a) Count the number of warrior types on the image, sum them up and determine how much gold they get combined.
b) Hagar is the one with big dress and orange nose looking prideful :)

a) \(A = 1\), \(S = 3\), \(B = 4\).
b) \(\mathbb{L} = {(5 - L, 2, L) \in \mathbb{R}^3 : \mathbb{L} \in \mathbb{R}}\)

a) We have
$$ \begin{array}{c c c c c c c} A & + & 2 S & + & L & = & 11 \\ A & + & S & + & 2 L & = & 12 \\ 2A & + & S & + & L & = & 9 \end{array} $$ such that the augmented matrix reads as
$$ \left( \begin{array}{ c c c | c } 1 & 2 & 1 & 11 \\ 1 & 1 & 2 & 12 \\ 2 & 1 & 1 & 9 \end{array} \right) $$ Gaussian elimination: $$ \begin{array}{c l} % \left( \begin{array}{ c c c | c} 1 & 2 & 1 & 11 \\ 1 & 1 & 2 & 12 \\ 2 & 1 & 1 & 9 \end{array} \right) & \begin{array}{l} \phantom{u} \\ \text{$|$ $(II) - (I)$}\\ \text{$|$ $(III) - 2 \cdot (I)$}\\ \end{array} \\[2mm] \Downarrow \\[2mm] % \left( \begin{array}{ c c c | c} 1 & 2 & 1 & 11 \\ 0 & -1 & 1 & 1 \\ 0 & -3 & -1 & -13 \end{array} \right) & \begin{array}{l} \phantom{u} \\ \phantom{u} \\ \text{$|$ $- 1 \times$}\\ \end{array} \\[2mm] \Downarrow \\[2mm] % \left( \begin{array}{ c c c | c} 1 & 2 & 1 & 11 \\ 0 & -1 & 1 & 1 \\ 0 & 3 & 1 & 13 \end{array} \right) & \begin{array}{l} \phantom{u} \\ \phantom{u} \\ \phantom{u} \\ \end{array} % \end{array} $$ $$ \begin{array}{c l} % \left( \begin{array}{ c c c | c} 1 & 2 & 1 & 11 \\ 0 & -1 & 1 & 1 \\ 0 & 3 & 1 & 13 \end{array} \right) & \begin{array}{l} \phantom{u} \\ \phantom{u} \\ \text{$|$ $(III) + 3 \cdot (II)$}\\ \end{array} \\[2mm] \Downarrow \\[2mm] % \left( \begin{array}{ c c c | c} 1 & 2 & 1 & 11 \\ 0 & -1 & 1 & 1 \\ 0 & 0 & 4 & 16 \end{array} \right) & \begin{array}{l} \phantom{u} \\ \text{$|$ $-1 \times$} \\ \text{$|$ $\tfrac{1}{4} \times$}\\ \end{array} \\[2mm] \Downarrow \\[2mm] % \left( \begin{array}{ c c c | c} 1 & 2 & 1 & 11 \\ 0 & 1 & -1 & -1 \\ 0 & 0 & 1 & 4 \end{array} \right) \\[2mm] \Downarrow % \end{array} $$ $$ \begin{array}{c l} % \left( \begin{array}{ c c c | c} 1 & 2 & 1 & 11 \\ 0 & 1 & -1 & -1 \\ 0 & 0 & 1 & 4 \end{array} \right) & \begin{array}{l} \text{$|$ $(I) - (III)$}\\ \text{$|$ $(II) + (III)$}\\ \phantom{u} \\ \end{array} \\[2mm] \Downarrow \\[2mm] % \left( \begin{array}{ c c c | c} 1 & 2 & 0 & 7 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & 4 \end{array} \right) & \begin{array}{l} \text{$|$ $(I) - 2 (II)$}\\ \phantom{u} \\ \phantom{u} \\ \end{array} \\[2mm] \Downarrow \\[2mm] \left( \begin{array}{ c c c | c} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & 4 \end{array} \right) % \end{array} $$ Finally,
\begin{eqnarray*} A & = & 1 \\ S & = & 3 \\ L & = & 4 \\ \end{eqnarray*} b) When we inspect the linear system of equation that results form the 'Hagar is the boss' assumption, we have $$ \begin{array}{c c c c c c c} A & + & 3 S & + & L & = & 11 \\ 2A & + & S & + & 2 L & = & 12 \\ 2A & + & S & + & 2 L & = & 9 \end{array} $$ One way to resolve the situation is to instead consider the underdetermined system $$ \begin{array}{c c c c c c c} A & + & 3 S & + & L & = & 11 \\ 2A & + & S & + & 2 L & = & 12 \end{array} $$ Hence, we have the situation of infintely many solutions (which is the case when we have an underdetermined system anyway).
Let us use \(L\) as the free variable, then we get $$ A \, \, = \, \, 5 - L \quad \text{and} \quad S \, \, = \, \, 2 $$ and thus $$ \mathbb{L} \, \, = \, \, \left\{ (5-L , 2, L) \in \mathbb{R}^3 \, : \, L \, \in \, \mathbb{R} \right\} $$ which is the equation of a line in 3D space. $$ \begin{array}{c l} % \left( \begin{array}{ c c c | c} 1 & 3 & 1 & 11 \\ 2 & 1 & 2 & 12 \end{array} \right) & \begin{array}{l} \phantom{u} \\ \text{$|$ $(II) - 2 \cdot (I)$}\\ \end{array} \\[2mm] \Downarrow \\[2mm] % \left( \begin{array}{ c c c | c} 1 & 3 & 1 & 11 \\ 0 & -5 & 0 & -10 \end{array} \right) & \begin{array}{l} \phantom{u} \\ \text{$|$ $-\tfrac{1}{5} \times$} \\ \end{array} \\[2mm] \Downarrow \\[2mm] % \left( \begin{array}{ c c c | c} 1 & 3 & 1 & 11 \\ 0 & 1 & 0 & 2 \end{array} \right) & \begin{array}{l} \text{$|$ $(I) - 3 \cdot (II)$} \\ \phantom{u} \\ \end{array} \\[2mm] \Downarrow \\[2mm] % \left( \begin{array}{ c c c | c} 1 & 0 & 1 & 5 \\ 0 & 1 & 0 & 2 \end{array} \right) \end{array} $$

Bring the systems of equations first to upper echelon form (for the augmented matrix) and then interpret the result: no solution, exactly one, or infinitely many solutions. In particular, we have no solution when \(0 \cdot x = a\) where \(a\) is non-zero, infinitely many solutions when \(0 \cdot x = 0 \) and one solution when \(a \cdot x = b\) where \(a\) and \(b\) are not \(0\).

a) No solution: \(a\) ≠ \(-5\) & \(ab = 2\), ∞ many solutions: \(a = 5\) & \(ab = 2\), Unique solution: \(ab\) ≠ \(2\)
b) No solution: \(a\) ≠ \(-5\) & \(b = 2a\), ∞ many solutions: \(a = 5\) & \(b = -2a\), Unique solution: \(b\) ≠ \(-2a\)

a) For \(x + by = -1\) and \(ax + 2y = 5\) we have with \((II) - a \cdot (I)\)

Bring the systems of equations first to upper echelon form (for the augmented matrix) and then perform a back-substitution to gain the results (make sure you write down the elementary operations you use). Finally, interpret the result geometrically.

a) \((-8, 3)\)
b) \(\mathbb{L} = {(5 , s-4, s) \in \mathbb{R}^3 : s \in \mathbb{R}}\)

a) We have $$ \begin{array}{c l} % \left( \begin{array}{ c c | c} 1 & 5 & 7 \\ -2 & -7 & -5 \end{array} \right) & \begin{array}{l} \phantom{u} \\ \text{$|$ $(II) + 2 \cdot (I)$}\\ \end{array} \\[2mm] \Downarrow \\[2mm] % \left( \begin{array}{ c c | c} 1 & 5 & 7 \\ 0 & 3 & 9 \end{array} \right) & \begin{array}{l} \phantom{u} \\ \text{$|$ $\frac{1}{3} \times$} \\ \end{array} \\[2mm] \Downarrow \\[2mm] % \left( \begin{array}{ c c | c} 1 & 5 & 7 \\ 0 & 1 & 3 \end{array} \right) & \begin{array}{l} \text{$|$ $(I) - 5 \cdot (II)$} \\ \phantom{u} \\ \end{array} \\[2mm] \Downarrow \\[2mm] % \left( \begin{array}{ c c | c} 1 & 0 & -8 \\ 0 & 1 & 3 \end{array} \right) \end{array} $$ Hence, we have \((x_1,x_2) = (-8,3)\) as a candidate for the solution.
We validate this candidate via matrix-column multiplication: \begin{eqnarray*} \begin{pmatrix} 1 & 5\\ -2 & -7 \end{pmatrix} \begin{pmatrix} -8 \\ 3 \end{pmatrix} & = & \begin{pmatrix} 1 \cdot (-8) + 5 \cdot 3 \\ -2 \cdot (-8) + (-7) \cdot 3 \end{pmatrix} \\ & = & \begin{pmatrix} -8 + 15 \\ 16 -21 \end{pmatrix} \\ & = & \begin{pmatrix} 7 \\ -5 \end{pmatrix} \end{eqnarray*}
Thus, \((-8,3)\) is indeed the unique solution.

b) We have $$ \begin{array}{c l} % \left( \begin{array}{ c c c | c} 1 & 1 & -1 & 1 \\ 2 & 1 & -1 & 6 \\ 3 & 7 & -7 & -13 \end{array} \right) & \begin{array}{l} \phantom{u} \\ \text{$|$ $(II) - 2 \cdot (I)$}\\ \text{$|$ $(III) - 3 \cdot (I)$}\\ \end{array} \\[2mm] \Downarrow \\[2mm] % \left( \begin{array}{ c c c | c} 1 & 1 & -1 & 1 \\ 0 & -1 & 1 & 4 \\ 0 & 4 & -4 & -16 \end{array} \right) & \begin{array}{l} \phantom{u} \\ \phantom{u} \\ \text{$|$ $(III) + 4 \cdot (II)$} \\ \end{array} \\[2mm] \Downarrow \\[2mm] % \left( \begin{array}{ c c c | c} 1 & 1 & -1 & 1 \\ 0 & -1 & 1 & 4 \\ 0 & 0 & 0 & 0 \end{array} \right) \end{array} $$ Hence, we have the situation of infintely many solutions.
Let us use \(x_3 = s\) as the free variable, then we get $$ x_1 \, \, = \, \, 5 \quad \text{and} \quad x_2 \, \, = \, \, s-4 $$ and thus $$ \mathbb{L} \, \, = \, \, \left\{ (5 , s-4, s) \in \mathbb{R}^3 \, : \, s \, \in \, \mathbb{R} \right\} $$ which is the equation of a line in 3D space.
Finally, we perform the matrix-column multiplication to verify the solution: \begin{eqnarray*} \begin{pmatrix} 1 & 1 & -1 \\ 2 & 1 & -1 \\ 3 & 7 & -7 \end{pmatrix} \begin{pmatrix} 5 \\ s-4 \\ s \end{pmatrix} & = & \begin{pmatrix} 1 \cdot 5 + 1 \cdot (s-4) + (-1) \cdot s \\ 2 \cdot 5 + 1 \cdot (s-4) + (-1) \cdot s \\ 3 \cdot 5 + 7 \cdot (s-4) + (-7) \cdot s \end{pmatrix}\\[2mm] & = & \begin{pmatrix} 5 + s - 4 - s \\ 10 + s - 4 - s \\ 15 + 7s - 28 - 7s \end{pmatrix} \\[2mm] & = & \begin{pmatrix} 1 \\ 6 \\ -13 \end{pmatrix} \end{eqnarray*}