Math 310: Hour Exam 1 (Solutions) Prof. S. Smith: 26 Sept 1994 Problem 1: Use row-reduced echelon form (Gauss-Jordan) to find all solutions of

( \begin{matrix}2 & 3 & 1 \\ 1 & 1 & 1 \\ 3 & 4 & 2 \end{matrix} ) ( \begin{matrix}x_{1} \\ x_{2} \\ x_{3} \end{matrix} ) = ( \begin{matrix}1 \\ 3 \\ 4 \end{matrix} ) .

→^{E_{1,2}} ( \begin{matrix}1 & 1 & 1 & 3 \\ 2 & 3 & 1 & 1 \\ 3 & 4 & 2 & 4 \end{matrix} ) →^{A_{2}^{−2×1},A_{3}^{−3×1}} ( \begin{matrix}1 & 1 & 1 & 3 \\ 0 & 1 & −1 & −5 \\ 0 & 1 & −1 & −5 \end{matrix} ) →^{A_{1}^{−1×2},A_{3}^{−1×2}} ( \begin{matrix}1 & 0 & 2 & 8 \\ 0 & 1 & −1 & −5 \\ 0 & 0 & 0 & 0 \end{matrix} )

Thus x_{3} is free variable; with x_{3} = 1 in Ax = 0, homogenous solutions are α(−2 1 1). With x_{3} = 0 in Ax = b, particular solution is (8 − 5 0). So general: (8 − 2α, −5 + α, α).

Problem 2:

(a) Is A = ( \begin{matrix}1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{matrix} ) invertible? If so, use elementary row operations to find A^{−1}.

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

Now A^{−1} is on the right.

(b) Use row operations to find the LU decomposition of A = ( \begin{matrix}1 & 0 & 0 \\ 1 & 1 & 1 \\ 0 & 0 & 1 \end{matrix} ).

→^{A_{2}^{−1×1}} ( \begin{matrix}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix} ) is echelon-form so gives U;

and from inverse of the row operation we get L = A_{2}^{1×1}(I_{3}) = ( \begin{matrix}1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} )

Problem 3:

(a) Find the determinant of A = ( \begin{matrix}1 & 2 & 0 \\ 1 & 3 & 1 \\ 3 & 1 & 1 \end{matrix} ). Is A singular? How many solutions does Ax = 0 have?

(top row) det(A) = 1(3 · 1 − 1 · 1) − 2(1 · 1 − 1 · 3) = 2 + 4 = 6.

Thus A is non-singular; and Ax = 0 has unique solution x = 0.

(b) Use Cramer’s rule (determinants) to solve ( \begin{matrix}2 & 3 \\ 3 & 2 \end{matrix} ) ( \begin{matrix}x_{1} \\ x_{2} \end{matrix} ) = ( \begin{matrix}2 \\ 5 \end{matrix} ).

First, det(A) = 2 · 2 − 3 · 3 = 4 − 9 = −5.

Next det ( \begin{matrix}2 & 3 \\ 5 & 2 \end{matrix} ) = 4 − 15 = −11 and det ( \begin{matrix}2 & 2 \\ 3 & 5 \end{matrix} ) = 10 − 6 = 4.

So x_{1} = \frac{11}{5} and x_{2} = −\frac{4}{5}.

Problem 4: Show that the set S consisting of 2x2 symmetric matrices (A^{T} = A) forms a subspace of R^{2×2}.

If A, B ∈ S then A^{T} = A, B^{T} = B.

Then (A + B)^{T} = A^{T} + B^{T} = A + B; so A + B is symmetric, and hence is also in S. Similarly if A ∈ S and c is scalar, then (cA)^{T} = c(A^{T}) = c(A) = cA,

so that cA is symmetric, and hence is also in S.