Problems 24-28
24 The linear equation
$$ \mathbf{Ax} = \begin{bmatrix} 2 & 3 \\ 4 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 1 \\ 17 \end{bmatrix} $$can be written as augmented matrix
$$ \left[ \begin{array}{c c | c} 2 & 3 & 1 \\ 4 & 1 & 17 \end{array} \right] $$that can be brought to upper triangle form by subtracting row one ($R_1$) from row two ($R_2$) twice. In other words,
$$ \begin{array}{l l} \xrightarrow{R_2 - 2R_1} & \left[ \begin{array}{c c | c} 2 & 3 & 1 \\ 0 & -5 & 15 \end{array} \right], \end{array} $$so
$$ \mathbf{Ax} = \begin{bmatrix} 2 & 3 \\ 0 & -5 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 1 \\ 15 \end{bmatrix} $$so $x_2$ $=$ $-3$ and $2x_1$ $+$ $3x_2$ $=$ $2x_1$ $+$ $3(-3)$ $=$ $2x_1$ $-$ $9$ $=$ $1$, which shows that $x_1$ $=$ $5$.
25 The matrix
$$ \mathbf{Ab} = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 5 & 7 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 6 \end{bmatrix} $$has no solution because $\mathbf{A}$ is singular. Row three is a sum of the first two. But when bottom-right $7$ is changed to $6$, then the system has solution. This can be shown by solving the augmented matrix
$$ \begin{array}{c l} \left[ \begin{array}{c | c} \mathbf{A} & \mathbf{b} \end{array} \right] = & \left[ \begin{array}{c c c | c} 1 & 2 & 3 & 1 \\ 2 & 3 & 4 & 2 \\ 3 & 5 & 6 & 6 \end{array} \right] \\[0.5em] \xrightarrow{R_2 - 2R_1} & \left[ \begin{array}{c c c | c} 1 & 2 & 3 & 1 \\ & -1 & -2 & 0 \\ 3 & 5 & 6 & 6 \end{array} \right] \\[0.5em] \xrightarrow{R_3 - 3R_1} & \left[ \begin{array}{c c c | c} 1 & 2 & 3 & 1 \\ & -1 & -2 & 0 \\ & -1 & -3 & 3 \end{array} \right], \\[0.5em] \xrightarrow{R_3 + R_2} & \left[ \begin{array}{c c c | c} 1 & 2 & 3 & 1 \\ & -1 & -2 & 0 \\ & & -1 & 3 \end{array} \right], \end{array} $$so $z$ $=$ $-3$, and substituting $z$ back to second row leads to $-y$ $-$ $2(-3)$ $=$ $-y$ $+$ $6$ $=$ $0$, revealing that $y$ $=$ $6$. Finally solving the first row shows that $x$ $+$ $2y$ $+$ $3z$ $=$ $x$ $+$ $2(6)$ $+$ $3(-3)$ $=$ $x$ $+$ $12$ $-$ $9$ $=$ $x$ $+$ $3$ $=$ $1$, so $x$ $=$ $-2$. Substituting $x$, $y$ and $z$ to $\mathbf{b}$ and taking the product shows that
$$ \mathbf{Ax} = \begin{bmatrix} 1 \\ 2 \\ 6 \end{bmatrix}. $$26 Solving the equations
$$ \begin{bmatrix} 1 & 4 \\ 2 & 7 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \qquad \begin{bmatrix} 1 & 4 \\ 2 & 7 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} $$at the same time can be used by solving by using “doubly augmented” matrix
$$ \left[ \begin{array}{c c | c c} 1 & 4 & 1 & 0 \\ 2 & 7 & 0 & 1 \end{array} \right]. $$The solution starts by subtracting the first row from the second twice. This means that the matrix becomes
$$ \left[ \begin{array}{c c | c c} 1 & 4 & 1 & 0 \\ 0 & -1 & -2 & 1 \end{array} \right]. $$revealing that $y$ $=$ $2$ and $v$ $=$ $-1$. Substituting $y$ and $v$ back to the first equations gives two equations:
$$ \begin{array}{ c l c c l } & x + 4y & \quad & & u + 4v \\ = & x + 4(2) & \quad & = & u + 4(-1) \\ = & x + 8 & \quad & = & u - 4 \\ = & 1 & \quad & = & 0 \end{array} $$From these outcomes it can be seen that $x$ $=$ $-7$ and $u$ $=$ $4$.
27 If augmented matrix is
$$ \left[ \begin{array}{ c c c c } 1 & 2 & 3 & a \\ 0 & 4 & 5 & b \\ 0 & 0 & d & c \end{array} \right], $$then (a) if $d$ $=$ $0$ anc $c$ $\neq$ $0$, the matrix of the augmented matrix does not have solutions. (b) If the $c$ something else than zero, the matrix has infinitely many solutions.
28 Since $\mathbf{AB}$ $=$ $\mathbf{I}$ and $\mathbf{BC}$ $=$ $\mathbf{I}$, then $\mathbf{A}$ $=$ $\mathbf{AI}$ $=$ $\mathbf{A}(\mathbf{BC})$ $=$ $(\mathbf{AB})\mathbf{C}$ $=$ $\mathbf{IC}$ $=$ $\mathbf{C}$.