# The elementary matrix corresponding to each row

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Chapter 7 / Exercise 5
Trigonometry
McKeague/Turner
Expert Verified
the elementary matrix corresponding to each row operation that you use, and hence in each case express both the matrix and the inverse as products of elementary matrices. (a) 1 3 4 3 (b) 1 0 0 0 0 1 1 2 3 (c) 0 0 1 1 1 1 0 1 2 Solution (a) 1 3 1 0 4 3 0 1 1 3 1 0 E 1 = 1 0 0 - 9 - 4 1 - 4 1 1 3 1 0 E 2 = 1 0 0 1 4 / 9 - 1 / 9 0 - 1 / 9 1 0 - 1 / 3 1 / 3 E 3 = 1 - 3 0 1 4 / 9 - 1 / 9 0 1 So A - 1 = - 1 / 3 1 / 3 4 / 9 - 1 / 9 . Now E 3 E 2 E 1 A = I , so A - 1 = E 3 E 2 E 1 = 1 - 3 0 1 1 0 0 - 1 / 9 1 0 - 4 1 , and A = E - 1 1 E - 1 2 E - 1 3 = 1 0 4 1 1 0 0 - 9 1 3 0 1 . 22
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Chapter 7 / Exercise 5
Trigonometry
McKeague/Turner
Expert Verified
(b) 1 0 0 1 0 0 0 0 1 0 1 0 1 2 3 0 0 1 1 0 0 1 0 0 E 1 = 1 0 0 0 0 1 0 1 0 0 1 0 0 2 3 - 1 0 1 - 1 0 1 1 0 0 1 0 0 E 2 = 1 0 0 0 2 3 - 1 0 1 0 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 0 E 3 = 1 0 0 0 1 3 / 2 - 1 / 2 0 1 / 2 0 1 / 2 0 0 0 1 0 1 0 0 0 1 1 0 0 1 0 0 E 4 = 1 0 0 0 1 0 - 1 / 2 - 3 / 2 1 / 2 0 1 - 3 / 2 0 0 1 0 1 0 0 0 1 so A - 1 = 1 0 0 - 1 / 2 - 3 / 2 1 / 2 0 1 0 . A - 1 = E 4 E 3 E 2 E 1 = 1 0 0 0 1 - 3 / 2 0 0 1 1 0 0 0 1 / 2 0 0 0 1 1 0 0 0 0 1 0 1 0 1 0 0 0 1 0 - 1 0 1 and A = E - 1 1 E - 1 2 E - 1 3 E - 1 4 = 1 0 0 0 1 0 1 0 1 1 0 0 0 0 1 0 1 0 1 0 0 0 2 0 0 0 1 1 0 0 0 1 3 / 2 0 0 1 . (c) 0 0 1 1 0 0 1 1 1 0 1 0 0 1 2 0 0 1 1 1 1 0 1 0 E 1 = 0 1 0 0 0 1 1 0 0 1 0 0 0 1 2 0 0 1 0 0 1 1 1 1 0 1 0 E 2 = 1 0 0 0 1 2 0 0 1 0 0 1 0 0 1 1 0 0 0 1 0 1 0 - 1 0 1 - 1 E 3 = 1 - 1 0 0 1 2 0 0 1 0 1 0 0 0 1 1 0 0 0 0 1 1 0 0 1 1 - 1 E 4 = 1 0 1 0 1 2 0 0 1 0 1 0 0 0 1 1 0 0 0 0 1 1 0 0 1 1 - 1 E 5 = 1 0 0 0 1 0 - 2 0 1 0 1 - 2 0 0 1 1 0 0 0 0 1 so A - 1 = 1 1 - 1 - 2 0 1 1 0 0 . 23
Then we have A - 1 = E 5 E 4 E 3 E 2 E 1 = 1 0 0 0 1 - 2 0 0 1 1 0 1 0 1 0 0 0 1 1 - 1 0 0 1 0 0 0 1 1 0 0 0 0 1 0 1 0 0 1 0 1 0 0 0 0 1 and A = E - 1 1 E - 1 2 E - 1 3 E - 1 4 E - 1 5 = 0 1 0 1 0 0 0 0 1 1 0 0 0 0 1 0 1 0 1 1 0 0 1 0 0 0 1 1 0 - 1 0 1 0 0 0 1 1 0 0 0 1 2 0 0 1 . 15. Find all values of α for which the following matrix is not invertible. A = 1 4 - 3 - 2 - 7 6 - 1 3 α Solution Row reducing A gives 1 0 - 3 0 1 0 0 0 α - 3 . To further row reduce to the identity we need α - 3 6 = 0, so α 6 = 3. Thus A is not invertible if α = 3. 16. Let A and B be matrices of the same size and suppose that B is invertible. (a) Prove that A is invertible if and only if AB is invertible. (b) Hence prove that either A + B and I + AB - 1 are both invertible or else both not invertible. Solution (a) This is an ‘if and only if’ statement. Therefore, we need to prove two things (with the assumption given, that B is invertible): (i) if A is invertible then AB is invertible, and (ii) if AB is invertible, then A is invertible. The key property of invertible matrices that we will use is the following fact: if X and Y are invertible matrices of the same size then XY is also invertible (in fact we saw in class that ( XY ) - 1 = Y - 1 X - 1 ). We prove (i). We are given that A is invertible and B is invertible. Therefore, AB is invertible by the key property above. We prove (ii). Now we are given that B is invertible and AB is invertible. Observe that ( AB ) B - 1 = A ( BB - 1 ) = AI = A . But this shows that A is the product of two invertible matrices, namely AB and B - 1 , and hence is invertible by the key property above (recall that we saw in lecture that B - 1 is invertible with ( B - 1 ) - 1 = B ).