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Hwk3 - HOMEWORK 3 SOLUTIONS Levandosky Linear Algebra 17.2...

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HOMEWORK 3 SOLUTIONS Levandosky, Linear Algebra 17.2. Using the standard formula for computing the determinant of a 2 × 2 matrix, we compute fl fl fl fl 4 6 6 9 fl fl fl fl = 4 · 9 - 6 · 6 = 0 . 17.4. We could use the standard formula for computing the determinant of a 3 × 3 matrix, and compute fl fl fl fl fl fl 1 2 5 1 2 3 1 2 1 fl fl fl fl fl fl = 1 · 2 · 1 + 2 · 3 · 1 + 5 · 1 · 2 - 5 · 2 · 1 - 3 · 2 · 1 - 1 · 2 · 1 = 0 . Alternatively, we could note that the columns are not linearly indepen- dent (since the second column is twice the first), and so the determinant is automatically zero. 17.5. Let’s compute this 3 × 3 determinant by expanding along the first column: fl fl fl fl fl fl 0 0 1 0 1 0 1 0 0 fl fl fl fl fl fl = 0 · fl fl fl fl 1 0 0 0 fl fl fl fl - 0 · fl fl fl fl 0 0 1 0 fl fl fl fl + 1 · fl fl fl fl 0 1 1 0 fl fl fl fl = fl fl fl fl 0 1 1 0 fl fl fl fl = - 1 . 17.7. Let’s compute this 4 × 4 determinant by expanding along the first column: fl fl fl fl fl fl fl fl 1 1 1 1 2 1 1 2 0 1 2 4 3 3 4 5 fl fl fl fl fl fl fl fl = 1 · fl fl fl fl fl fl 1 1 2 1 2 4 3 4 5 fl fl fl fl fl fl - 2 · fl fl fl fl fl fl 1 1 1 1 2 4 3 4 5 fl fl fl fl fl fl + 0 · fl fl fl fl fl fl 1 1 1 1 1 2 3 4 5 fl fl fl fl fl fl - 3 · fl fl fl fl fl fl 1 1 1 1 1 2 1 2 4 fl fl fl fl fl fl = (10 + 12 + 8 - 12 - 16 - 5) - 2(10 + 12 + 4 - 6 - 16 - 5) - 3(4 + 2 + 2 - 1 - 4 - 4) = - 3 - 2( - 1) - 3( - 1) = 2 . 17.10. This is just like we did in class on Wednesday. Suppose A = a 1 b 1 c 1 d 1 , B = a 2 b 2 c 2 d 2 1
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2 HOMEWORK 3 SOLUTIONS are two matrices. Then AB = a 1 a 2 + b 1 c 2 a 1 b 2 + b 1 d 2 a 2 c 1 + c 2 d 1 b 2 c 1 + d 1 d 2 . So, we have det( AB ) = fl fl fl fl a 1 a 2 + b 1 c 2 a 1 b 2 + b 1 d 2 a 2 c 1 + c 2 d 1 b 2 c 1 + d 1 d 2 fl fl fl fl = ( a 1 a 2 + b 1 c 2 )( b 2 c 1 + d 1 d 2 ) - ( a 1 b 2 + b 1 d 2 )( a 2 c 1 + c 2 d 1 ) = a 1 a 2 b 2 c 1 + a 1 a 2 d 1 d 2 + b 1 b 2 c 1 c 2 + b
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