# Explanation if a has n pivot positions it has a pivot

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6. x 1 = 1 + t, t arbitrary Explanation: By row reduction B = 1 1 1 2 0 0 3 9 0 0 1 5 1 1 1 2 0 0 3 9 0 0 0 2 , which is now in echelon form. But the system x 1 x 2 + x 3 = 2 , 3 x 3 = 9 ,
Explanation: If A has n pivot positions, it has a pivot in each of its n columns and in each of its n rows. The reduced echelon form has a 1 in each pivot position, so the reduced echelon form is the n × n identity matrix. Consequently, the statement is TRUE . 009 10.0points If the augmented matrix for a system of linear equations in variables x 1 , x 2 , and x 3 is row equivalent to the matrix B = 1 1 1 2 0 0 3 9 0 0 1 5 , determine x 1 , if possible. 1. x 1 = 2 2. no solution x 1 exists correct 3. x 1 = 0 4. x 1 = 1 5. x 1 = 1 + t, t arbitrary 0 x 3 = 5 , associated with this matrix is inconsistent be- cause there is no solution x 3 . Consequently, no solution x 1 exists because the original sys- tem is inconsistent . 010 10.0points The matrices A = bracketleftbigg 1 0 a 1 bracketrightbigg , B = bracketleftbigg 1 0 b 1 bracketrightbigg , have the property AB = BA, for all values of a and b . True or False? 1. FALSE 2. TRUE correct Explanation: When A = bracketleftbigg 1 0 a 1 bracketrightbigg , B = bracketleftbigg 1 0 b 1 bracketrightbigg ,
Version 023 – EXAM01 – gilbert – (57245) 5 then AB = bracketleftbigg 1 0 a 1 bracketrightbigg bracketleftbigg 1 0 b 1 bracketrightbigg = bracketleftbigg 1 0 a + b 1 bracketrightbigg , while BA = bracketleftbigg 1 0 b 1 bracketrightbigg bracketleftbigg 1 0 a 1 bracketrightbigg = bracketleftbigg 1 0 b + a 1 bracketrightbigg . But a + b = b + a for all a, b . Consequently, the statement is TRUE . 011 10.0points Determine the 3 × 3 matrix E such that EA = B when A 3 R 1 + R 3 R 3 −−−−−−−−→ A 1 4 R 2 R 2 −−−−−−→ A 2 - 2 R 2 + R 1 R 1 −−−−−−−−−→ B. 1. E = 1 8 0 0 1 4 0 3 0 1 2. E = 1 8 0 0 4 0 3 0 1 3. E = 1 8 0 0 4 0 3 0 1 4. E = 1 8 0 0 1 4 0 3 0 1 5. E = 1 8 0 0 1 4 0 3 0 1 6. E = 1 8 0 0 4 0 3 0 1 correct Explanation: Since A 1 = 1 0 0 0 1 0 3 0 1 A, A 2 = 1 0 0 0 4 0 0 0 1 A 1 , and B = 1 2 0 0 1 0 0 0 1 A 2 , we see that E = 1 2 0 0 1 0 0 0 1 1 0 0 0 4 0 0 0 1 1 0 0 0 1 0 3 0 1 = 1 8 0 0 4 0 0 0 1 1 0 0 0 1 0 3 0 1 . Consequently, E = 1 8 0 0 4 0 3 0 1 . 012 10.0points If the columns of an m × n matrix A = [ a 1 a 2 ... a n ] span R m , then the equation A x = b is consis- tent for each b in R m . True or False? 1. FALSE 2. TRUE correct Explanation: When x = x 1 x 2 . . . x n ,
Version 023 – EXAM01 – gilbert – (57245) 6 then A x = [ a 1 a 2 ... a n ] x 1 x 2 . . . x n = x 1 a 1 + x 2 a 2 + · · · + x n a n . But if R m = Span { a 1 , a 2 , ..., a n } , then any vector b in R m can be represented by a linear combination b = c 1 a 1 + c 2 a 2 + ... + c n a n for suitable scalars c 1 , c 2 , ..., c n . Thus A x = b has a solution for each b in R m with x j = c j .
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