Math 310: Hour Exam 1 (Solutions) Prof. S. Smith: Fri 17 October 2003

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

lowing augmented matrix: (A|b) = ( \begin{matrix}1 & −7 & 0 & 6 & 5 \\ 0 & 0 & 1 & −2 & −3 \\ −1 & 7 & −4 & 2 & 7 \end{matrix} ). Show the STEPS you use.

→^{A_{3}^{1×1}} ( \begin{matrix}1 & −7 & 0 & 6 & 5 \\ 0 & 0 & 1 & −2 & −3 \\ 0 & 0 & −4 & 8 & 12 \end{matrix} ) →^{A_{3}^{4×2}} ( \begin{matrix}1 & −7 & 0 & 6 & 5 \\ 0 & 0 & 1 & −2 & −3 \\ 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 2 and 4 have no pivot, so those variables are free. Then infinite number of solutions: (5 + 7α − 6β, α, −3 + 2β, β)^{T} for all real α, β

Problem 2: (a) Find the LU-decomposition (show method) of the matrix A = ( \begin{matrix}6 & 9 \\ 4 & 5 \end{matrix} ).

To row-reduce (Gaussian elmination) ( \begin{matrix}6 & 9 \\ 4 & 5 \end{matrix} ), we apply A_{2}^{− \frac{2}{3} ×1} to obtain ( \begin{matrix}6 & 9 \\ 0 & −1 \end{matrix} ) as U,

and take as L the inverse of the matrix for A_{2}^{\frac{2}{3} ×1}, namely ( \begin{matrix}1 & 0 \\ \frac{2}{3} & 1 \end{matrix} )

(b) Find the inverse (by any method—but show work) of the matrix A = ( \begin{matrix}5 & 10 \\ 4 & 7 \end{matrix} ).

By adjoint method, A^{−1} = −\frac{1}{5} ( \begin{matrix}7 & −10 \\ −4 & 5 \end{matrix} ).

Problem 3: (a) 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_{1} − 5x_{2} − 2x_{3} = 0. Show that S is a subSPACE of R^{3}.

Vectors in S have the general form form (5a + 2b, a, b)^{T}. (+) Take two general vectors in S: (5a + 2b, a, b)^{T}, (5c + 2d, c, d)^{T} and add; their sum is (5a + 2b + 5c + 2d, a + c, b + d)^{T} = (5(a + c) + 2(b + d), a + c, b + d)^{T}, which is also in S. (sc.mult.) For a general scalar c, and general vector (5a + 2b, a, b)^{T} in S, the scalar multiple is ((5a + 2b)c, ac, bc)^{T} = (5(ac) + 2(bc), ac, bc)^{T}, which is also in S.

(b) Give a basis for the nullspace of the matrix A = ( \begin{matrix}1 & 3 & 5 & 0 \\ 0 & 1 & 4 & −2 \end{matrix} ).

What is the dimension of this space ?

Using A_{1}^{−3×2} we obtain the row-reduced echelon form of A ( \begin{matrix}1 & 0 & −7 & 6 \\ 0 & 1 & 4 & −2 \end{matrix} ).

The solutions of Ax = 0 are of form (7α − 6β, −4α + 2β, α, β)^{T}, so choosing α = 1, β = 0 and then α = 0, β = 1, we get a basis given by (7, −4, 1, 0)^{T} and (−6, 2, 0, 1)^{T}. Therefore the dimension of the nullspace is 2.

Problem 4: (a) Recall that P_{3} is the space of polynomials of degree less than three (that is, quadratic polynomials). Are the three “vectors” 1 + x + x^{2}, 2 − 2x + 2x^{2}, and 4 + 4x^{2} a spanning set for this space ? (Why/why not ?)

Set a linear combination equal to a general vector of the space: a(1 + x + x^{2}) + b(2 − 2x + 2x^{2}) + c(4 + 4x^{2}) = d + ex + fx^{2}. Get an equation for each power of x: (1:) a + 2b + 4c = d (x:) a − 2b = e (x^{2}:) a + 2b + 4c = f

To the augmented matrix ( \begin{matrix}1 & 2 & 4 & d \\ 1 & −2 & 0 & e \\ 1 & 2 & 4 & f \end{matrix} ) we apply A_{3}^{−1×1} to get ( \begin{matrix}1 & 2 & 4 & d \\ 1 & −2 & 0 & e \\ 0 & 0 & 0 & f − d \end{matrix} ).

From the third row, we see that ONLY polynomials with d = f (that is, of the form d+ex+dx^{2}) are in the span our our set of 3 vectors; so not all vectors of P_{3} are in their span: that is, they are NOT a spanning set for P_{3}.

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

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

Problem 5: (a) If A is a 3 × 5 matrix of rank 2, what is the nullity (dimension of the nullspace) of A ?

rank + nullity = 5, so nullity = 3.

(b) Use the method of the (classical) adjoint to find the inverse of the matrix A = ( \begin{matrix}1 & 2 & 4 \\ 0 & −3 & 1 \\ 0 & 0 & 3 \end{matrix} ).

First compute det(A) = −9; so we get A^{−1} = −\frac{1}{9} ( \begin{matrix}−9 & −6 & 14 \\ 0 & 3 & −1 \\ 0 & 0 & −3 \end{matrix} ).