The value of x, y and z will be 1, 5 and 2 respectively
An augmented matrix in linear algebra is a matrix created by joining the columns of two supplied matrices, often so that the same basic row operations may be applied to each of the given matrices individually.
Lets write the augmented matrix by writing the coefficients of all the variables:
3 -4 -5 -27 Row 1
5 2 -2 11 Row 2
5 -4 4 -7 Row 3
We need to get
1 0 0
0 1 0
0 0 1
then the values of x, y, and z will be in the last column.
The row operation (R1=R1/3) is used to get the identity matrix.
1 -4/3 -5/3 -9 Row 1
5 2 -2 11 Row 2
5 -4 4 -7 Row 3
Add row 2 to row 1 and multiply by 5 (R2=R2(5)R1).
1 -4/3 -5/3 -9 Row 1
0 26/3 19/3 56 Row 2
5 -4 4 -7 Row 3
Add row 3 to row 1 and multiply by 5 (R3=R3(5)R1).
1 -4/3 -5/3 -9 Row 1
0 26/3 19/3 56 Row 2
0 8/3 37/3 38 Row 3
Multiply row 2 by 326 (R2=(3/26)R2)
1 -4/3 -5/3 -9 Row 1
0 1 19/26 84/13 Row 2
0 8/3 37/3 38 Row 3
Add row 2 multiplied by 4/3 to row 1 (R1=R1+(4/3)R2)
1 0 -9/13 -5/13 Row 1
0 1 19/26 84/13 Row 2
0 8/3 37/3 38 Row 3
Add row 3 to row 2 and multiply the result by 8/3 (R3=R3(8/3)R2).
1 0 -9/13 -5/13 Row 1
0 1 19/26 84/13 Row 2
0 0 135/13 270/13 Row 3
Multiply row 3 by 13/135 (R3=(13/135)R3)
1 0 -9/13 -5/13 Row 1
0 1 19/26 84/13 Row 2
0 0 1 2 Row 3
Add row 3 multiplied by 9/13 to row 1 (R1=R1+(9/13)R3)
1 0 0 1 Row 1
0 1 19/26 84/13 Row 2
0 0 1 2 Row 3
Row 2 is reduced by row 3 multiplied by 19/26 (R2=R2(19/26)R3).
1 0 0 1 Row 1
0 1 0 5 Row 2
0 0 1 2 Row 3
Hence the value of x, y and z will be 1, 5 and 2 respectively
Learn more about augmented matrix here:
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