Question

Prove the following lemma.

Let B be an n ✕ n matrix and let E be an n ✕ n elementary matrix. Then det(EB) = det(E) det(B)

1. Write the proof and submit as a free response. (Submit a file with a maximum size of 1 MB.)

2. Which of the following could begin a direct proof of the statement?

If E interchanges two rows, then det(E) = 1 by Theorem 4.4. Also, EB is the same as B but with two rows interchanged, so by Theorem 4.3(b), det(EB) = det(B) = det(E) det(B).If E interchanges two rows, then det(E) = −1 by Theorem 4.4. Also, EB is the same as B but with two rows interchanged, so by Theorem 4.3(b), det(EB) = −det(B) = det(E) det(B).    If E interchanges two rows, then det(E) = 1 by Theorem 4.4. Also, EB is different from B, so by Theorem 4.3(b), det(EB) = −det(B) = det(E) det(B).If E interchanges two rows, then det(E) = k by Theorem 4.4. Also, EB is the same as B but with two rows interchanged, so by Theorem 4.3(b), det(EB) = −det(B) = −det(E).If E interchanges two rows, then det(E) = k by Theorem 4.4. Also, EB is the same as B but with two rows interchanged, so by Theorem 4.3(b), det(EB) = −det(B) = −det(E) det(B).

Consider the following. Theorem 4.3: Let A = [a, be a square matrix. a. If A has a zero row (column), then det(A) = 0. b. If B is obtained by interchanging two rows (columns) of A, then det(B) = -det(A). C. If A has two identical rows (columns), then det(A) = 0. d. If B is obtained by multiplying a row (column) of A by k, then det(B) = k det(A). e. If A, B, and C are identical except that the ith row (column) of C is the sum of the ith rows (columns) of A and B, then det(C) = det(A) + det(B). f. If B is obtained by adding a multiple of one row (column) of A to another row (column), then det(B) = det(A). Theorem 4.4: Let E be an nxn elementary matrix. a. If E results from interchanging two rows of 1, then det(E) = -1. b. If E results from multiplying one row of I, by k, then det(E) = k. C. If E results from adding a multiple of one row of I, to another row, then det(E) = 1. Prove the following lemma. Let B be an nxn matrix and let E be an nxn elementary matrix. Then det(EB) = det(E) det(B) Which of the following could begin a direct proof of the statement? If E interchanges two rows, then det(E) = 1 by Theorem 4.4. Also, EB is the same as B but with two rows interchanged, so by Theorem 4.3(b), det(EB) = det(B) = det(E) det(B). If E interchanges two rows, then det(E) = -1 by Theorem 4.4. Also, EB is the same as B but with two rows interchanged, so by Theorem 4.3(b), det(EB) = -det(B) = det(E) det(B). If E interchanges two rows, then det(E) = 1 by Theorem 4.4. Also, EB is different from B, so by Theorem 4.3(b), det(EB) = -det(B) = det(E) det(B). If E interchanges two rows, then det(E) = k by Theorem 4.4. Also, EB is the same as B but with two rows interchanged, so by Theorem 4.3(b), det(EB) = -det(B) = -det(E). If E interchanges two rows, then det(E) = k by Theorem 4.4. Also, EB is the same as B but with two rows interchanged, so by Theorem 4.3(b), det(EB) = -det(B) = -det(E) det(B). Write the proof and submit as a free response. (Submit a file with a maximum size of 1 MB.) 1 Choose File No file chosen This answer has not been graded yet. Need Help? Read It Talk to a Tutor

Answer 1

Stalement! and Let nxn let B be an nxn matrix elementary matrix. E be an Then, det LEB) = det (E) det (6) Proof : Elementary matrices can be obtained by o E Interchanges 2 rows of identity maslin. 2 E multiples a row of In by non-zero consta ③ E mulkplies row i by c & adds it to another row. We shall CASE 1 take cases CInterchanging 2 vows) (Th 4.4) If E interchanges 2 rows, then det CE)=-1 •EB = B but with 2 rows merchanged. . Thm 4.3) det (EB) = - det (B) lusing = +det(E).det (B) Case 2: Multiplies a row byk. B one row is the but with multiplied by EB same as 4.4(d) 4.4 (6) det (EB) = k det (B) From Thm det (E) = k From Thm det (EB) = det (E) det (B). Page No

Date _/_ Case 3: E multiplies now By i and adds it to rowi row i multiplied EB is the same as with k and added to ß but with now j. Thm. From Thm 4.3 (4) det (EB) = 4.4 (c) det LE) = det (B) from Thm 1 . det CEB)= 1. det (B) dot(Ed. det (B) Hence Proved