Math 310: Hour Exam 1 (Solutions) Prof. S. Smith: Wed 18 October 2006

Problem 1: (a) Using Gaussian elimination, find the row echelon form

of the following augmented matrix: (A|b) = ( \begin{matrix}1 & 1 & 2 & 5 \\ 1 & 2 & 0 & 5 \\ 0 & −1 & 2 & 0 \end{matrix} ). Show the STEPS you use.

→^{A_{2}^{−1×1}} ( \begin{matrix}1 & 1 & 2 & 5 \\ 0 & 1 & −2 & 0 \\ 0 & −1 & 2 & 0 \end{matrix} ) →^{A_{3}^{1×2}} ( \begin{matrix}1 & 1 & 2 & 5 \\ 0 & 1 & −2 & 0 \\ 0 & 0 & 0 & 0 \end{matrix} ).

(b) Using your answer in (a), perform back substitution to give all SOLUTIONS (if any) of the linear equation system Ax = b determined by the augmented matrix (A|b).

Notice column 3 has no pivot, so that variable x_{3} is free; call it “α”.

Row 2 says x_{2} − 2x_{3} = 0, so x_{2} = 2x_{3} = 2α.

Row 1 says x_{1} + x_{2} + 2x_{3} = 5, so x_{1} = 5 − 2α − 2α = 5 − 4α.

So infinite number of solutions (5 − 4α, 2α, α)^{T} = (5, 0, 0)^{T} + α(−4, 2, 1)^{T} for all real α.

Problem 2: (a) Find the LU-decomposition (show STEPS) of the matrix A = ( \begin{matrix}1 & 0 & 2 \\ −2 & 4 & −4 \\ 0 & 4 & 6 \end{matrix} ).

We apply A_{2}^{2×1} to obtain ( \begin{matrix}1 & 0 & 2 \\ 0 & 4 & 0 \\ 0 & 4 & 6 \end{matrix} ) and then A_{3}^{−1×2} to obtain ( \begin{matrix}1 & 0 & 2 \\ 0 & 4 & 0 \\ 0 & 0 & 6 \end{matrix} ) as U.

The product of the matrices for the inverses, namely A_{2}^{−2×1} and A_{3}^{1×2}, gives L = ( \begin{matrix}1 & 0 & 0 \\ −2 & 1 & 0 \\ 0 & 1 & 1 \end{matrix} ).

(b) Find the inverse (any method, but show STEPS) of the matrix A = ( \begin{matrix}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix} ).

Probably quickest is via chapter 1, row-reduce the “big” augmented matrix (A|I) to (I|A^{−1}):

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

So A^{−1} = ( \begin{matrix}1 & −1 & 0 \\ 0 & 1 & −1 \\ 0 & 0 & 1 \end{matrix} ) .

Problem 3: (a) Let S be the subSET of R^{3} consisting of all 3-vectors ( \begin{matrix}x_{1} \\ x_{2} \\ x_{3} \end{matrix} ) satisfying the condition x_{3} = x_{1} + x_{2}. Show that S is a subSPACE of R^{3}.

A general vector in S has the form ( \begin{matrix}a \\ b \\ a + b \end{matrix} ). (closure, +) Take two general vectors in S: ( \begin{matrix}a \\ b \\ a + b \end{matrix} ), ( \begin{matrix}c \\ d \\ c + d \end{matrix} ), and add;

their sum is ( \begin{matrix}a + c \\ b + d \\ (a + b) + (c + d) \end{matrix} ) = ( \begin{matrix}a + c \\ b + d \\ (a + c) + (b + d) \end{matrix} ) which is also in S.

(closure, sc.mult.) For a general scalar c, and general vector ( \begin{matrix}a \\ b \\ a + b \end{matrix} ) in S, the scalar multiple is ( \begin{matrix}ca \\ cb \\ c(a + b) \end{matrix} ) = ( \begin{matrix}ca \\ cb \\ ca + cb \end{matrix} ), which is also in S.

(b) What is the rank of the matrix A = ( \begin{matrix}1 & 0 & 1 \\ 2 & 1 & 3 \\ 0 & 2 & 2 \end{matrix} )? Give a basis for the span of the columns of A (column space).

The rank is 2; for example the row-reduced echelon form of A is ( \begin{matrix}1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{matrix} ) with 2 pivots. Just need two LI columns. In this example, in fact any pair of columns from A will work.

Problem 4: (a) Are the vectors

v_{1} = ( \begin{matrix}1 \\ 2 \\ 2 \end{matrix} ), v_{2} = ( \begin{matrix}2 \\ 5 \\ 4 \end{matrix} ), and v_{3} = ( \begin{matrix}1 \\ 3 \\ 2 \end{matrix} ) linearly independent in R^{3} ? (Why/why not?)

NO: (short) Notice v_{3} = v_{2} − v_{1}.

(longer) Set a linear combination equal to zero, get system: ( \begin{matrix}1 & 2 & 1 & 0 \\ 2 & 5 & 3 & 0 \\ 2 & 4 & 2 & 0 \end{matrix} )

Row-reduce to ( \begin{matrix}1 & 0 & −1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{matrix} ); no pivot in 3rd column, so infinite solutions, hence lin.dep.

(b) Inside the space R^{2×2} of 2 × 2 matrices,

let S denote the subspace consisting of symmetric matrices, which have the general form ( \begin{matrix}a & b \\ b & c \end{matrix} ).

Do ( \begin{matrix}1 & 0 \\ 0 & 0 \end{matrix} ), ( \begin{matrix}1 & 1 \\ 1 & 0 \end{matrix} ), and ( \begin{matrix}1 & 1 \\ 1 & 1 \end{matrix} ) give a spanning set for S? (Why/why not)

YES: Set an unknown linear combination equal to the general vector:

c_{1} ( \begin{matrix}1 & 0 \\ 0 & 0 \end{matrix} ) + c_{2} ( \begin{matrix}1 & 1 \\ 1 & 0 \end{matrix} ) + c_{3} ( \begin{matrix}1 & 1 \\ 1 & 1 \end{matrix} ) = ( \begin{matrix}a & b \\ b & c \end{matrix} ), so ( \begin{matrix}c_{1} + c_{2} + c_{3} & c_{2} + c_{3} \\ c_{2} + c_{3} & c_{3} \end{matrix} ) = ( \begin{matrix}a & b \\ b & c \end{matrix} ).

We obtain an equation from each of the four matrix positions:

c_{1} + c_{2} + c_{3} = a
c_{2} + c_{3} = b (twice, from (1, 2) and (2, 1) positions)
c_{3} = c

And we CAN solve. Just quickly by hand: c_{3} = c; c_{2} = b − c; and c_{1} = a − (b − c) − c = a − b.

Or by the usual full method: ( \begin{matrix}1 & 1 & 1 & a \\ 0 & 1 & 1 & b \\ 0 & 0 & 1 & c \end{matrix} ) has row-reduced echelon form ( \begin{matrix}1 & 0 & 0 & a − b \\ 0 & 1 & 0 & b − c \\ 0 & 0 & 1 & c \end{matrix} ).

Problem 5: (a) Recall that the space P_{3} consists of polynomials of degree less than 3. Find a basis for the subspace S of P_{3} given by polynomials with constant term equal to zero. What is the dimension of S ? (Say WHY your choice is a basis.)

A general polynomial with zero constant term has form ax + bx^{2}. So choosing a, b successively in the “standard basis” way, we get 2 “vectors” in a basis: a = 1, b = 0 gives x; and b = 1, a = 0 gives x^{2}. Now, why basis? (spanning set:) the lin.comb. with coefficients a, b gives the general ax + bx^{2} above. (linearly independent:) Setting that lin.comb., namely ax + bx^{2}, equal to 0 forces a = b = 0.

(b) Find the change-of-basis matrix (transition matrix)

from old basis given by the standard basis ( \begin{matrix}1 \\ 0 \end{matrix} ) , ( \begin{matrix}0 \\ 1 \end{matrix} ) to the new basis given by ( \begin{matrix}2 \\ 1 \end{matrix} ) , ( \begin{matrix}3 \\ 1 \end{matrix} ).

(You need to find coordinates of the old basis in terms of the new basis.)

Directly: just solve ( \begin{matrix}2 & 3 & 1 \\ 1 & 1 & 0 \end{matrix} ) and ( \begin{matrix}2 & 3 & 0 \\ 1 & 1 & 1 \end{matrix} ) to get the two columns of ( \begin{matrix}−1 & 3 \\ 1 & −2 \end{matrix} ).

Or: invert the transition matrix in the other (“easy”) direction: ( \begin{matrix}2 & 3 \\ 1 & 1 \end{matrix} )^{−1} = ( \begin{matrix}−1 & 3 \\ 1 & −2 \end{matrix} ).