Solutions to ExercisesCollege of the RedwoodsMathematics DepartmentMath Math 45 — Linear AlgebraQuiz #3—Linear AlgebraDavid ArnoldCopyrightc 2000 [email protected] Revision Date: September 19, 2001 Version 1.002Directions: Place your solution in the space provided. No calcula-tors allowed!Exercise 1. Let A be a 3 × 3 matrix. If you want to multiply row 1by 2, row 2 by 3, and row 3 by 4, then multiply on thebythe matrix.Exercise 2. Let A be a 3 × 3 matrix. If you want to interchangecolumns 1 and 3 of matrix A, then multiply on theby thematrix.Exercise 3. If the 4 × 4 matrix A has row 3 = 2(row 1) + 3(row 2),find a solution of yA =0000.Exercise 4. If the 5 × 5 matrix A has col 3 = 2(col 1) + 3(col 2), finda solution of Ax = 0.Exercise 5. Multiply:1234567=3Exercise 6. What is the inverse of the following elementary matrix?10−301 000 1Exercise 7. Use block multiplication to show an easy way of findingthe inverse of the following matrix. Note: Don’t bother with Gauss–Jordan elimination. You’ll receive no credit for this technique. Thinkblocks!1000010041103101Exercise 8. Use Gauss–Jordan elimination to find the inverse of thematrix1234.Solutions to Exercises 4Solutions to ExercisesExercise 1. Let A be a 3 × 3 matrix. If you want to multiply row1 by 2, row 2 by 3, and row 3 by 4, then multiply on the left by thematrix200030004Exercise 1Solutions to Exercises 5Exercise 2. Let A be a 3 × 3 matrix. If you want to interchangecolumns 1 and 3 of matrix A, then multiply on the right by the matrix001010100Exercise 2Solutions to Exercises 6Exercise 3. Because 2(row1) + 3(row2) = 1(row3), we can write2(row1) + 3(row2) − row3=−→0 .Thus, if A is 4 × 4, then−→yA=23−10row1row2row3row4=2(row1) + 3(row2) − 1(row3) + 0(row4)=0000=−→0Exercise 3Solutions to Exercises 7Exercise 4. If 2(col1) + 3(col2) = col3, then we can write2(col1) + 3(col2) − 1(col3) =−→0Thus, if A is 5 × 5,A−→x =col1 col2 col3 col4 col523−100=2(col1) + 3(col2) − 1(col3) + 0(col4) + 0(col5)=00000=−→0Exercise 4Solutions to Exercises 8Exercise 5. Multiply each column of the second matrix by123.1234567=45678 10121412 15 18 21Exercise 5Solutions to Exercises 9Exercise 6. If we subtract 3 times row3fromrow1, then we getA =10−301 000 1,Thus,A−1=103010001Exercise 6Solutions to Exercises 10Exercise 7. Break the matrix into 4 2 × 2blocks.B =1000010041103101=I 0AIBecauseI 0AIadds A times row1torow2, its inverse isI 0−AIChecking,Solutions to Exercises 11100001004110310110000100−4 −110−3 −101=1000010000100001Exercise 7Solutions to Exercises 12Exercise 8. Use elimination to sendAItoIA−1AI=12103401⇒12 100 −2 −31⇒10−210 −2 −31⇒10−21013/2 −1/2=IA−1Therefore,A−1=−213/2 −1/2Exercise
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