Mathematics 244, section 1 – Linear Algebra Solutions for Exam 1 – February 16, 2007 If A, B are n × n matrices and A is invertible, then there is a unique n × n matrix X

with AXA = B.

I. A) (10) For which value(s) of c ∈ R is the system of linear equations represented by the

following augmented matrix consistent?

( \begin{matrix}−1 & 2 & 0 & c & | & 1 \\ 0 & 1 & 2 & 0 & | & −1 \\ 2 & −4 & 0 & 8 & | & 5 \end{matrix} )

Solution : Perform the row operation R_{3} ↦ R_{3} + 2R_{1}:

( \begin{matrix}−1 & 2 & 0 & c & | & 1 \\ 0 & 1 & 2 & 0 & | & −1 \\ 0 & 0 & 0 & 8 + 2c & | & 7 \end{matrix} )

If 8 + 2c = 0, then the system is inconsistent, otherwise it is consistent. So the answer here is that the system is consistent for all c ∈ R, c ≠ −4.

B) (10) Give a parametrization of the set of solutions for a value of c (your choice) that makes the system consistent.

Solution : We will take c = 0 for simplicity. Continuing from the matrix in the solution for part A, we obtain the row-reduced echelon form

( \begin{matrix}1 & 0 & 4 & 0 & | & −3 \\ 0 & 1 & 2 & 0 & | & −1 \\ 0 & 0 & 0 & 1 & | & 7/8 \end{matrix} )

This gives a system where x_{3} is a free variable. Set of solutions is

\{( \begin{matrix}x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{matrix} ) = ( \begin{matrix}−3 \\ −1 \\ 0 \\ 7/8 \end{matrix} ) + t ( \begin{matrix}−4 \\ −2 \\ 1 \\ 0 \end{matrix} ) : t ∈ R\}

II. A) (5) Define: The set S of vectors in R^{n} is linearly independent.

S is linearly independent if whenever a linear combination

c_{1}v_{1} + c_{2}v_{2} + ⋯ + c_{k}v_{k} = 0

for v_{i} ∈ S, then c_{1} = c_{2} = ⋯ = c_{k} = 0. B) (10) For which value of a ∈ R is

S = \{( \begin{matrix}3 \\ 2 \\ 1 \end{matrix} ) , ( \begin{matrix}−1 \\ 1 \\ 0 \end{matrix} ) , ( \begin{matrix}1 \\ 0 \\ a \end{matrix} )\}

a linearly dependent subset of R^{3}? Solution : We do row operations on the matrix with columns equal to the vectors in S:

( \begin{matrix}3 & −1 & 1 \\ 2 & 1 & 0 \\ 1 & 0 & a \end{matrix} ) ↦ ( \begin{matrix}1 & 0 & a \\ 2 & 1 & 0 \\ 3 & −1 & 1 \end{matrix} )

↦ ( \begin{matrix}1 & 0 & a \\ 0 & 1 & −2a \\ 0 & −1 & 1 − 3a \end{matrix} )

↦ ( \begin{matrix}1 & 0 & a \\ 0 & 1 & −2a \\ 0 & 0 & 1 − 5a \end{matrix} )

In order for the set to be linearly dependent, the matrix must have fewer than 3 pivot columns, which means 1 − 5a = 0, so a = 1/5. C) (15) Let b = ( \begin{matrix}3 \\ 3 \\ 4 \end{matrix} ). Using your value of a from part B, is b ∈ Span(S)? Why or why

not? Solution : We do the row operations as in the previous part, but on the augmented matrix:

( \begin{matrix}3 & −1 & 1 & | & 3 \\ 2 & 1 & 0 & | & 3 \\ 1 & 0 & 1/5 & | & 4 \end{matrix} ) ↦ ( \begin{matrix}1 & 0 & 1/5 & | & 4 \\ 2 & 1 & 0 & | & 3 \\ 3 & −1 & 1 & | & 3 \end{matrix} )

↦ ( \begin{matrix}1 & 0 & 1/5 & | & 4 \\ 0 & 1 & −2/5 & | & −5 \\ 0 & −1 & 2/5 & | & −9 \end{matrix} )

↦ ( \begin{matrix}1 & 0 & 1/5 & | & 4 \\ 0 & 1 & −2/5 & | & −5 \\ 0 & 0 & 0 & | & −14 \end{matrix} )

This says the system is inconsistent, so b is not in the span of S.

III. Let T : R^{3} → R^{3} be the mapping defined by

T (x) = ( \begin{matrix}1 & 2 & 1 \\ −1 & 4 & 0 \\ 1 & 0 & 1 \end{matrix} ) x

A) (10) Show that T is a linear mapping by verifying the properties in the definition. Solution : For any 3 × 3 (or m × n) matrix, A(x + y) = Ax + Ay and A(cx) = cAx. This says T (x + y) = T (x) + T (y) and T (cx) = cT (x), so T is linear. B) (10) Is T an onto mapping? Why or why not? Solution : There are a number of ways to answer this. One of the most direct is just to see whether the matrix is invertible. If so, the Invertible Matrix Theorem says T is onto. We follow our procedure for computing the matrix inverse:

( \begin{matrix}1 & 2 & 1 & 1 & 0 & 0 \\ −1 & 4 & 0 & 0 & 1 & 0 \\ 1 & 0 & 1 & 0 & 0 & 1 \end{matrix} ) ↦ ( \begin{matrix}1 & 2 & 1 & 1 & 0 & 0 \\ 0 & 6 & 1 & 1 & 1 & 0 \\ 0 & −2 & 0 & −1 & 0 & 1 \end{matrix} )

↦ ( \begin{matrix}1 & 2 & 1 & 1 & 0 & 0 \\ 0 & −2 & 0 & −1 & 0 & 1 \\ 0 & 6 & 1 & 1 & 1 & 0 \end{matrix} )

↦ ( \begin{matrix}1 & 0 & 1 & 0 & 0 & 1 \\ 0 & −2 & 0 & −1 & 0 & 1 \\ 0 & 6 & 1 & 1 & 1 & 0 \end{matrix} )

↦ ( \begin{matrix}1 & 0 & 1 & 0 & 0 & 1 \\ 0 & −2 & 0 & −1 & 0 & 1 \\ 0 & 0 & 1 & −2 & 1 & 3 \end{matrix} )

↦ ( \begin{matrix}1 & 0 & 0 & 2 & −1 & −2 \\ 0 & 1 & 0 & 1/2 & 0 & −1/2 \\ 0 & 0 & 1 & −2 & 1 & 3 \end{matrix} )

Since A is invertible the answer is yes – the mapping T is onto. IV. (True-False) Determine whether each of the following statements is true or false. For the ones that are true, give short proofs; for those that are false, give counterexamples. A) (10) If the homogeneous system Ax = 0 has two free variables, then the set of solutions

of Ax = b is a plane for all b. Solution : This is FALSE because the system Ax = b could also be inconsistent for some b. An example is given by

[A|b] = ( \begin{matrix}1 & 2 & 3 & | & 1 \\ 2 & 4 & 6 & | & −1 \end{matrix} )

B) (10) If x ∈ Span(S) (and x ∉ S), then the set S ∪ {x} is linearly dependent. Solution : This is TRUE. If x = c_{1}v_{1} + ⋯ + c_{k}v_{k}, where v_{i} ∈ S for all i, then

0 = (−1)x + c_{1}v_{1} + ⋯ + c_{k}v_{k}.

Since the coefficient −1 ≠ 0 in this linear combination, the set S∪{x} is linearly dependent. Comment: Without the added statement x ∉ S, this would actually be FALSE(!) Counterexample: S = {e_{1}, e_{2}} in R^{n} for any n ≥ 2. e_{1} = 1 · e_{1} + 0 · e_{2} is in Span(S), but S ∪ {e_{1}} = {e_{1}, e_{2}} is still linearly independent. C) (10) Let A be an m × n matrix that defines an onto mapping T_{A} : R^{n} → R^{m}

(T_{A}(x) = Ax for x ∈ R^{n}). Then the span of the set of columns of A is R^{m}. Solution : This is TRUE. Since T is onto, for every y ∈ R^{m}, there is some x ∈ R^{n} such that T (x) = Ax = y. If x = ( x_{1} ; ⋮ ; x_{n} ), then

y = Ax = x_{1}a_{1} + ⋯ + x_{n}a_{n}

where a_{1}, …, a_{n} are the columns of A . This shows that every y ∈ R^{m} is a linear combination of the columns of A.