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# F10HW5 - F2010 M341 Homework 5 Section 2.4 2.b 3 2 1(number...

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F2010- M341- Homework 5 Section 2.4. 2.b) 3 - 1 4 2 - 2 1 - 1 3 2 1 - 3 - 2 0 4 5 0 8 10 1 - 3 - 2 0 4 5 0 0 0 . The rank of this matrix is 2 (number of non all zero rows). The matrix is singular (not invertible). 3.b) The matrix ( 10 - 5 - 4 2 ) is singular ( ad - bc = 10 × 2 - ( - 5)( - 4) = 0). Other (better) method: By the row operation < 2 > < 2 > +4 / 10 < 1 > we have: ( 10 - 5 - 4 2 ) ( 10 - 5 0 0 ) . So the rank is only 1 which means that the matrix is singular. 4.b) 5 7 - 6 | 1 0 0 - 3 1 - 2 | 0 1 0 1 - 5 2 | 0 0 1 1 7 / 5 - 6 / 5 | 1 / 5 0 0 - 3 1 - 2 | 0 1 0 1 - 5 2 | 0 0 1 1 7 / 5 - 6 / 5 | 1 / 5 0 0 0 - 16 / 5 - 8 / 5 | - 3 / 5 1 0 0 - 32 / 5 16 / 5 | - 1 / 5 0 1 1 7 / 5 - 6 / 5 | 1 / 5 0 0 0 1 - 1 / 2 | 3 / 16 - 5 / 16 0 0 - 32 / 5 16 / 5 | - 1 / 5 0 1 1 7 / 5 - 6 / 5 | 1 / 5 0 0 0 1 - 1 / 2 | 3 / 16 - 5 / 16 0 0 0 0 | 1 - 2 1 Unless I did a mistake (always possible with these heavy computations, the rank is 2, less than 3 and the matrix is singular. 7 b) the matrix of coeﬃcients of the system is A = - 5 3 6 3 - 1 - 7 - 2 1 2 . By row reducing [

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F10HW5 - F2010 M341 Homework 5 Section 2.4 2.b 3 2 1(number...

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