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Problems #8 and #12 from pg.10 are graded.
Problem 8: Use row operations to solve the system of equations whose augmented matrix is [1−5400010100030000020].
Solution: We will reduce this matrix to reduced echelon form. Compute
[1−5400010100030000020]r∗2=r2−12r4=[1−5400010000030000020]r∗1=r1−43r3=[1−5000010000030000020]r∗1=r1+5r2=[10000010000030000020]r∗3=13r3r∗4=12r4=[10000010000010000010].
Therefore we see that the system associated with the original augmented matrix has solution x1=x2=x3=x4=0.
Problem #12: Solve the system
{x1−5x2+4x3=−32x1−7x2+3x3=−2−2x1+x2+7x3=−1.
Solution: We will solve this by writing its associated augmented matrix and then reducing that matrix to reduced ecehlon form. The associated augmented matrix is [1−54−32−73−2−217−1]. Now compute
[1−54−32−73−2−217−1]r∗2=r2−2r1r∗3=r3+2r1=[1−54−303−540−915−7]r∗3=r3+3r2=[1−54−303−540005]r∗3=15r3=[1−54−303−540001]r∗1=r1+3r3r∗2=r2−4r3=[1−54003−500001]r∗1=r1+53r2=[10−133003−500001]r∗2=13r2=[10−133001−5300001].
Now it is clear that the system has no solution because the system of equations associated with the reduced echelon form is
{x1−133x3=0x2−53x3=00=1,
which has no solution because no choice of x1,x2, and x3 can make the third equation true.
