Defined row operations for making a matrix ,containing varibel coefficients, into RREF
I am trying to solve for the inverse for a given $3x3$ matrix. The matrix looks like this:
begin{pmatrix}1&0&1\ :a&0&1\ :1&a&0end{pmatrix}
I have tried solving it through carrying out row operations on the matrix above whilest in parallel carrying out the same operations on a $3x3$ Identity matrix.
I am worried that I might be preforming "undefined" or "non-approved" row operations on $A$ and that it, is the cause of the incorrect result. What i got so far looks like this:
begin{pmatrix}1&0&1\ :::::a&0&1\ :::::1&a&0end{pmatrix} $R_2to R_2 -aR_1$
$R_3to R_3-R_1$
begin{pmatrix}1&0&1\ ::::::0&0&1-a\ ::::::0&a&-1end{pmatrix}
Assuming $aneq 1$
Switch $R_2leftrightarrow R_3$
begin{pmatrix}1&0&1\ :::::0&a&-1\ :::::0&0&1-aend{pmatrix}
$R_3to frac{1}{1-a}R_3$begin{pmatrix}1&0&1\ ::::::0&a&-1\ ::::::0&0&1end{pmatrix}
$R_2to R_2+R_3$
$R_1to R_1-R_3$
begin{pmatrix}1&0&0\ :::::::0&a&0\ :::::::0&0&1end{pmatrix}
$R_2to R_2cdot(1/a)$
begin{pmatrix}1&0&0\ :::::::0&1&0\ :::::::0&0&1end{pmatrix}
This it what i am getting, if i do the same elementry row operations in the same sequence to the Identity matrix i get an incorrect result. What am i doing wrong?
(Sorry about the messy notation, first time poster)
Kind regards and many thanks!
linear-algebra matrices
edited Nov 30 at 17:11
Thomas Shelby
1,202116
asked Nov 30 at 16:43
Student123451
1
add a comment |
I am trying to solve for the inverse for a given $3x3$ matrix. The matrix looks like this:
begin{pmatrix}1&0&1\ :a&0&1\ :1&a&0end{pmatrix}
I have tried solving it through carrying out row operations on the matrix above whilest in parallel carrying out the same operations on a $3x3$ Identity matrix.
I am worried that I might be preforming "undefined" or "non-approved" row operations on $A$ and that it, is the cause of the incorrect result. What i got so far looks like this:
begin{pmatrix}1&0&1\ :::::a&0&1\ :::::1&a&0end{pmatrix} $R_2to R_2 -aR_1$
$R_3to R_3-R_1$
begin{pmatrix}1&0&1\ ::::::0&0&1-a\ ::::::0&a&-1end{pmatrix}
Assuming $aneq 1$
Switch $R_2leftrightarrow R_3$
begin{pmatrix}1&0&1\ :::::0&a&-1\ :::::0&0&1-aend{pmatrix}
$R_3to frac{1}{1-a}R_3$begin{pmatrix}1&0&1\ ::::::0&a&-1\ ::::::0&0&1end{pmatrix}
$R_2to R_2+R_3$
$R_1to R_1-R_3$
begin{pmatrix}1&0&0\ :::::::0&a&0\ :::::::0&0&1end{pmatrix}
$R_2to R_2cdot(1/a)$
begin{pmatrix}1&0&0\ :::::::0&1&0\ :::::::0&0&1end{pmatrix}
This it what i am getting, if i do the same elementry row operations in the same sequence to the Identity matrix i get an incorrect result. What am i doing wrong?
(Sorry about the messy notation, first time poster)
Kind regards and many thanks!
linear-algebra matrices
edited Nov 30 at 17:11
Thomas Shelby
1,202116
asked Nov 30 at 16:43
Student123451
1
add a comment |
I am trying to solve for the inverse for a given $3x3$ matrix. The matrix looks like this:
begin{pmatrix}1&0&1\ :a&0&1\ :1&a&0end{pmatrix}
I have tried solving it through carrying out row operations on the matrix above whilest in parallel carrying out the same operations on a $3x3$ Identity matrix.
I am worried that I might be preforming "undefined" or "non-approved" row operations on $A$ and that it, is the cause of the incorrect result. What i got so far looks like this:
begin{pmatrix}1&0&1\ :::::a&0&1\ :::::1&a&0end{pmatrix} $R_2to R_2 -aR_1$
$R_3to R_3-R_1$
begin{pmatrix}1&0&1\ ::::::0&0&1-a\ ::::::0&a&-1end{pmatrix}
Assuming $aneq 1$
Switch $R_2leftrightarrow R_3$
begin{pmatrix}1&0&1\ :::::0&a&-1\ :::::0&0&1-aend{pmatrix}
$R_3to frac{1}{1-a}R_3$begin{pmatrix}1&0&1\ ::::::0&a&-1\ ::::::0&0&1end{pmatrix}
$R_2to R_2+R_3$
$R_1to R_1-R_3$
begin{pmatrix}1&0&0\ :::::::0&a&0\ :::::::0&0&1end{pmatrix}
$R_2to R_2cdot(1/a)$
begin{pmatrix}1&0&0\ :::::::0&1&0\ :::::::0&0&1end{pmatrix}
This it what i am getting, if i do the same elementry row operations in the same sequence to the Identity matrix i get an incorrect result. What am i doing wrong?
(Sorry about the messy notation, first time poster)
Kind regards and many thanks!
linear-algebra matrices
edited Nov 30 at 17:11
Thomas Shelby
1,202116
asked Nov 30 at 16:43
Student123451
1
I am trying to solve for the inverse for a given $3x3$ matrix. The matrix looks like this:
begin{pmatrix}1&0&1\ :a&0&1\ :1&a&0end{pmatrix}
I have tried solving it through carrying out row operations on the matrix above whilest in parallel carrying out the same operations on a $3x3$ Identity matrix.
I am worried that I might be preforming "undefined" or "non-approved" row operations on $A$ and that it, is the cause of the incorrect result. What i got so far looks like this:
begin{pmatrix}1&0&1\ :::::a&0&1\ :::::1&a&0end{pmatrix} $R_2to R_2 -aR_1$
$R_3to R_3-R_1$
begin{pmatrix}1&0&1\ ::::::0&0&1-a\ ::::::0&a&-1end{pmatrix}
Assuming $aneq 1$
Switch $R_2leftrightarrow R_3$
begin{pmatrix}1&0&1\ :::::0&a&-1\ :::::0&0&1-aend{pmatrix}
$R_3to frac{1}{1-a}R_3$begin{pmatrix}1&0&1\ ::::::0&a&-1\ ::::::0&0&1end{pmatrix}
$R_2to R_2+R_3$
$R_1to R_1-R_3$
begin{pmatrix}1&0&0\ :::::::0&a&0\ :::::::0&0&1end{pmatrix}
$R_2to R_2cdot(1/a)$
begin{pmatrix}1&0&0\ :::::::0&1&0\ :::::::0&0&1end{pmatrix}
This it what i am getting, if i do the same elementry row operations in the same sequence to the Identity matrix i get an incorrect result. What am i doing wrong?
(Sorry about the messy notation, first time poster)
Kind regards and many thanks!
linear-algebra matrices
linear-algebra matrices
edited Nov 30 at 17:11
Thomas Shelby
1,202116
asked Nov 30 at 16:43
Student123451
1
edited Nov 30 at 17:11
Thomas Shelby
1,202116
asked Nov 30 at 16:43
Student123451
1
edited Nov 30 at 17:11
Thomas Shelby
1,202116
edited Nov 30 at 17:11
Thomas Shelby
1,202116
edited Nov 30 at 17:11
Thomas Shelby
1,202116
1,202116
asked Nov 30 at 16:43
Student123451
1
asked Nov 30 at 16:43
Student123451
1
asked Nov 30 at 16:43
Student123451
1
1
add a comment |
add a comment |
1 Answer
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$a≠0,1$ or the matrix will be singular. The row transformations look okay. The inverse comes out to be $begin{pmatrix}frac1{1-a}&frac1{a-1}&0\ :frac1{a(a-1)}&frac1{a(1-a)}&frac1a\ :frac a{a-1}&frac1{1-a}&0end{pmatrix}$.
answered Nov 30 at 17:17
Shubham Johri
3,766716
add a comment |
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$a≠0,1$ or the matrix will be singular. The row transformations look okay. The inverse comes out to be $begin{pmatrix}frac1{1-a}&frac1{a-1}&0\ :frac1{a(a-1)}&frac1{a(1-a)}&frac1a\ :frac a{a-1}&frac1{1-a}&0end{pmatrix}$.
answered Nov 30 at 17:17
Shubham Johri
3,766716
add a comment |
$a≠0,1$ or the matrix will be singular. The row transformations look okay. The inverse comes out to be $begin{pmatrix}frac1{1-a}&frac1{a-1}&0\ :frac1{a(a-1)}&frac1{a(1-a)}&frac1a\ :frac a{a-1}&frac1{1-a}&0end{pmatrix}$.
answered Nov 30 at 17:17
Shubham Johri
3,766716
add a comment |
$a≠0,1$ or the matrix will be singular. The row transformations look okay. The inverse comes out to be $begin{pmatrix}frac1{1-a}&frac1{a-1}&0\ :frac1{a(a-1)}&frac1{a(1-a)}&frac1a\ :frac a{a-1}&frac1{1-a}&0end{pmatrix}$.
answered Nov 30 at 17:17
Shubham Johri
3,766716
$a≠0,1$ or the matrix will be singular. The row transformations look okay. The inverse comes out to be $begin{pmatrix}frac1{1-a}&frac1{a-1}&0\ :frac1{a(a-1)}&frac1{a(1-a)}&frac1a\ :frac a{a-1}&frac1{1-a}&0end{pmatrix}$.
answered Nov 30 at 17:17
Shubham Johri
3,766716
answered Nov 30 at 17:17
Shubham Johri
3,766716
answered Nov 30 at 17:17
Shubham Johri
3,766716
answered Nov 30 at 17:17
Shubham Johri
3,766716
3,766716
add a comment |
add a comment |
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