[2] (8 points) Let B and C be 4 x 4 matrices such that det B = 2 and det C = -5, and let A be the matrix

[ 0 4 5 1 ] (a) Determine det (3(B^{-3}C^{2})^{T}). [ 1 0 7 1 ] [ 1 0 2 1 ] [ 0 1 3 0 ]. (b) Use Cramer's rule to find x_{2}in the equation Ax=b, whereb^{T}= ( 5, 7, 2, 3). (This can be done in your head, but show how you obtain your answer.) (c) Why can't you always use Cramer's rule to solve linear equations? Give an example of a linear system of equations for which you can't use Cramer's rule, and explain why Cramer's rule doesn't work for your example.

- (a) det (3(B
^{-3}C^{2})^{T}) = det (3(B^{-3}C^{2})) = 3^{4}(det B)^{-3}(det C)^{2}= 81 x (1/8) x 25 = 2025/8. - (b) Cramer's rule tells us x
_{2}= (det E)/(det A), where E is the matrix[ 0 5 5 1 ] [ 1 7 7 1 ] [ 1 2 2 1 ] [ 0 3 3 0 ].

Since two of the columns of E are the same, we know that E is not invertible, so det E = 0. Thus x_{2}= 0. - (c) Cramer's rule applies only to systems of linear equations whose coefficient matrix is invertible. Thus it can't be used to solve 3x + 4y = 7, for example.

[4] (9 points) For this problem, you may assume that B is the reduced row echelon form of A, where A and B are the following matrices:

[ 3 1 -2 0 1 2 1 ] [ 1 0 -1 0 0 -2 -3 ] (a) Find a basis for the [ 1 1 0 -1 1 2 2 ] [ 0 1 1 0 0 2 3 ] column space of A. [ 3 2 -1 1 1 8 9 ] [ 0 0 0 1 0 4 5 ] [ 0 2 2 -1 1 6 8 ] [ 0 0 0 0 1 6 7 ] (b) Find a basis for the [ 0 3 3 3 -3 0 3 ] [ 0 0 0 0 0 0 0 ] null space of A.

- (a) From B we see that columns 1, 2, 4 and 5 of A are the pivot columns of A, thus columns 1, 2, 4 and 5 of A give a basis for Col A.
- (b) The parametric vector form of the solution to A
**x**=**0**is[ x

So the vectors [1, -1, 1, 0, 0, 0, 0]_{1}]_{ }[ 1]_{ }[ 2]_{ }[ 3] [ x_{2}]_{ }[-1]_{ }[-2]_{ }[-3] [ x_{3}]_{ }[ 1]_{ }[ 0]_{ }[ 0] [ x_{4}] = x_{3}[ 0] + x_{6}[-4] + x_{7}[-5] . [ x_{5}]_{ }[ 0]_{ }[-6]_{ }[-7] [ x_{6}]_{ }[ 0]_{ }[ 1]_{ }[ 0] [ x_{7}]_{ }[ 0]_{ }[ 0]_{ }[ 1]^{T}, [2, -2, 0, -4, -6, 1, 0]^{T}, and [3, -3, 0, -5, -7, 0, 1]^{T}give a basis of Nul A.

[6] (6 points): Say A is a 5x7 matrix with dim Nul(A)=4. Find the dimension of the row space of A and of the null space of A

ANSWER: Since dim Nul(A)=4, we know rank(A) = 7 - 4 = 3. But rank(A) = dim Row A, so dim Row A = 3. But rank(A

[7] (6 points): Let A be the matrix

[ 1 -4 3 ] [ 4 -4 0 ] [ 2 -4 2 ].In each case determine if the given vector is an eigenvector of A and if so find its eigenvalue.

- (a)
**v**= [1,1,1]^{T}

ANSWER: A**v**= 0**v**, so**v**is an eigenvector of eigenvalue 0. - (b)
**v**= [1,0,1]^{T}

ANSWER: A**v**= 4**v**, so**v**is an eigenvector of eigenvalue 4. - (c)
**v**= [0,0,0]^{T}

ANSWER: The zero vector is never an eigenvector, so**v**cannot be an eigenvector.

[ 1 1 1 ] [ 0 2 1 ] [ 0 0 3 ].This is a diagonalizable matrix; thus P

- (a) Find each eigenvalue l of A,
and a basis for each eigenspace E
_{l}.

ANSWER: Since A is triangular, the eigenvalues are on the diagonal: 1, 2, and 3. Since we have three different eigenvalues for a 3x3 matrix, we know each eigenspace will be 1-dimensional, so now we just need to find a single eigenvector for each eigenvalue. One way to do this is to solve (A - lI)**x**=**0**, for each eigenvalue l. It is also sometimes possible to find them by inspection. In any case, [1, 0, 0]^{T}works for l = 1, [1, 1, 0]^{T}works for l = 2, and [1, 1, 1]^{T}works for l = 3. - (b) Find P and D.

ANSWER: D is the diagonal matrix whose diagonal entries are 1, 2 and 3, and P is the matrix whose columns are the corresponding eigenvectors, [1, 0, 0]^{T}, [1, 1, 0]^{T}and [1, 1, 1]^{T}.

(a) The 5x5 identity matrix. (b) [ 1 1 0 0 0 ] ANSWER: This is not diagonalizable, ANSWER: This is diagonal, [ 0 1 1 0 0 ] since the 0 eigenspace has algebraic hence diagonalizable. [ 0 0 1 0 0 ] multiplicity 2 but dimension 1. [ 0 0 0 0 1 ] (Also, the 1 eigenspace has algebraic [ 0 0 0 0 0 ] multiplicity 3 but dimension 1.) (c) A 5x5 matrix A with eigenvalues (d) A matrix of the form PDP[10] (8 points): Let A be the matrix^{-1}, where 0, 2, and 3, and such that the P is invertible and D is diagonal. dimension of the null space is 2 ANSWER: The matrix PDP^{-1}is diagonalizable for both A and A-3I. by definition. ANSWER: Since the algebraic multiplicity and dimension of each eigenspace are equal, we know this matrix is diagonalizable.

[ 0 -1 ] [ 2 3 ].

- (a) Find the eigenvalues of A.

ANSWER: From the characteristic equation we find the eigenvalues to be 1 and 2. - (b) Find a basis for the eigenspace corresponding
to each eigenvalue of A.

ANSWER: A basis for the 1 eigenspace is given by [1, -1]^{T}and for the 2 eigenspace a basis is given by [1, -2]^{T}. - (c) Find A
^{10}**v**, where**v**= [1, 0]^{T}.

ANSWER: Since [1, 0]^{T}= 2[1, -1]^{T}- [1, -2]^{T}, A^{10}[1, 0]^{T}= 2(1^{10})[1, -1]^{T}- 2^{10}[1, -2]^{T}= [ -1022, 2046]^{T}.