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Invert a 3x3 matrix using row operations
Draft
A a 3×3 matrix. Using row operations on the augmented matrix (A|I3) reduce to (I3|A−1).
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England schools
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England university
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Scotland schools
Taxonomy: mathcentre
Taxonomy: Kind of activity
Taxonomy: Context
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History
Newcastle University Mathematics and Statistics 9 years, 2 months ago
Created this.| Name | Status | Author | Last Modified | |
|---|---|---|---|---|
| Invert a 3x3 matrix using row operations | draft | Newcastle University Mathematics and Statistics | 20/11/2019 14:50 | |
| Hollie's copy of Invert a 3x3 matrix using row operations | draft | Hollie Tarr | 01/12/2016 12:33 | |
| Hollie's copy of Invert a 3x3 matrix using row operations | draft | Hollie Tarr | 22/02/2017 14:22 | |
| Harry's copy of Invert a 3x3 matrix using row operations | draft | Harry Flynn | 09/03/2018 14:46 | |
| Maria's copy of Invert a 3x3 matrix using row operations | draft | Maria Aneiros | 23/05/2019 07:00 | |
| Simon's copy of Invert a 3x3 matrix using row operations | draft | Simon Thomas | 12/06/2019 13:51 |
There are 18 other versions that do you not have access to.
| Name | Type | Generated Value |
|---|
| a | integer |
4
|
||||
| a11 | integer |
1
|
||||
| a12 | integer |
-2
|
||||
| a13 | integer |
-6
|
||||
| a21 | integer |
-4
|
||||
| a22 | integer |
7
|
||||
| a23 | number |
20
|
||||
| a31 | integer |
-4
|
||||
| a32 | integer |
2
|
||||
| a33 | integer |
1
|
||||
| b | integer |
2
|
||||
| b24 | integer |
76
|
||||
| b25 | integer |
23
|
||||
| c | integer |
1
|
||||
| c1 | integer |
5
|
||||
| c2 | integer |
1
|
||||
| c3 | integer |
1
|
||||
| f1 | integer |
-1
|
||||
| f2 | integer |
1
|
||||
| f3 | integer |
1
|
||||
| g1 | integer |
-1
|
||||
| g2 | integer |
1
|
||||
| g3 | integer |
1
|
||||
| s | integer |
1
|
Generated value: integer
4
This variable doesn't seem to be used anywhere.
Parts
Gap-fill
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Part 1.
Introduce zeros in the first column below the first entry by adding suitable multiples of the first row to rows 2 and 3.
Input all numbers as fractions or integers and not as decimals.
| (.... |
{a11} | {a12} | {a13} | 1 | 0 | 0 | ).... |
| 0 | 1 | 0 | |||||
| 0 | 0 | 1 |
Now, if necessary, multiply the second row by a suitable number so that the second entry in the second row is 1.
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