Math 310: Hour Exam 1 (Solutions) Prof. S. Smith: Mon 29 Sept 1997

Problem 1: Find the row-reduced echelon form of the following augmented matrix. INDICATE the row operations you use.

(A|b) = ( \begin{matrix}1 & 2 & 3 & 4 \\ 2 & 3 & 4 & 6 \\ 3 & 4 & 5 & 8 \end{matrix} ) .

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

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

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

So solutions are (r, 2 − 2r, r) for free variable x_{3} = r.

Problem 2: Give the LU-decomposition of

A = ( \begin{matrix}4 & 5 \\ 2 & 3 \end{matrix} );

that is, find lower-triangular L and upper-triangular U, so that A = LU.

Get U from →^{A_{2}^{− \frac{1}{2} ×1}} ( \begin{matrix}4 & 5 \\ 0 & \frac{1}{2} \end{matrix} ). So L from inverse operation A_{2}^{+ \frac{1}{2} ×1} is

( \begin{matrix}1 & 0 \\ \frac{1}{2} & 1 \end{matrix} )

Problem 3: Find the inverse (any method) of:

A = ( \begin{matrix}1 & 2 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 2 \end{matrix} ) .

Quick via adjoint: det (by bottom row) is 0 − 0 + 2(−2) = −4

so inverse is −\frac{1}{4} ( \begin{matrix}−1 & −3 & 2 \\ −2 & 2 & 0 \\ 1 & −1 & −2 \end{matrix} )

Problem 4: Find the determinant of

A = ( \begin{matrix}1 & 1 & 0 \\ 3 & 4 & 2 \\ 1 & 3 & 1 \end{matrix} ) .

What is the determinant of A^{−1} ?

Top row: 1(4.1−3.2)−1(3.1−1.2) = (−2)−(1) = −3. Then det(A^{−1}) = \frac{1}{det(A)} = −\frac{1}{3}.

Problem 5: (a) Is (2, 5, 4) a linear combination of (1, 1, 2) and (1, 3, 2)? Either give coefficients, or explain why not possible.

Yes: Set up augmented matrix (A|b) = ( \begin{matrix}1 & 1 & 2 \\ 1 & 3 & 5 \\ 2 & 2 & 4 \end{matrix} ) .

Row-reduction quickly produces coefficients \frac{1}{2} and \frac{3}{2}.

(b) Let V be the space of 2 × 2 matrices, and W the subSET of lower triangular matrices. Show that the W is a subSPACE of V.

A ∈ W says lower triangular, meaning A_{2,1} = 0, so A has form ( \begin{matrix}a & 0 \\ b & c \end{matrix} ).

Similarly if B ∈ W it has form ( \begin{matrix}d & 0 \\ e & f \end{matrix} )

Is A + B ∈ W? It is ( \begin{matrix}a + d & 0 \\ b + e & c + f \end{matrix} ), also lower-triangular, so yes.

For scalar r, is rA ∈ W? It is ( \begin{matrix}ra & 0 \\ rb & rc \end{matrix} ), also lower-triangular, so yes.