Math 310: Hour Exam 1 (Solutions) Prof. S. Smith: Fri 19 Feb 1999

Problem 1:

(a) Using Gauss-Jordan, find the row-reduced echelon form of the following augmented matrix. (INDICATE the row operations you use).

(A|b) = ( \begin{matrix}1 & 4 & 7 & 10 \\ 2 & 5 & 8 & 11 \\ 3 & 6 & 9 & 12 \end{matrix} ) .

→^{A_{2}^{−2×1},A_{3}^{−3×1}} ( \begin{matrix}1 & 4 & 7 & 10 \\ 0 & −3 & −6 & −9 \\ 0 & −6 & −12 & −18 \end{matrix} ) →^{M^{− \frac{1}{3} ×2}} ( \begin{matrix}1 & 4 & 7 & 10 \\ 0 & 1 & 2 & 3 \\ 0 & −6 & −12 & −18 \end{matrix} ) →^{A_{1}^{−4×2},A_{3}^{6×2}} ( \begin{matrix}1 & 0 & −1 & −2 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 0 \end{matrix} )

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

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

Problem 2: (a) Given matrices

A = ( \begin{matrix}2 & 4 \\ 6 & 8 \end{matrix} ) and B = ( \begin{matrix}2 & 4 \\ 0 & −4 \end{matrix} ) ,

find an elementary 2 × 2 row matrix E such that left multiplication by E will convert A to B: that is, EA = B.

row operation A_{2}^{−3×1} : add −2 times 1st row to 2nd.

So use E = ( \begin{matrix}1 & 0 \\ −3 & 1 \end{matrix} ) : (obtained by doing that operation to the identity matrix).

(b) Give the LU-decomposition of

A = ( \begin{matrix}2 & 4 \\ 6 & 8 \end{matrix} );

that is, find lower-triangular L and upper-triangular U, so that A = LU.

As in (a) A_{2}^{−3×1} to get U = ( \begin{matrix}2 & 4 \\ 0 & −4 \end{matrix} ). So L from inverse operation A_{2}^{+3×1} is ( \begin{matrix}1 & 0 \\ 3 & 1 \end{matrix} )

Problem 3: (a) Suppose a population is initially evenly distributed between two states, as represented by the column vector v = (.5, .5)^{T}. Assume the distribution vector after one time unit

is given by Av, where A is the transition matrix ( \begin{matrix}1 & .5 \\ 0 & .5 \end{matrix} ). Compute A^{2}, and give the distribution

vector after 2 time units.

A^{2} = ( \begin{matrix}1 & .5 \\ 0 & .5 \end{matrix} ) ( \begin{matrix}1 & .5 \\ 0 & .5 \end{matrix} ) = ( \begin{matrix}1 & .75 \\ 0 & .25 \end{matrix} ), so A^{2}v = A(Av) = ( \begin{matrix}1 & .75 \\ 0 & .25 \end{matrix} ) ( \begin{matrix}.5 \\ .5 \end{matrix} ) = ( \begin{matrix}.875 \\ .125 \end{matrix} ).

(b) Find the determinant of

A = ( \begin{matrix}1 & 5 & 0 \\ 2 & 4 & −1 \\ 1 & 0 & 1 \end{matrix} ) .

Via top row: 1(4.1 − 0.1) − 5(2.1 − 1.(−1)) + 0 = 1(4) − 5(3) = −11.

Problem 4:

(a) Find the inverse (any method) of:

A = ( \begin{matrix}3 & −1 & −2 \\ 0 & 1 & 3 \\ 0 & 0 & 2 \end{matrix} ) .

Using classcal-adjoint method:

det(A) = 6 (triangular) and matrix of cofactors is ( \begin{matrix}2 & 0 & 0 \\ 2 & 6 & 0 \\ −1 & −9 & 3 \end{matrix} ) so A^{−1} = \frac{1}{6} ( \begin{matrix}2 & 2 & −1 \\ 0 & 6 & −9 \\ 0 & 0 & 3 \end{matrix} )

(b) Given matrices A = ( \begin{matrix}2 & 1 \\ 5 & 3 \end{matrix} ) and Y = ( \begin{matrix}1 & 2 \\ 3 & 4 \end{matrix} ) and B = ( \begin{matrix}3 & 4 \\ 5 & 6 \end{matrix} ), and also A^{−1} =

( \begin{matrix}3 & −1 \\ −5 & 2 \end{matrix} ); solve the matrix equation Y = AX + B—that is, find the 2 × 2 matrix X.

We see X = A^{−1}(Y − B) = ( \begin{matrix}3 & −1 \\ −5 & 2 \end{matrix} ) ( \begin{matrix}−2 & −2 \\ −2 & −2 \end{matrix} ) = ( \begin{matrix}−4 & −4 \\ 6 & 6 \end{matrix} )

Problem 5: (a) Is (7, 4, −2) in the span of (1, −2, −5) and (2, 5, 6)? Either give coefficients in a linear combination, or explain why it is not possible.

(No): Set up augmented matrix (A|b) = ( \begin{matrix}1 & 2 & 7 \\ −2 & 5 & 4 \\ −5 & 6 & −2 \end{matrix} ) . Row operations A_{2}^{2×1} and A_{3}^{5×1} give

( \begin{matrix}1 & 2 & 7 \\ 0 & 9 & 18 \\ 0 & 16 & 33 \end{matrix} ) . Now M_{\frac{1}{9} ×2} gives ( \begin{matrix}1 & 2 & 7 \\ 0 & 1 & 2 \\ 0 & 16 & 33 \end{matrix} ) . And then A_{3}^{−16×2} gives ( \begin{matrix}1 & 2 & 7 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{matrix} ) .

So it is not possible to find coefficients in a linear combination, and then the indicated vector is not in the span. (There are other ways of solving this problem...)

(b) Let P_{4} be the space of polynomials of degree less than 4, and W the subSET of polynomials of form a + bx + (a + b)x^{2} (for all real numbers a, b). Show that W is a subSPACE of P_{4}.

Shortcut: notice W is just the span of 1 + x^{2} and x + x^{2}, and the span of a set of vectors is always a subspace.

For the full method: add general vectors a + bx + (a + b)x^{2} and c + dx + (c + d)x^{2} from W to get (a + c) + (b + d)x + (a + b + c + d)x^{2}. We see the coefficient for x^{2} is the sum of those for 1 and x, so the sum is also in W.

Similarly a scalar multiple f(a+bx+(a+b)x^{2} = fa+(fb)x+(fa+fb)x^{2}. Again the coefficient for x^{2} is the sum of those for 1 and x, so the scalar multiple is also in W.

We have closure under both operations, so W is a subspace.