PracticeTest - R n n , is the equation ( A + B ) 2 = A 2 +2...

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Linear Algebra Midterm Summer 2011 1. Consider the system of equations x 1 + 2 x 2 + 3 x 3 + 2 x 4 = 5 4 x 1 + 8 x 2 + x 3 + x 4 + 6 x 5 = 10 3 x 1 + 6 x 2 + x 3 + 2 x 4 + 5 x 5 = 15 2 x 1 + 4 x 2 + x 3 + 9 x 4 + 10 x 5 = 30 This can be succintly written as A x = b and represented in matrix form as [ A | b ] R 4 × 6 . Given that rref([ A | b ]) = 1 2 0 0 0 27 0 0 1 0 0 - 18 0 0 0 1 0 16 0 0 0 0 1 - 16 (a) Find ker A as a span of vectors. (You’ll need to consider rref([ A | 0 ]) here.) (b) Find im A as a span of vectors. (c) Find a particular solution s of the system A x = b . 1
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(d) Use parts (a) and (c) to find the set of all possible solutions K = s + ker A , and write this in the form K = { s + t 1 v 1 + ··· + t k v k | t 1 ,...,t k R } where the vectors v i are the basis vectors of ker A . (e) Use part (d) with t 1 = ··· = t k = 3 to find a second solution s 2 of A x = b and verify that this is indeed a solution. 2. Let A R n × n satisfy A 2 = A . Show that if all entries of A are nonzero, then A is not invertible. [Hint: It’s easier to prove the contrapositive of this statement.] 2
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3. If a and b are real numbers, we know that ( a + b ) 2 = a 2 + 2 ab + b 2 . If A,B
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Unformatted text preview: R n n , is the equation ( A + B ) 2 = A 2 +2 AB + B 2 still true? If so, prove it, if not, nd examples of matrices A and B for which this fails. 4. Let A = 2 4 8 4 5 1 7 9 3 . (a) Find ker A and im A as spans of vectors. 3 (b) Find all solutions of the system A x = b , where b = 6 9 16 . 5. Consider the basis = 1 1 , 1 2 for R 2 . Find the representation [ x ] of the vector x = -2 5 in this basis. 6. Let = { e 1 , e 2 } be the standard basis for R 2 and let = 1 1 , 1 be another basis for R 2 . If T A L ( R 2 , R 2 ) is the counterclockwise rotation through / 4, with associated matrix A = " 1 2-1 2 1 2 1 2 # nd the matrix representation [ T A ] , of T A in these bases. 4...
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PracticeTest - R n n , is the equation ( A + B ) 2 = A 2 +2...

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