Example of Solving a System of Linear Equations by Cramer's Rule.
This solution was made using the calculator presented on the site.
It is necessary to solve the system of linear equations using Cramer's rule.
 | - | 2 | x1 | + | x2 | + | x3 | = | -13 | ||
| - | x1 | + | x2 | + | 2 | x3 | = | -9 | |||
| 3 | x1 | + | x2 | + | x3 | = | 12 |
Let's write the Cramer's rule:
x1 = det A1 / det A
x2 = det A2 / det A
x3 = det A3 / det A
It is impossible to divide by zero. Therefore, if the determinant of A is zero, then it is impossible to use Cramer's rule.
Let's calculate the determinant A. more info
The determinant A consists of the coefficients of the left side of the system.
| | - | 2 | x1 | + | x2 | + | x3 | = | -13 | ||
| - | x1 | + | x2 | + | 2 | x3 | = | -9 | |||
| 3 | x1 | + | x2 | + | x3 | = | 12 |
| det A = | -2 | 1 | 1 | = | ||
| -1 | 1 | 2 | ||||
| 3 | 1 | 1 |
The elements of row 1 multiplied by -1 are added to the corresponding elements of row 3. more info
| -2 | 1 | 1 | ||
| -1 | 1 | 2 | ||
| 3 + ( -2) * ( -1) | 1 + 1 * ( -1) | 1 + 1 * ( -1) |
This elementary transformation does not change the value of the determinant.
| = | -2 | 1 | 1 | = | ||
| -1 | 1 | 2 | ||||
| 5 | 0 | 0 |
Expand the determinant along the row 3. more info
|
Row number 3 Column number 1 |
Element | Row 3 and column 1 have been deleted |
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| ( -1) 3 + 1 | * | 5 | * |
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Row number 3 Column number 2 |
Element | Row 3 and column 2 have been deleted |
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| ( -1) 3 + 2 | * | 0 | * |
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Row number 3 Column number 3 |
Element | Row 3 and column 3 have been deleted |
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| ( -1) 3 + 3 | * | 0 | * |
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Products are summed. If the element is zero than product is zero too.
| = ( -1) 3 + 1 * 5 * | 1 | 1 | = | ||
| 1 | 2 |
| = 5 * | 1 | 1 | = | ||
| 1 | 2 |
= 5 * ( 1 * 2 - 1 * 1 ) =
= 5 * ( 2 - 1 ) =
= 5
The determinant A is not zero. It is possible to use the Cramer's rule.
Let's calculate the determinant A1. more info
It is necessary to change column 1 in determinant A to the column of the right side of the system.
| System | det A | det A1 | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
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| det A1 = | -13 | 1 | 1 | = | ||
| -9 | 1 | 2 | ||||
| 12 | 1 | 1 |
The elements of row 1 multiplied by -1 are added to the corresponding elements of row 3. more info
| -13 | 1 | 1 | ||
| -9 | 1 | 2 | ||
| 12 + ( -13) * ( -1) | 1 + 1 * ( -1) | 1 + 1 * ( -1) |
This elementary transformation does not change the value of the determinant.
| = | -13 | 1 | 1 | = | ||
| -9 | 1 | 2 | ||||
| 25 | 0 | 0 |
Expand the determinant along the row 3. more info
|
Row number 3 Column number 1 |
Element | Row 3 and column 1 have been deleted |
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| ( -1) 3 + 1 | * | 25 | * |
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Row number 3 Column number 2 |
Element | Row 3 and column 2 have been deleted |
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| ( -1) 3 + 2 | * | 0 | * |
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|
Row number 3 Column number 3 |
Element | Row 3 and column 3 have been deleted |
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| ( -1) 3 + 3 | * | 0 | * |
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Products are summed. If the element is zero than product is zero too.
| = ( -1) 3 + 1 * 25 * | 1 | 1 | = | ||
| 1 | 2 |
| = 25 * | 1 | 1 | = | ||
| 1 | 2 |
= 25 * ( 1 * 2 - 1 * 1 ) =
= 25 * ( 2 - 1 ) =
= 25
Let's calculate the determinant A2. more info
It is necessary to change column 2 in determinant A to the column of the right side of the system.
| System | det A | det A2 | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
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| det A2 = | -2 | -13 | 1 | = | ||
| -1 | -9 | 2 | ||||
| 3 | 12 | 1 |
The elements of row 1 are added to the corresponding elements of row 3. more info
| -2 | -13 | 1 | ||
| -1 | -9 | 2 | ||
| 3 + ( -2) | 12 + ( -13) | 1 + 1 |
This elementary transformation does not change the value of the determinant.
| = | -2 | -13 | 1 | = | ||
| -1 | -9 | 2 | ||||
| 1 | -1 | 2 |
The elements of row 3 multiplied by 2 are added to the corresponding elements of row 1. more info
| -2 + 1 * 2 | -13 + ( -1) * 2 | 1 + 2 * 2 | ||
| -1 | -9 | 2 | ||
| 1 | -1 | 2 |
This elementary transformation does not change the value of the determinant.
| = | 0 | -15 | 5 | = | ||
| -1 | -9 | 2 | ||||
| 1 | -1 | 2 |
The elements of row 3 are added to the corresponding elements of row 2. more info
| 0 | -15 | 5 | ||
| -1 + 1 | -9 + ( -1) | 2 + 2 | ||
| 1 | -1 | 2 |
This elementary transformation does not change the value of the determinant.
| = | 0 | -15 | 5 | = | ||
| 0 | -10 | 4 | ||||
| 1 | -1 | 2 |
Expand the determinant along the column 1. more info
|
Row number 1 Column number 1 |
Element | Row 1 and column 1 have been deleted |
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| ( -1) 1 + 1 | * | 0 | * |
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Row number 2 Column number 1 |
Element | Row 2 and column 1 have been deleted |
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| ( -1) 2 + 1 | * | 0 | * |
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Row number 3 Column number 1 |
Element | Row 3 and column 1 have been deleted |
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| ( -1) 3 + 1 | * | 1 | * |
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Products are summed. If the element is zero than product is zero too.
| = ( -1) 3 + 1 * 1 * | -15 | 5 | = | ||
| -10 | 4 |
| = | -15 | 5 | = | ||
| -10 | 4 |
= -15 * 4 - 5 * ( -10) =
= -60 + 50 =
= -10
Let's calculate the determinant A3. more info
It is necessary to change column 3 in determinant A to the column of the right side of the system.
| System | det A | det A3 | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
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| det A3 = | -2 | 1 | -13 | = | ||
| -1 | 1 | -9 | ||||
| 3 | 1 | 12 |
The elements of row 2 multiplied by -1 are added to the corresponding elements of row 1. more info
| -2 + ( -1) * ( -1) | 1 + 1 * ( -1) | -13 + ( -9) * ( -1) | ||
| -1 | 1 | -9 | ||
| 3 | 1 | 12 |
This elementary transformation does not change the value of the determinant.
| = | -1 | 0 | -4 | = | ||
| -1 | 1 | -9 | ||||
| 3 | 1 | 12 |
The elements of row 2 multiplied by -1 are added to the corresponding elements of row 3. more info
| -1 | 0 | -4 | ||
| -1 | 1 | -9 | ||
| 3 + ( -1) * ( -1) | 1 + 1 * ( -1) | 12 + ( -9) * ( -1) |
This elementary transformation does not change the value of the determinant.
| = | -1 | 0 | -4 | = | ||
| -1 | 1 | -9 | ||||
| 4 | 0 | 21 |
Expand the determinant along the column 2. more info
|
Row number 1 Column number 2 |
Element | Row 1 and column 2 have been deleted |
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| ( -1) 1 + 2 | * | 0 | * |
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Row number 2 Column number 2 |
Element | Row 2 and column 2 have been deleted |
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| ( -1) 2 + 2 | * | 1 | * |
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Row number 3 Column number 2 |
Element | Row 3 and column 2 have been deleted |
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| ( -1) 3 + 2 | * | 0 | * |
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Products are summed. If the element is zero than product is zero too.
| = ( -1) 2 + 2 * 1 * | -1 | -4 | = | ||
| 4 | 21 |
| = | -1 | -4 | = | ||
| 4 | 21 |
= -1 * 21 - ( -4) * 4 =
= -21 + 16 =
= -5
Result:
x1 = det A1 / det A = 25/5 = 5
x2 = det A2 / det A = -10/5 = -2
x3 = det A3 / det A = -5/5 = -1