final-review

# final-review - Solutions to practice problems for the Final...

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Solutions to practice problems for the Final Systems of linear equations 1. Find the general solution of the following system of equations. x 1 + 3 x 2 + x 3 + 5 x 4 + x 5 = 5 + x 2 + x 3 + 2 x 4 + x 5 = 4 2 x 1 + 4 x 2 + 7 x 4 + x 5 = 3 Solution. 1 3 1 5 1 5 0 1 1 2 1 4 2 4 0 7 1 3 1 3 1 5 1 5 0 1 1 2 1 4 0 2 2 3 1 7 1 3 1 5 1 5 0 1 1 2 1 4 0 0 0 1 1 1 1 3 1 0 4 0 0 1 1 0 1 2 0 0 0 1 1 1 1 0 2 0 1 6 0 1 1 0 1 2 0 0 0 1 1 1 x 1 = 6 + 2 s + t, x 2 = 2 s + t, x 3 = s, x 4 = 1 t, x 5 = t 2. Solve the following system of linear equations. 2 x 1 + x 2 + 3 x 3 = 1 4 x 1 + 3 x 2 + 5 x 3 = 1 6 x 1 + 5 x 2 + 5 x 3 = 3 Solution. 2 1 3 1 4 3 5 1 6 5 5 3 2 1 3 1 0 1 1 1 0 2 4 6 2 1 3 1 0 1 1 1 0 0 2 4 1 1 2 3 2 1 2 0 1 1 1 0 0 1 2 1 1 2 0 5 2 0 1 0 1 0 0 1 2 1 0 0 3 0 1 0 1 0 0 1 2 x 1 = 3 , x 2 = 1 , x 3 = 2 3. Consider the following system of linear equations. x 1 + x 2 + 3 x 3 = a 2 x 1 + x 2 + 4 x 3 = b 3 x 1 + x 2 + 5 x 3 = c For any fixed values of a , b , and c we obtain a system of 3 equations in 3 unkowns. (a) Find a set of values for a , b , and c so that the system is inconsistent. (b) For a = 0, b = 1, c = 2, find the general form of the solution. 1

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Solution. Consider the following row reduction 1 1 3 a 2 1 4 b 3 1 5 c 1 1 3 a 0 1 2 2 a + b 0 2 4 3 a + c 1 1 3 a 0 1 2 2 a + b 0 0 0 a 2 b + c 1 1 3 a 0 1 2 2 a b 0 0 0 a 2 b + c 1 0 1 a + b 0 1 2 2 a b 0 0 0 a 2 b + c (a) From the last line of the last matrix above we see that any choice of a, b, c with a 2 b + c ̸ = 0 makes the system inconsistent. For a specific example we can take a = 1, b = c = 0. (b) For a = 0, b = 1, c = 2 the system is consistent and the general form of the solution of the system is x 1 = 1 t , x 2 = 1 2 t , x 3 = t . 4. Explain why a homogeneous system of linear equations with more unkowns than equations always has nontrivial solutions. Solution. Matrix algebra 5. Either compute the inverse of the matrix A below, or explain why A is not invertible. A = 2 5 8 5 1 2 3 1 2 4 7 2 1 3 5 3 Solution. We row reduce the matrix [ A | I ] . 2 5 8 5 1 0 0 0 1 2 3 1 0 1 0 0 2 4 7 2 0 0 1 0 1 3 5 3 0 0 0 1 1 2 3 1 0 1 0 0 2 5 8 5 1 0 0 0 2 4 7 2 0 0 1 0 1 3 5 3 0 0 0 1 1 2 3 1 0 1 0 0 0 1 2 3 1 2 0 0 0 0 1 0 0 2 1 0 0 1 2 2 0 1 0 1 1 2 3 1 0 1 0 0 0 1 2 3 1 2 0 0 0 0 1 0 0 2 1 0 0 0 0 1 1 1 0 1 1 2 3 1 0 1 0 0 0 1 2 3 1 2 0 0 0 0 1 0 0 2 1 0 0 0 0 1 1 1 0 1 1 2 3 0 1 2 0 1 0 1 2 0 2 1 0 3 0 0 1 0 0 2 1 0 0 0 0 1 1 1 0 1
1 2 0 0 1 8 3 1 0 1 0 0 2 5 2 3 0 0 1 0 0 2 1 0 0 0 0 1 1 1 0 1 1 0 0 0 3 2 1 5 0 1 0 0 2 5 2 3 0 0 1 0 0 2 1 0 0 0 0 1 1 1 0 1 So A 1 = 3 2 1 5

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