Math 310: PRACTICE Hour Exam 1 (Solutions)

Prof. S. Smith: (Fall 94: orig. given 27 Sept 1993)

Problem 1: Find the general solution of ( \begin{matrix}1 & 3 & 3 \\ 2 & 6 & 9 \\ −1 & −3 & 3 \end{matrix} ) ( \begin{matrix}x_{1} \\ x_{2} \\ x_{3} \end{matrix} ) = ( \begin{matrix}1 \\ 5 \\ 5 \end{matrix} ) .

→^{A_{2}^{−2×1},A_{3}^{1×1}} ( \begin{matrix}1 & 3 & 3 & 1 \\ 0 & 0 & 3 & 3 \\ 0 & 0 & 6 & 6 \end{matrix} ) →^{M_{\frac{1}{3} ×2}} ( \begin{matrix}1 & 3 & 3 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 6 & 6 \end{matrix} ) →^{A_{1}^{−3×2},A_{3}^{−6×2}} ( \begin{matrix}1 & 3 & 0 & −2 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{matrix} )

Thus x_{2} is free variable; with x_{2} = 1 in Ax = 0, special solution is (−3 1 0).

With x_{2} = 0 in Ax = b, particular solution is (−2 0 1). So general: (−2 0 1) + a(−3 1 0). Problem 2:

(a) Is A = ( \begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 2 & 3 & 1 \end{matrix} ) invertible? If so, find A^{−1}.

(b) Give the LDU decomposition of ( \begin{matrix}2 & 1 & 0 \\ 0 & 4 & 2 \\ 6 & 3 & 5 \end{matrix} ).

(a) This is product of elementary matrices A_{3}^{2×1}A_{3}^{3×2}, so inverse is

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

(b) →^{A_{3}^{−3×1}} ( \begin{matrix}2 & 1 & 0 \\ 0 & 4 & 2 \\ 0 & 0 & 5 \end{matrix} ), so D = ( \begin{matrix}2 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 5 \end{matrix} ) and U = ( \begin{matrix}1 & \frac{1}{2} & 0 \\ 0 & 1 & \frac{1}{2} \\ 0 & 0 & 1 \end{matrix} ) .

And L = (A_{3}^{−3×1})^{−1} = A_{3}^{3×1} = ( \begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 3 & 0 & 1 \end{matrix} )

Problem 3: Let V be the space of real-valued functions of x. Show the solution set S of f′(x) = xf(x) is a subspace of V.

If f, g ∈ S then f′ = xf, g′ = xg.

So (f + g)′ = f′ + g′ = xf + xg = x(f + g); and also f + g ∈ S. Similarly if f ∈ S and c is scalar, (cf)′ = c(f′) = c(xf) = x(cf) so that also cf ∈ S.

Problem 4:

(a) For A = ( \begin{matrix}1 & 1 & 0 & 2 \\ 1 & 3 & 1 & 2 \\ 3 & 1 & −1 & 6 \end{matrix} ) find a basis for the row space and the column space.

Is Ax = b solvable for all b?

(b) Let V be the space of polynomials in x of degree at most 3. Give a basis for the subspace W of such polynomials with f(2) = 0.

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

Then A_{3}^{1×2} converts third row to 0, so (1 1 0 2) and (0 2 1 0) give basis.

(columns) Work on A^{T} = ( \begin{matrix}1 & 1 & 3 \\ 1 & 3 & 1 \\ 0 & 1 & −1 \\ 2 & 2 & 6 \end{matrix} ) →^{A_{2}^{−1×1},A_{4}^{−2×1}} ( \begin{matrix}1 & 1 & 3 \\ 0 & 2 & −2 \\ 0 & 1 & −1 \\ 0 & 0 & 0 \end{matrix} ).

This time A_{3}^{− \frac{1}{2} ×2} kills third row; so basis is (1 1 3)^{T} and (0 2 − 2)^{T}. And Ax = b solvable only for b in this column space—here, NOT all b.

(b) Easiest to note that V is 4-dimensional with basis {1, x, x^{2}, x^{3}}, and W is 3-dimensional subspace with x − 2 dividing f(x); so a nice basis is {x − 2, x(x − 2), x^{2}(x − 2)}, that is, {x − 2, x^{2} − 2x, x^{3} − 2x^{2}}