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Problems #12, pg.40 and the additional problem (A) are graded.
Problem #12, pg.40: Write the augmented matrix for the linear system that corresponds to the matrix equation A→x=→b for A=[12−1−3−42523] and →b=[−2412]. Then solve the system and write the solution as a vector.
Solution: The augmented matrix we seek is [12−11−3−422523−3]. We will find a solution by row reduction. So compute
[12−11−3−422523−3]r∗2=r2+3r1r∗3=r3−5r1∼[12−1102−150−88−8]r∗3=r3+4r2∼[12−1102−1500412]r∗3=14r3∼[12−1102−150013]r∗1=r1+r3r∗2=r2+r3∼[120402080013]r∗1=r1−r2∼[100−402080013]r∗2=12r2∼[100−401040013].
Therefore we see that the solution is x1=−4,x2=4,x3=3. So we now write the solution vector →x:
→x=[x1x2x3]=[−443].
Additional Problem (A): Solve the matrix equation A→x=→b where A=[1010−1100−1] and →b=[b1b2b3].
Solution: We know the solution of this equation to be equivalent to the solution of the system whose augmented matrix is
[101b10−11b200−1b3]r∗1=r1+r3r∗2=r2+r3∼[100b1+b30−10b2+b300−1b3]r∗2=−r2r∗3=−r3∼[100b1+b3010−b2−b3001−b3]
hence we see that the solution is given by
→x=[b1+b3−b2−b3−b3.]
