Math 310: Hour Exam 1 (Solutions) Prof. S. Smith: Mon 27 Sept 1999

Problem 1: (a) Using Gauss-Jordan, find the row-reduced echelon form of the following aug-

mented matrix: (A|b) = ( \begin{matrix}1 & 2 & 1 & 0 \\ 0 & 2 & 2 & −2 \\ 2 & 6 & 4 & −2 \end{matrix} ).

→^{A_{3}^{−2×1}} ( \begin{matrix}1 & 2 & 1 & 0 \\ 0 & 2 & 2 & −2 \\ 0 & 2 & 2 & −2 \end{matrix} ) →^{M_{\frac{1}{2} ×1}} ( \begin{matrix}1 & 2 & 1 & 0 \\ 0 & 1 & 1 & −1 \\ 0 & 2 & 2 & −2 \end{matrix} ) →^{A_{1}^{−2×2},A_{3}^{−2×2}} ( \begin{matrix}1 & 0 & −1 & 2 \\ 0 & 1 & 1 & −1 \\ 0 & 0 & 0 & 0 \end{matrix} )

(b) Then give the solutions of the corresponding linear system Ax = b.

So solutions are (2 + r, −1 − r, r) for free variable x_{3} = r.

Problem 2: (a) Using Gaussian elimination (best to use only “add” operations), determine an

upper-triangular form U (row-echelon, not reduced) for A = ( \begin{matrix}1 & 1 & 2 \\ 2 & 4 & 2 \\ 3 & 7 & 10 \end{matrix} ).

U = ( \begin{matrix}1 & 1 & 2 \\ 0 & 2 & −2 \\ 0 & 0 & 8 \end{matrix} )

(b) Indicate the ROW OPERATIONS you used to transform A to U.

A_{2}^{−2×1}, A_{3}^{−3×1}, A_{3}^{−2×2} corrected last, 3/2/05

(c) What ELEMENTARY ROW MATRICES E_{i} will, by left muliplication, perform the same operations? (that is, E_{3}E_{2}E_{1}A = U)

E_{1} = ( \begin{matrix}1 & 0 & 0 \\ −2 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} ), E_{2} = ( \begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ −3 & 0 & 1 \end{matrix} ), E_{1} = ( \begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & −2 & 1 \end{matrix} )

(d) Give the DETERMINANT of each matrix in (c), and det(U). How is det(U) related to det(A)? det(U) = 16, and each det(E_{i}) = 1; so as E_{3}E_{2}E_{1}A = U, 1.1.1. det(A) = det(U) = 16.

Problem 3: (a) Find the inverse (any method) of: A = ( \begin{matrix}1 & 2 \\ 3 & 5 \end{matrix} ).

Using “adjoint” method: det(A) = −1, cofactor matrix is ( \begin{matrix}5 & −3 \\ −2 & 1 \end{matrix} ); so inverse ( \begin{matrix}−5 & 2 \\ 3 & −1 \end{matrix} ).

(b) Suppose a country has two political parties D and R, and that the “Markov” matrix A =

( \begin{matrix}1 & \frac{1}{2} \\ 0 & \frac{1}{2} \end{matrix} ) gives the probability of transition among the parties in one year; that is, if v_{0} = (D_{0}, R_{0})^{T}

is the original distribution, then one year later the distribution is Av_{0}.

If the original distribution is v_{0} = (\frac{1}{2}, \frac{1}{2})^{T}, what is the distribution after 2 years?

Either compute A(Av_{0}); or A^{2}v_{0} with A^{2} = ( \begin{matrix}1 & \frac{3}{4} \\ 0 & \frac{1}{4} \end{matrix} ). So final distribution is (\frac{7}{8}, \frac{1}{8})^{T}.

Problem 4: (a) Find the determinant (show work) of A = ( \begin{matrix}1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 3 & 9 \end{matrix} ).

Top row: 1(2.9 − 3.4) − 1(1.9 − 1.4) + 1(1.3 − 1.2) = 6 − 5 + 1 = 2.

(b) Use Cramer’s rule to solve ( \begin{matrix}1 & 1 & 3 \\ 1 & 3 & 6 \end{matrix} ).

det(A) = 1.3 − 1.1 = 2, so x_{1} = \frac{1}{2}(3.3 − 6.1) = \frac{3}{2} and x_{2} = \frac{1}{2}(1.6 − 1.3) = \frac{3}{2}. Problem 5: (a) Determine the nullspace of the matrix A = ( \begin{matrix}1 & 0 & −3 \\ 0 & 2 & 2 \end{matrix} ).

Just solutions of Ax = 0, namely (3r, −r, r)^{t}—that is, multiples of (3, −1, 1)^{T}.

(b) Let V be the space of 3×3 matrices, and W the subSET of “skew-symmetric” matrices: These satisfy A^{T} = −A, and so they have general form ( \begin{matrix}0 & a & b \\ −a & 0 & c \\ −b & −c & 0 \end{matrix} ). Show W is a subSPACE of V.

(addition:) Take A and B in W, compute A + B:

( \begin{matrix}0 & a & b \\ −a & 0 & c \\ −b & −e & 0 \end{matrix} ) + ( \begin{matrix}0 & d & e \\ −d & 0 & f \\ −e & −f & 0 \end{matrix} ) = ( \begin{matrix}0 & a + d & b + e \\ −a − d & 0 & c + f \\ −b − e & −c − f & 0 \end{matrix} ).

The result A + B is also skew-symmetric, so A + B also lies in W. (scalar multiplication:) For scalar r and A ∈ W, compute rA ∈ W:

r ( \begin{matrix}0 & a & b \\ −a & 0 & c \\ −b & −e & 0 \end{matrix} ) = ( \begin{matrix}0 & ra & rb \\ −ra & 0 & rc \\ −rb & −rc & 0 \end{matrix} ).

The result rA is also skew-symmetric, so rA also lies in W. We see “yes”, W is a subspace of V.