Math 310: Hour Exam 1 (Solutions)

Prof. S. Smith: Fall 1995

Problem 1: (a) Find the row-reduced echelon form of

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

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

(b) What are the solutions of the system Ax = 0 ? (Check!)

Third variable is free, so solutions x_{3}(1, −2, 1)^{T}. Problem 2: Give the LU-decomposition of

A = ( \begin{matrix}1 & 1 \\ 1 & 3 \end{matrix} );

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

Get U from →^{A_{2}^{−1×1}} ( \begin{matrix}1 & 1 \\ 0 & 2 \end{matrix} ). So L from inverse operation A_{2}^{+1×1} = ( \begin{matrix}1 & 0 \\ 1 & 1 \end{matrix} )

Problem 3: Use Cramer’s rule (determinants) to solve Ax = b given by

(A|b) = ( \begin{matrix}1 & 1 & 5 \\ 1 & 2 & 7 \end{matrix} ) .

First det ( \begin{matrix}1 & 1 \\ 1 & 2 \end{matrix} ) = 1, so x_{1} = (\frac{1}{1}) det ( \begin{matrix}5 & 1 \\ 7 & 2 \end{matrix} ) = 3 and x_{2} = det ( \begin{matrix}1 & 5 \\ 1 & 7 \end{matrix} ) = 2.

Problem 4: (a) Find the inverse (by any method) of

A = ( \begin{matrix}1 & 2 \\ 3 & 5 \end{matrix} ) .

Quick via adjoint: det ( \begin{matrix}1 & 2 \\ 3 & 5 \end{matrix} ) = −1, so inverse is − ( \begin{matrix}5 & −2 \\ −3 & 1 \end{matrix} ) = ( \begin{matrix}−5 & 2 \\ 3 & −1 \end{matrix} ).

(b) Use the above to express the solutions of Ax = b in terms of the constants b_{1} and b_{2}. By A^{−1}b, namely x_{1} = −5b_{1} + 2b_{2} and x_{2} = 3b_{1} − b_{2}.

Problem 5: (a) Is (1, 2, 3) in the span of (4, 0, 5) and (6, 0, 7) ?

No—for example, any linear combination of the latter two vectors has 0 in second entry.

(b) Let V be the space of all functions (with at least 2 derivatives). Let W be the subSET of all functions f which are solutions of the differential equation

f′′ + 5f = 0.

Show that the solution set W is a subSPACE of V.

Take f, g ∈ W, and scalar c: Then we have f′′ + 5f = 0 = g′′ + 5g.

Is f + g ∈ W? (f + g)′′ + 5(f + g) = (f′′ + 5f) + (g′′ + 5g) = 0 + 0 = 0, so OK.

Is cf ∈ W? (cf)′′ + 5(cf) = c(f′′ + 5f) = c(0) = 0, also OK.