Math 310: Hour Exam 1 (Solutions) Prof. S. Smith: Wed 2 March 2005

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

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

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

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

Notice columns 3 has no pivot, so that variable is free.

Then infinite number of solutions: (2 − α, −α, α)^{T} for all real α.

Problem 2: (a) Find the LU-decomposition (show STEPS) of the matrix A = ( \begin{matrix}2 & 3 \\ 4 & 11 \end{matrix} ).

To row-reduce (Gaussian elimination) we apply A_{2}^{−2×1} to obtain ( \begin{matrix}2 & 3 \\ 0 & 5 \end{matrix} ) as U,

and take as L the inverse of the matrix for A_{2}^{−2×1}, namely ( \begin{matrix}1 & 0 \\ 2 & 1 \end{matrix} )

(b) Use the row-operations method (show STEPS) to find the inverse of the matrix A = ( \begin{matrix}1 & 2 \\ 3 & 5 \end{matrix} ).

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

so A^{−1} = ( \begin{matrix}−5 & 2 \\ 3 & −1 \end{matrix} )

Problem 3: (a) Use Cramer’s rule to solve the system with augmented matrix (A|b) = ( \begin{matrix}2 & 1 & 4 \\ 3 & 1 & 5 \end{matrix} ).

det A = −1, so x_{1} = det ( \begin{matrix}4 & 1 \\ 5 & 1 \end{matrix} ) / − 1 = 1 and x_{2} = det ( \begin{matrix}2 & 4 \\ 3 & 5 \end{matrix} ) / − 1 = 2.

(b) Let S be the subSET of R^{3} consisting of all 3-vectors (x_{1}, x_{2}, x_{3})^{T} which satisfy the condition x_{2} = 2x_{3}. Show that S is a subSPACE of R^{3}.

Vectors in S have the general form form (a, 2b, b)^{T}.

(closure, +) Take two general vectors in S: (a, 2b, b)^{T}, (c, 2d, d)^{T} and add; their sum is (a + c, 2b + 2d, b + d)^{T} = (a + c, 2(b + d), b + d)^{T}, which is also in S. (closure, sc.mult.) For a general scalar c, and general vector (a, 2b, b)^{T} in S, the scalar multiple is (ac, (2b)c, bc)^{T} = (ac, 2(bc), bc)^{T}, which is also in S.

Problem 4: (a) Do the vectors v_{1} = (1, 0, 1)^{T}, v_{2} = (0, 2, 0)^{T}, v_{3} = (3, −2, 3)^{T} span R^{3} ? (Why/why not?)

NO: (short) Notice v_{3} = 3v_{1} − 2v_{2}; so their span is 2-dimensional, but R^{3} is 3-dimensional.

(longer) Put the v_{i} as columns of a matrix augmented by a general vector (a, b, c)^{T}. ( \begin{matrix}1 & 0 & 3 & a \\ 0 & 2 & −2 & b \\ 1 & 0 & 3 & c \end{matrix} )

Row-reduce to ( \begin{matrix}1 & 0 & 3 & a \\ 0 & 1 & −1 & \frac{b}{2} \\ 0 & 0 & 0 & c − a \end{matrix} ) and see NOT solvable for every a, b, c (only for c = a)

(b) Recall that P_{3} is the space of polynomials of degree less than three (i.e., quadratic polynomials). Are the three “vectors” 1, 1 + x, 1 + x + x^{2} linearly independent in this space ? (Why/why not ?)

YES: Set an unknown linear combination equal to 0: a(1) + b(1 + x) + c(1 + x + x^{2}) = 0 = 0.1 + 0.x + 0.x^{2}. Get an equation for each power of x: (1:) a + b + c = 0 (x:) b + c = 0 (x^{2}:) c = 0 and check that we get only the zero solution a = b = c = 0.

E.g., ( \begin{matrix}1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0 \end{matrix} ) has row-reduced echelon form ( \begin{matrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{matrix} ).

Problem 5: (a) Find a basis for the subspace S of R^{2×2} given by the upper triangular matrices. What is the dimension of S ? (Say WHY your choice is a basis.)

A general upper triangular matrix has form A = ( \begin{matrix}a & b \\ 0 & c \end{matrix} ).

So choosing a, b, c successively in the “standard basis” way, we get 3 “vectors” in a basis:

( \begin{matrix}1 & 0 \\ 0 & 0 \end{matrix} ), ( \begin{matrix}0 & 1 \\ 0 & 0 \end{matrix} ), ( \begin{matrix}0 & 0 \\ 0 & 1 \end{matrix} ). So dim S = 3.

Why basis? (spanning set:) the linear combination of these with coefficients a, b, c gives the general matrix A above.

(linearly independent:) Setting that linear combination equal to the zero-matrix gives only the zero solution a = b = c = 0.

(b) Find the coordinates of the vector (7, 4)^{T} in the basis of R^{2} given by (1, 2)^{T} and (3, 5)^{T}.

Solve ( \begin{matrix}1 & 3 & 7 \\ 2 & 5 & 4 \end{matrix} ) to get (−23, 10)^{T}.