MATHEMATICS 2 (MA10193) EXAMPLES SHEET 2: SOLUTIONS

1.

\begin{matrix}( \begin{matrix}1 & 2 & −8 \\ 2 & −3 & 5 \\ 3 & 2 & −12 \end{matrix} ) & ⟶^{R_{2}−2R_{1}} ( \begin{matrix}1 & 2 & −8 \\ 0 & −7 & −11 \\ 3 & 2 & −12 \end{matrix} ) \\ ⟶^{R_{3}−3R_{1}} & ( \begin{matrix}1 & 2 & −8 \\ 0 & −7 & −11 \\ 0 & −4 & −36 \end{matrix} ) \\ ⟶^{R_{3}− \frac{4}{7} R_{1}} & ( \begin{matrix}1 & 2 & −8 \\ 0 & −7 & −11 \\ 0 & 0 & −208/7 \end{matrix} ) \end{matrix}

and the bottom row gives \frac{−208}{7}z = so z = 0. Then the next row gives −7y − 11z = 0 so y = 0 and the top row gives x + 2y + 8z = 0 so x = 0 also.

2.

\begin{matrix}( \begin{matrix}9 & 4 & 3 & −1 \\ 5 & 1 & 2 & 1 \\ 7 & 3 & 4 & 1 \end{matrix} ) & ⟶^{R_{2}− \frac{5}{9} R_{1}} ( \begin{matrix}9 & 4 & 3 & −1 \\ 0 & \frac{−11}{9} & \frac{−3}{9} & \frac{14}{9} \\ 7 & 3 & 4 & 1 \end{matrix} ) \\ ⟶^{R_{2}×9} & ( \begin{matrix}9 & 4 & 3 & −1 \\ 0 & −11 & 3 & 14 \\ 7 & 3 & 4 & 1 \end{matrix} ) \\ ⟶^{R_{3}− \frac{7}{9} R_{1}} & ( \begin{matrix}9 & 4 & 3 & −1 \\ 0 & −11 & 3 & 14 \\ 0 & \frac{−1}{9} & \frac{15}{9} & \frac{16}{9} \end{matrix} ) \\ ⟶^{R_{3}×9} & ( \begin{matrix}9 & 4 & 3 & −1 \\ 0 & −11 & 3 & 14 \\ 0 & −1 & 15 & 16 \end{matrix} ) \\ ⟶^{R_{3}− \frac{1}{11} R_{2}} & ( \begin{matrix}9 & 4 & 3 & −1 \\ 0 & −11 & 3 & 14 \\ 0 & 0 & \frac{162}{11} & \frac{162}{11} \end{matrix} ) \\ ⟶^{R_{3}× \frac{11}{162}} & ( \begin{matrix}9 & 4 & 3 & −1 \\ 0 & −11 & 3 & 14 \\ 0 & 0 & 1 & 1 \end{matrix} ) \\ ⟶^{R_{2}−3R_{3}} & ( \begin{matrix}9 & 4 & 3 & −1 \\ 0 & −11 & 0 & 11 \\ 0 & 0 & 1 & 1 \end{matrix} ) \end{matrix}

\begin{matrix}⟶^{R_{1}−3R_{3}} & ( \begin{matrix}9 & 4 & 0 & −4 \\ 0 & −11 & 0 & 11 \\ 0 & 0 & 1 & 1 \end{matrix} ) \\ ⟶^{R_{2}× \frac{1}{11}} & ( \begin{matrix}9 & 4 & 0 & −4 \\ 0 & −1 & 0 & 1 \\ 0 & 0 & 1 & 1 \end{matrix} ) \\ ⟶^{R_{1}+4R_{2}} & ( \begin{matrix}9 & 0 & 0 & 0 \\ 0 & −1 & 0 & 1 \\ 0 & 0 & 1 & 1 \end{matrix} ) \end{matrix}

so the solution is z = 1, y = −1 and x = 0.

3. We may assume that a_{11} ≠ 0 (otherwise, swap the rows or columns round).

( \begin{matrix}a_{11} & a_{12} & b_{1} \\ a_{21} & a_{22} & b_{2} \end{matrix} ) ⟶^{R_{2}− \frac{a_{21}}{a_{11}} R_{1}} ( \begin{matrix}a_{11} & a_{12} & b_{1} \\ 0 & a_{22} − \frac{a_{12}a_{21}}{a_{11}} & b_{2} − \frac{b_{1}a_{21}}{a_{1}1} \end{matrix} )
⟶^{R_{2}×a_{11}} ( \begin{matrix}a_{11} & a_{12} & b_{1} \\ 0 & a_{11}a_{22} − a_{12}a_{21} & b_{2}a_{11} − b_{1}a_{21} \end{matrix} )

so y = \frac{b_{2}a_{11}−b_{1}a_{21}}{a_{11}a_{22}−a_{12}a_{21}} and hence y = \frac{b_{2}a_{12}−b_{1}a_{22}}{a_{11}a_{22}−a_{12}a_{21}} , which is Cramer’s Rule.

4. Taking twice the second row away from the third gives

( \begin{matrix}0 & 2 & −1 \\ 1 & 0 & 3 \\ 0 & −4 & 4 \end{matrix} )

which has determinant −4. This is not zero, so the rank of A is 3 and the equations therefore have exactly one solution.

5. The determinant is (x + 1)(1 × 1 + 5 × 4) − x(2 × 1 + 3 × 4) + (x − 4)(2 × 5 − 3 × 1) = 21x + 21 − 14x + 7x − 28 = 14x − 7, which is zero when x = \frac{1}{2}.

6. det A = −2x^{2} − 5x + 4 so if det A = 0 then −2x^{2} − 5x + 4 = 0. In that case x = (−5 ± \sqrt{57})/4, i.e. x ≈ −3.14 or x ≈ 0.64. If x = 2 then det A ≠ 0 so rank A = 3. GKS, 9/05/05