13:01
Shadow ZShadow Z
396
asked Jan 5 at 13:01
Shadow ZShadow Z
396
396
closed as off-topic by Henrik, KM101, darij grinberg, max_zorn, Ali Caglayan Jan 5 at 19:27
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Henrik, KM101, darij grinberg, max_zorn, Ali Caglayan
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Henrik, KM101, darij grinberg, max_zorn, Ali Caglayan Jan 5 at 19:27
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Henrik, KM101, darij grinberg, max_zorn, Ali Caglayan
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Welcome to the website. I have done it for you this time, but in future, please typeset your equations using Mathjax for better presentation and readability.
$endgroup$
– Shubham Johri
Jan 5 at 14:43
add a comment |
$begingroup$
Welcome to the website. I have done it for you this time, but in future, please typeset your equations using Mathjax for better presentation and readability.
$endgroup$
– Shubham Johri
Jan 5 at 14:43
$begingroup$
Welcome to the website. I have done it for you this time, but in future, please typeset your equations using Mathjax for better presentation and readability.
$endgroup$
– Shubham Johri
Jan 5 at 14:43
$begingroup$
Welcome to the website. I have done it for you this time, but in future, please typeset your equations using Mathjax for better presentation and readability.
$endgroup$
– Shubham Johri
Jan 5 at 14:43
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Since the coefficient matrix for both systems is $A=begin{bmatrix}1&2\3&7end{bmatrix}$, we can solve them simultaneously. This is because the row transformations required to be done on $A$ to convert it into its row echelon form will be the same for both systems.
$begin{bmatrix}1&2&Big|&2&1\3&7&Big|&8&7end{bmatrix}xrightarrow{R_2to R_2-3R_1}begin{bmatrix}1&2&Big|&2&1\0&1&Big|&2&4end{bmatrix}xrightarrow{R_1to R_1-2R_2}begin{bmatrix}1&0&Big|&-2&-7\0&1&Big|&2&4end{bmatrix}$
Now the solution for the first system is given $$begin{bmatrix}1&0&Big|&-2\0&1&Big|&2end{bmatrix}$$ and the solution of the second system is given by $$begin{bmatrix}1&0&Big|&-7\0&1&Big|&4end{bmatrix}$$
answered Jan 5 at 14:58
Shubham JohriShubham Johri
5,525818
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since the coefficient matrix for both systems is $A=begin{bmatrix}1&2\3&7end{bmatrix}$, we can solve them simultaneously. This is because the row transformations required to be done on $A$ to convert it into its row echelon form will be the same for both systems.
$begin{bmatrix}1&2&Big|&2&1\3&7&Big|&8&7end{bmatrix}xrightarrow{R_2to R_2-3R_1}begin{bmatrix}1&2&Big|&2&1\0&1&Big|&2&4end{bmatrix}xrightarrow{R_1to R_1-2R_2}begin{bmatrix}1&0&Big|&-2&-7\0&1&Big|&2&4end{bmatrix}$
Now the solution for the first system is given $$begin{bmatrix}1&0&Big|&-2\0&1&Big|&2end{bmatrix}$$ and the solution of the second system is given by $$begin{bmatrix}1&0&Big|&-7\0&1&Big|&4end{bmatrix}$$
answered Jan 5 at 14:58
Shubham JohriShubham Johri
5,525818
$endgroup$
add a comment |
$begingroup$
Since the coefficient matrix for both systems is $A=begin{bmatrix}1&2\3&7end{bmatrix}$, we can solve them simultaneously. This is because the row transformations required to be done on $A$ to convert it into its row echelon form will be the same for both systems.
$begin{bmatrix}1&2&Big|&2&1\3&7&Big|&8&7end{bmatrix}xrightarrow{R_2to R_2-3R_1}begin{bmatrix}1&2&Big|&2&1\0&1&Big|&2&4end{bmatrix}xrightarrow{R_1to R_1-2R_2}begin{bmatrix}1&0&Big|&-2&-7\0&1&Big|&2&4end{bmatrix}$
Now the solution for the first system is given $$begin{bmatrix}1&0&Big|&-2\0&1&Big|&2end{bmatrix}$$ and the solution of the second system is given by $$begin{bmatrix}1&0&Big|&-7\0&1&Big|&4end{bmatrix}$$
answered Jan 5 at 14:58
Shubham JohriShubham Johri
5,525818
$endgroup$
add a comment |
$begingroup$
Since the coefficient matrix for both systems is $A=begin{bmatrix}1&2\3&7end{bmatrix}$, we can solve them simultaneously. This is because the row transformations required to be done on $A$ to convert it into its row echelon form will be the same for both systems.
$begin{bmatrix}1&2&Big|&2&1\3&7&Big|&8&7end{bmatrix}xrightarrow{R_2to R_2-3R_1}begin{bmatrix}1&2&Big|&2&1\0&1&Big|&2&4end{bmatrix}xrightarrow{R_1to R_1-2R_2}begin{bmatrix}1&0&Big|&-2&-7\0&1&Big|&2&4end{bmatrix}$
Now the solution for the first system is given $$begin{bmatrix}1&0&Big|&-2\0&1&Big|&2end{bmatrix}$$ and the solution of the second system is given by $$begin{bmatrix}1&0&Big|&-7\0&1&Big|&4end{bmatrix}$$
answered Jan 5 at 14:58
Shubham JohriShubham Johri
5,525818
$endgroup$
Since the coefficient matrix for both systems is $A=begin{bmatrix}1&2\3&7end{bmatrix}$, we can solve them simultaneously. This is because the row transformations required to be done on $A$ to convert it into its row echelon form will be the same for both systems.
$begin{bmatrix}1&2&Big|&2&1\3&7&Big|&8&7end{bmatrix}xrightarrow{R_2to R_2-3R_1}begin{bmatrix}1&2&Big|&2&1\0&1&Big|&2&4end{bmatrix}xrightarrow{R_1to R_1-2R_2}begin{bmatrix}1&0&Big|&-2&-7\0&1&Big|&2&4end{bmatrix}$
Now the solution for the first system is given $$begin{bmatrix}1&0&Big|&-2\0&1&Big|&2end{bmatrix}$$ and the solution of the second system is given by $$begin{bmatrix}1&0&Big|&-7\0&1&Big|&4end{bmatrix}$$
answered Jan 5 at 14:58
Shubham JohriShubham Johri
5,525818
answered Jan 5 at 14:58
Shubham JohriShubham Johri
5,525818
answered Jan 5 at 14:58
Shubham JohriShubham Johri
5,525818
answered Jan 5 at 14:58
Shubham JohriShubham Johri
5,525818
5,525818
add a comment |
add a comment |
$begingroup$
Welcome to the website. I have done it for you this time, but in future, please typeset your equations using Mathjax for better presentation and readability.
$endgroup$
– Shubham Johri
Jan 5 at 14:43