addpr - Therefore we get A = E-1 1 E-1 2 E-1 3(or A = E-1 1...

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Practice Problems for Midterm 1 8. Consider the invertible matrix A = 0 1 0 1 1 0 4 0 1 . (a) Express A as a product of elementary matrices. (b) Using the result of the part (a), compute det( A ). Solution. (a) Three consecutive row operations on A R 1 ←→ R 2 R 1 R 1 - R 2 R 3 R 3 - 4 R 1 . lead into the identity matrix. Let E 1 , E 2 and E 3 be the elementary matrices that corre- sponds the above three operations, respectively. Then the above row operations suggest E 3 E 2 E 1 A = I where E 1 = 0 1 0 1 0 0 0 0 1 , E 2 = 1 - 1 0 0 1 0 0 0 1 , E 3 = 1 0 0 0 1 0 - 4 0 1 . Now, we observe ( E 3 E 2 E 1 ) A = I ( E 3 E 2 E 1 ) - 1 ( E 3 E 2 E 1 ) A = ( E 3 E 2 E 1 ) - 1 I IA = ( E 3 E 2 E 1 ) - 1 A = E - 1 1 E - 1 2 E - 1 3 . Note that the inverse matrices of the elementary matrices are also elementary matrices.
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Unformatted text preview: Therefore we get A = E-1 1 E-1 2 E-1 3 (or A = E-1 1 E-1 2 E-1 3 I ) as a solution where E-1 1 = 0 1 0 1 0 0 0 0 1 , E-1 2 = 1 1 0 0 1 0 0 0 1 , E-1 3 = 1 0 0 0 1 0 4 0 1 . (b) We have det( A ) =det( E-1 1 E-1 2 E-1 3 I ). Here E-1 2 and E-1 3 are type III and E-1 1 is type I. Since type III operations do change the determinants while type I change the signs, we get det( A ) =-det( I ) =-1....
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This note was uploaded on 07/17/2008 for the course MATH 571 taught by Professor Kim during the Spring '08 term at Ohio State.

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