# We find this by projecting ar onto n where a is any

• Test Prep
• jxscarlettgao
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. We find this by projecting --→ AR onto n , where A is any point on the plane. This is because this projection is the vector from R to the plane which is perpendicular to the plane. Since O (0 , 0 , 0) is on the plane, we may take A = O , and so --→ AR = 1 - 2 3 . We calculated in part (b) that: proj n 1 - 2 3 = 4 3 1 - 2 1 So the shortest distance is therefore: proj n 1 - 2 3 = 4 3 1 - 2 1 = 4 3 p 1 2 + ( - 2) 2 + 1 2 = 4 6 3 7
2. Let A = - 2 - 1 6 - 1 2 - 2 (a) Find elementary matrices E 1 , E 2 , E 3 , E 4 such that E 4 E 3 E 2 E 1 A = R , where R is the reduced row-echelon form of A . [6 marks] 1 1 - 4 .
(b) Can A be expressed as a product of elementary matrices? Why or why not? [2 marks]
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3 (a) Define what it means for a set of vectors { x 1 , x 2 , . . . , x k } in R n to be linearly dependent . [2 marks] 3 (b) For what values of c R is the set c linearly dependent? [6 marks]