Spring 2007 - Mengi's Class - Quiz 3

# Spring 2007 - Mengi's Class - Quiz 3 - A = 0 and A is not...

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Quiz 3, Math 20F - Lecture B (Spring 2007) 1. Let u = 1 2 3 , v = 1 1 1 , A = uv T and B = 0 1 2 1 2 3 2 3 5 . a) (4.5 points) What are the rank of A and the dimension of the null space of A ? Is A invertible? Justify your answers. Solution: In general if a matrix can be written in the form uv T its column space is spanned by u . Therefore unless u = 0, the dimension of the column space is one or equivalently rank( uv T ) = 1. Speciﬁcally A = 1 1 1 2 2 2 3 3 3 , and the column space of A is spanned by 1 2 3 , that is the rank of A is one. But the rank of A is also equal to the dimension of its row space, which is equal to the number of pivot columns. On the other hand the dimension of the null space of A is equal to the number of free variables or the non-pivot columns meaning rank( A ) + dim(Null( A )) = 3 . Therefore the dimension of the null space of A is two. Since A is not of full rank and its columns are linearly dependent, det(
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Unformatted text preview: A ) = 0 and A is not invertible. b) (3 points) Calculate the determinants of B , AB and B-1 . Solution: You only need to calculate the determinant of B using a cofactor expansion. det( B ) = ± ± ± ± ± ± 0 1 2 1 2 3 2 3 5 ± ± ± ± ± ± = ± ± ± ± ± ± 1 2 1 2 3-1-1 ± ± ± ± ± ± = (1)(-1) 1+2 ± ± ± ± 1 2-1-1 ± ± ± ± =-1(-1 + 2) =-1 Above the second equality holds, because the matrix on the right of the equality is obtained from the matrix on the left by subtracting two times the second row from the third row. The other determinants can be deduced by the product rule det( A 1 A 2 ) = det( A 1 ) det( A 2 ) as follows det( AB ) = det( A ) det( B ) = (0)(-1) = 0 det( I ) = det( BB-1 ) = det( B ) det( B-1 ) ⇒ det( B-1 ) = 1 det B =-1 ....
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