Math 310: Hour Exam 1 (Solutions) Prof. S. Smith: Fri 3 March 2000

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

mented matrix: (A|b) = ( \begin{matrix}1 & −1 & 0 & −1 \\ 2 & −1 & 1 & 3 \\ 4 & −3 & 1 & 1 \end{matrix} ).

Use row operations →^{A_{2}^{−2×1} A_{3}^{−4×1}} ( \begin{matrix}1 & −1 & 0 & −1 \\ 0 & 1 & 1 & 5 \\ 0 & 1 & 1 & 5 \end{matrix} ) →^{A_{1}^{1×2} A_{3}^{−1×2}} ( \begin{matrix}1 & 0 & 1 & 4 \\ 0 & 1 & 1 & 5 \\ 0 & 0 & 0 & 0 \end{matrix} )

(b) Then give the solutions of the corresponding linear system Ax = b.

So solutions are (4 − r, 5 − r, r) for free variable x_{3} = r.

Problem 2: (a) Give the LU-decomposition of the matrix A = ( \begin{matrix}1 & 1 \\ 3 & 7 \end{matrix} ).

Use A_{2}^{−3×1} to get U = ( \begin{matrix}1 & 1 \\ 0 & 4 \end{matrix} ); and then the inverse of that operation gives L = ( \begin{matrix}1 & 0 \\ 3 & 1 \end{matrix} )

(b) Find the inverse (by any method) of the matrix A = ( \begin{matrix}11 & 2 \\ 6 & 1 \end{matrix} ),

Quickly, by classical-adjoint method: A^{−1} = − ( \begin{matrix}1 & −2 \\ −6 & 11 \end{matrix} ) = ( \begin{matrix}−1 & 2 \\ 6 & −11 \end{matrix} ),

Problem 3: (a) Find the determinant of: A = ( \begin{matrix}1 & 2 & 0 \\ 3 & 5 & 2 \\ 6 & 4 & 0 \end{matrix} ).

Down 2nd column: (−2)(1.4 − 6.2) = 16.

(b) Use Cramer’s rule to solve ( \begin{matrix}1 & 2 & 4 \\ 1 & 5 & 2 \end{matrix} ).

det(A) = 1.5 − 1.2 = 3, so x_{1} = \frac{1}{3}(4.5 − 2.2) = \frac{16}{3} and x_{2} = \frac{1}{3}(1.2 − 1.4) = −\frac{2}{3}.

Problem 4: (a) In the space P_{4} of polynomials of degree at most 3, let S be the subSET of polynomials of degree at most 2 (that is, of form a + bx + cx^{2} for a, b, c real numbers). Show that S is a subSPACE of P_{4}.

(+)Add two general polynomials in S: (a+bx+cx^{2})+(d+ex+fx^{2}) = (a+d)+(b+e)x+(c+f)x^{2}; the result still has degree ≤ 2, so is also in S. (sc.mult.) Multiply general polynomial in S by scalar d: d(a + bx + cx^{2}) = da + (db)x + (dc)x^{2}; the result still has degree ≤ 2, so is also in S.

(b) Are the columns of A = ( \begin{matrix}1 & 1 & 3 & 0 \\ 2 & 1 & 5 & 1 \\ 1 & 1 & 3 & 0 \end{matrix} ) a spanning set for R^{3}? (Show why/why not).

No. One way: row-reduce to ( \begin{matrix}1 & 1 & 3 & 0 \\ 0 & −1 & −1 & 1 \\ 0 & 0 & 0 & 0 \end{matrix} ).

Only 2 pivots, so dim(col.space) = 2, which is less than the dimension 3 of R^{3}.

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

General solution: (−2a + b, −3a − 5b, a, b)^{T}. Choosing a = 1, b = 0 gives (−2, −3, 1, 0)^{T}, choosing a = 0, b = 1 gives (1, −5, 0, 1)^{T}. (Now CHECK this is a basis...this works out easily...)

(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, 1)^{T} and (1, −1)^{T}.

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

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

What are the coordinates of (4, 6)^{T} in this new basis?

One way: Multiply transition matrix by old coordinates (4, 6)^{T} to get new coordinates (5, −1)^{T}.