Recall the definition vectors v 1 v 2 v 3 are linearly independent if one can

# Recall the definition vectors v 1 v 2 v 3 are

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Recall the definition: vectors v 1 , v 2 , v 3 are linearly independent if one can find a nontrivial linear combination c 1 v 1 + c 2 v 2 , + c 3 v 3 that is equal to 0 . The system c 1 3 0 - 6 1 + c 2 - 6 0 4 - 3 + c 3 9 0 h 3 = 0 0 0 0 has nontrivial solution when the matrix 3 - 6 9 0 0 0 - 6 4 h 1 - 3 - 3 row-reduces to an echelon form with at least one non-pivot column. Here are the first steps in the Gaussian elimination: 3 - 6 9 0 0 0 - 6 4 h 1 - 3 - 3 7→ 1 - 2 3 - 6 4 h 1 - 3 3 0 0 0 7→ 1 - 2 3 0 - 8 h + 18 0 - 1 0 0 0 0 7→ 1 - 2 3 0 1 0 0 0 h + 18 0 0 0 The matrix 1 - 2 3 0 1 0 0 0 h + 18 0 0 0 is in echelon form and has a pivot in each column, except when h + 18 = 0 , as then the last column has no pivot! The vectors are linearly dependent when h = - 18 2. Consider linear transformation T defined by T ( x ) = A x with A = 1 2 3 4 3 4 5 6 2 3 4 5 . Find vector x whose image under T is vector b = 0 4 2 , and determine whether x is unique. We row-reduce the augmented matrix 1 2 3 4 0 3 4 5 6 4 2 3 4 5 2 7→ 1 0 - 1 - 2 4 0 1 2 3 - 2 0 0 0 0 0 Solving for basic variables x 1 , x 2 we get x 1 = 4 + x 3 + 2 x 4 , x 2 = - 2 - 2 x 3 - 3 x 4 with infinite number of answers x = 4 - 2 0 0 + s 1 - 2 1 0 + t 2 - 3 0 1 Printed: February 1, 2020 Name
Linear Algebra MATH 2076 Quiz-3 B Answer: Key Be sure to show your work. No credit for inspired answers! 1. Find the value(s) of h for which the vectors v 1 = 2 0 - 2 4 , v 2 = 4 0 - 6 7 , and v 3 = - 2 0 2 h are linearly dependent . Justify your answer. Answer: This problems is similar to problems 11, 12 in Section 1.7. Recall the definition: vectors v 1 , v 2 , v 3 are linearly independent if one can find a nontrivial linear combination c 1 v 1 + c 2 v 2 , + c 3 v 3 that is equal to 0 . The system c 1 2 0 - 2 4 + c 2 4 0 - 6 7 + c 3 - 2 0 2 h = 0 0 0 0 has nontrivial solution when the matrix 2 4 - 2 0 0 0 - 2 - 6 2 4 7 h row-reduces to an echelon form with at least one non-pivot column. Here are the first steps in the Gaussian elimination: 2 4 - 2 - 2 - 6 2 0 0 0 4 7 h 7→ 2 4 - 2 0 - 2 0 0 - 1 h + 4 0 0 0 7→ 2 4 - 2 0 1 0 0 0 h + 4 0 0 0 7→ 1 2 - 1 0 1 0 0 0 h + 4 0 0 0 The matrix 1 2 - 1 0 1 0 0 0 h + 4 0 0 0

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