Problems 11-20
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12 Given the system of equations
$$ \left\{ \ \ \begin{array}{ r c r c r c r } 2x & + & 3y & + & z & = & 8 \\ 4x & + & 7y & + & 5z & = & 20 \\ & & -2y & + & 2z & = & 0 \end{array} \right. \ \ , $$the system can be reduced into upper triangular form with two operations by subtracting first row ($R_1$) multiplied by $l_{21} / l_{11}$ $=$ $4 / 2$ $=$ $2$ from the second row ($R_2$). That is
$$ \begin{array}{c l} \xrightarrow{R_2 - 2R_1} & \left\{ \ \ \begin{array}{ r c r c r c r } 2x & + & 3y & + & z & = & 8 \\ & & y & + & 3z & = & 4 \\ & & -2y & + & 2z & = & 0 \end{array} \right. \ \ . \end{array} $$Then subtracting the multiple $l_{32} / l_{22}$ $=$ $-2/1$, i.e. adding the multiple of $2$ of $R_2$ to the third row $(R_3)$ leads to
$$ \begin{array}{c l} \xrightarrow{R_2 - 2R_1} & \left\{ \ \ \begin{array}{ r c r c r c r } \mathbf{2}x & + & 3y & + & z & = & 8 \\ & & \mathbf{1}y & + & 3z & = & 4 \\ & & & & \mathbf{8}z & = & 8 \end{array} \right. \ \ , \end{array} $$where the pivots are boldened.
From $8z = 8$ it can be seen that $z = 1$, so substituing this to $R_2$ and solving for $y$ shows that it’s equal to $1$. Substituting these values to $R_1$, the row beocmes
$$ \phantom{,} \ 2x + 3(1) + 1 = 2x + 4 = 8 \ , $$from where it can be seen that $x = 2$. Then the solution vector for this system is
$$ \phantom{.} \ \ \mathbf{b} = \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix} \ \ . $$13 Subtracting first row from the second after multiplying it by $l_{21}/l_{11}$ leads to system of equations
$$ \phantom{.} \quad \left\{ \begin{array}{ r r r r r } 2x & - & 3y & & & = & 3 \\ & & y & + & z & = & 1 \\ 2x & - & y & - & 3z & = & 7 \end{array} \right. \quad . $$Subtracting the first row from the third once leads to
$$ \phantom{.} \quad \left\{ \begin{array}{ r r r r r r r } 2x & - & 3y & & & = & 3 \\ & & y & + & z & = & 1 \\ & & 2y & - & 3z & = & 4 \end{array} \right. \quad $$and subtracting the second row from the third twice further leads to
$$ \phantom{.} \quad \left\{ \begin{array}{ r r r r r r r } 2x & - & 3y & & & = & 3 \\ & & y & + & z & = & 1 \\ & & & - & 5z & = & 2 \end{array} \right. \quad , $$so $z = -\frac{2}{5}$ and so $y = \frac{7}{5}$. Since the multiplier of $z$ is $0$, only the $y$ needs to be substituted to the first equation which leads to
$$ 2x - 3 \left( \dfrac{7}{5} \right) = 2x - \dfrac{21}{5} = 3 $$so
$$ \phantom{,} \ 2x = 3 + \dfrac{21}{5} = \dfrac{15}{5} + \dfrac{21}{5} = \dfrac{36}{5} \ , $$which shows that
$$ \phantom{.} \ x = \dfrac{36/5}{2} = \dfrac{36}{10} \ . $$So
$$ \phantom{,} \quad \left\{ \begin{array}{ r r r } x & = & \frac{36}{10} \\ y & = & \frac{7}{5} \\ z & = & -\frac{2}{5} \end{array} \right. \quad . $$Substituting these values with, say, the third row of the given system, it can be seen that
$$ \begin{array}{r c l} 2x - y - 3z & = & 2 \left( \dfrac{36}{10} \right) - \left( \dfrac{7}{5} \right) + 3 \left( \dfrac{2}{5} \right) \\[0.5em] & = & \dfrac{72}{10} - \dfrac{14}{10} + \dfrac{12}{10} \\[0.5em] & = & \dfrac{70}{10} \\[0.5em] & = & 7, \end{array} $$14
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16 (a) When $\mathbf{A}$ is a $3 \times 3$ matrix, then a matrix that needs two row exchanges in order to reach a triangular form and a solution is
$$ \phantom{.} \begin{bmatrix} 0 & 0 & \gamma \\ 0 & \beta & \gamma \\ \alpha & \beta & \gamma \\ \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \mathbf{b} , $$because without row exchanges, trying to pivot first row ($R_1$) with the second row ($R_2$), and then the same between $R_2$ and the third row ($R_3$) leads to multiplication by $0$ which is not allowed, so the elimination breaks.
However, exchanging $R_3$ with $R_1$, that is the first row exchange, and then exchanging $R_2$ with $R_1$ leads to a system of equations of the form
$$ \phantom{.} \begin{bmatrix} \alpha & \beta & \gamma \\ 0 & \beta & \gamma \\ 0 & 0 & \gamma \end{bmatrix} \begin{bmatrix} z \\ y \\ x \end{bmatrix} = \mathbf{b} . $$(b) Example of a matrix that needs a one row exchange and later breaks down is
$$ \phantom{.} \begin{bmatrix} 0 & \beta & \gamma \\ \alpha & \beta & \gamma \\ 0 & \beta & \gamma \\ \end{bmatrix} = \mathbf{b} . $$17 In the example case, if no rows are exchanged, one can eliminate all $x$’s, $y$’s and $z$’s from the system of equations. The steps are subtract the first row ($R_1$) from the second row ($R_2$) once, and $R_1$ from the third row ($R_3$) two times. This leads to a solution
$$ \phantom{.} \quad \left\{ \begin{array}{ r r r } 0 & = & 0 \\ 0 & = & 0 \\ 0 & = & 2 \\ \end{array} \right. \quad , $$No row exchanges are needed.
Same goes for the second systems, that is, subtracting $R_1$ two times from $R_2$ and $R_1$ three times from $R_3$. This results in the same system as the one above.
18 Such a matrix is can be constructed by arbitrarily a equation with coeffiecents
$$ \phantom{,} \ \alpha x + \beta y + \delta z = \gamma \ , $$such that $\alpha \neq \beta \neq \delta$, and then multiplying the values with some, say, integers like $3$ and $5$, so that the system takes the form
$$ \phantom{.} \quad \left\{ \begin{array}{ r r r r r r r } \alpha x & + & \beta y & + & \gamma z & = & \delta \\ 3 \alpha x & + & 3 \beta y & + & 3 \gamma z & = & 3\delta \\ 5 \alpha x & + & 5 \beta y & + & 5 \gamma z & = & 5\delta \end{array} \right. \quad . $$Now subtracting the first row multiplied by $3$ from the second and the first row multiplied by $5$ from the scoend leads to
$$ \left\{ \begin{array}{ r r r r r r r } \alpha x & + & \beta y & + & \gamma z & = & \delta \\ 0 & + & 0 & + & 0 & = & 0 \\ 0 & + & 0 & + & 0 & = & 0 \end{array} \right. $$after elimination.
When $\mathbf{b} = (1, 10, 100)$, the system becomes
$$ \left\{ \begin{array}{ r r r r r r r } \alpha x & + & \beta y & + & \gamma z & = & 1 \\ 3 \alpha x & + & 3 \beta y & + & 3 \gamma z & = & 10 \\ 5 \alpha x & + & 5 \beta y & + & 5 \gamma z & = & 100 \\ \end{array} \right. $$and applying the same elimination steps as above leads to
$$ \phantom{.} \quad \left\{ \begin{array}{ r r r r r r r } \alpha x & + & \beta y & + & \gamma z & = & 1 \\ & & & & 0 & = & 7 \\ & & & & 0 & = & 95 \\ \end{array} \right. \quad . $$When $\mathbf{b} = (0, 0, 0)$, the system becomes
$$ \left\{ \begin{array}{ r r r r r r r } \alpha x & + & \beta y & + & \gamma z & = & 0 \\ 3 \alpha x & + & 3 \beta y & + & 3 \gamma z & = & 0 \\ 5 \alpha x & + & 5 \beta y & + & 5 \gamma z & = & 0 \\ \end{array} \right. $$and applying the same elimination steps as above leads to
$$ \phantom{.} \quad \left\{ \begin{array}{ r r r r r r r } \alpha x & + & \beta y & + & \gamma z & = & 0 \\ & & & & 0 & = & 0 \\ & & & & 0 & = & 0 \\ \end{array} \right. \quad . $$19 Subtracting the first row ($R_1$) once from the second ($R_2$), leads to $R_2$ being equal to $3y$ $-$ $4z$ $=$ $5$. Subtracting $R_2$ once from the third row ($R_3$) means that $R_3$ becomes $(q - 6)z$ $=$ $t - 5$. If $q = 5$, the third row becomes $0 = t - 5$, and so the system becomes singular with infinitely many solutions. If $t$ too is equal to $5$, then the third row becomes $0 = 0$, so the system remains singular and has infinitely many solutions. If $q = t = 6$, then $R_3$ is $z = 1$.
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