#7 please show work. 3. Answer each of the following with the fewest number of c
ID: 1720895 • Letter: #
Question
#7 please show work. 3. Answer each of the following with the fewest number of computations (by hand) as possible. Also state which conditions from The Invertible Matrix Theorem you have used in each case. 1 0 2 5 0 -23 0 7 () Are the col7 (a) Are the cols of 0 0 7 3 9 linearly independent? (b) Is the transformation 121 11 3] IZg (c) Let A be the matrix given in (a). Does there exist a vector b such that Ax b has many solutions? Suppose that A is a square matrix. Prove that if A is invertible, then the columns of AT are linearly independent. Suppose that A is a square matrix. Prove that if the rows of A span R", then Ax- 0 has only the trivial solution. 6. The Invertible Matrix Theorem is powerless in the face of a non-square matrix! We illustrate this fact bere. (a) Give an example of a matrix A whose columns span R2 but are not linearly independent. (b) Give an example of a matrix A whose columns are linearly independent, but whose rows are linearly dependent. (c) Give an example of a matrix transformation x Ax (that is, define A) which is onto but not one-to-one. 7. In this problem, you will show that the transformation T(x1 )-(-5zt9n, 7n) is invertible, and find its inverse transformation (a) Find the standard matrix A of T (b) Explain why A is invertible using the Invertible Matrix Theorem. (c) Find A-1 (d) Write (in function notation, just as T is described above) the transformation whose standard matrix is A-1 Call it S (e) Show that both TS(zi, z2)s (zi, ra) and So T(zi )-(zi,n). What does this imply about S?Explanation / Answer
a) Find the determinant value as zero since ii row is 0
Hence linearly dependent
b) Since determinant value = 6 there is unique x1, x2 for all solutions
Hence onto
c) Yes when b is a zero vector
4) A is invertible if A is non singular
i.e. A is invertible if columns are linearly independent
5) If Ax =0 has a trivial solution then it follows x=0 and A non zero
Hence rows of A are linearly independent and span vector space
7) T = -5 9
4 -7
|T|= -1 not 0 hence invertible
S = Inverse = 7 9
4 5
SoT (x1,x2) = ToS (x1,x2) = I (x1,x2)
Since I is identity matrix we get back (x1, x2)
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