Math 310: Hour Exam 1 (Solutions) Prof. S. Smith: Fri 18 October 2002

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

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

Use row operations (Gauss-Jordan)

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

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

Notice column 3 has no pivot, so that variable is free. Infinite number of solutions: (2 + α, −1 − α, α)^{T} for real α.

Problem 2: (a) Find the LU-decomposition (show method) of the matrix A = ( \begin{matrix}3 & 4 \\ 6 & 7 \end{matrix} ).

To row-reduce (Gaussian elmination) ( \begin{matrix}3 & 4 \\ 6 & 7 \end{matrix} ), we apply A_{2}^{−2×1} to obtain ( \begin{matrix}3 & 4 \\ 0 & −1 \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) Find the inverse (by any method—but show work) of the matrix A = ( \begin{matrix}3 & 4 \\ 6 & 7 \end{matrix} ).

By adjoint method, A^{−1} = −\frac{1}{3} ( \begin{matrix}7 & −4 \\ −6 & 3 \end{matrix} ).

Problem 3: (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} in this space linearly independent ? (Why/why not ?)

Set a linear combination equal to zero: a(1 + x + x^{2}) + b(2 − 2x + 2x^{2}) + c(4 + 4x^{2}) = 0. Get an equation for each power of x: (1:) a + 2b + 4c = 0 (x:) a − 2b = 0 (x^{2}:) a + 2b + 4c = 0

When we row-reduce the corresponding matrix A = ( \begin{matrix}1 & 2 & 4 \\ 1 & −2 & 0 \\ 1 & 2 & 4 \end{matrix} ) to ( \begin{matrix}1 & 0 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{matrix} ),

the third column has no pivot: so ”No”, linearly dependent rather than independent.

(b) Solve Ax = b by Cramer’s Rule for the augmented matrix (A|b): ( \begin{matrix}2 & 4 & 6 \\ 3 & 5 & 7 \end{matrix} ).

det(A) = 2·5−3·4 = −2, so x_{1} = det ( \begin{matrix}6 & 4 \\ 7 & 5 \end{matrix} ) /(−2) = −1, and x_{2} = det ( \begin{matrix}2 & 6 \\ 3 & 7 \end{matrix} ) /(−2) = 2.

Problem 4: (a) Let S be the subspace of R^{3} consisting of all 3-vectors whose coordinates sum to zero: that is, x_{1} + x_{2} + x_{3} = 0. Show that S is a subpace of R^{3}.

(+) If (x_{1}, x_{2}, x_{3}) satisfies x_{1} + x_{2} + x_{3} = 0, and (y_{1}, y_{2}, y_{3}) satisfies y_{1} + y_{2} + y_{3} = 0, then their sum (x_{1} + y_{1}, x_{2} + y_{2}, x_{3} + y_{3}) satisfies

(x_{1} + y_{1}) + (x_{2} + y_{2}) + (x_{3} + y_{3}) = (x_{1} + x_{2} + x_{3}) + (y_{1} + y_{2} + y_{3}) = 0 + 0 = 0, and so also lies in S. (sc.mult.) For a scalar c, the scalar multiple c(x_{1}, x_{2}, x_{3}) = (cx_{1}, cx_{2}, cx_{3}) satsifies cx_{1} + cx_{2} + cx_{2} = c(x_{1} + x_{2} + x_{3}) = c · 0 = 0, and so also lies in S.

(b) Give a basis for the row space of the matrix A = ( \begin{matrix}1 & 2 & −1 \\ 1 & 9 & −1 \\ −3 & 8 & 3 \\ −2 & 3 & 2 \end{matrix} ).

What is the dimension of this space ?

We compute the row-reduced echelon form of A as ( \begin{matrix}1 & 0 & −1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} ).

Therefore the first two rows give a basis, and the dimension is 2.

Problem 5: (a) Find a basis for the subspace S of polynomials in P_{2} of form ax^{2} +bx+(3a+2b); show that your basis IS a basis.

Best to choose the “free variables” a, b in the “standard basis” way: Thus a = 1, b = 0 gives x^{2} + 3, a = 0, b = 1 gives x + 2. Spanning set: ax^{2} + bx + (3a + 2b) = a(x^{2} + 3) + b(x + 2). linearly independent: 0 = ax^{2} + bx + (3a + 2b) gives equations for each power of x, including a = 0, b = 0; that is, only the zero-solution.

(b) Find the matrix of transition from the “old” basis given by the standard basis of R^{2} (namely (1, 0)^{T} and (0, 1)^{T}) to the “new” basis given by (1, 2)^{T} and (2, 3)^{T}.

One way: The matrix [new]_{old} is given by ( \begin{matrix}1 & 2 \\ 2 & 3 \end{matrix} ),

so the transition matrix [old]_{new} from old to new is given by its inverse, namely ( \begin{matrix}−3 & 2 \\ 2 & −1 \end{matrix} ).

What are the coordinates of (5, 7)^{T} in this new basis?

Can multiply transition matrix by “old” coordinates (5, 7)^{T} to get new coordinates (−1, 3)^{T}.