1. **Problem statement:** Determine for which values of $k \in \mathbb{R}$ the matrix $$ A = \begin{pmatrix} 3 & -2 \\ 2 & 2 \end{pmatrix} $$ belongs to the span $$ W = \text{Span}\left( M_1 = \begin{pmatrix} 2 & 0 \\ 2 & 0 \end{pmatrix}, M_2 = \begin{pmatrix} 1 & k \\ 0 & -k \end{pmatrix}, M_3 = \begin{pmatrix} k & 6 \\ 1 & -6 \end{pmatrix} \right). $$ 2. **Formula and approach:** To check if $A \in W$, we want scalars $a,b,c \in \mathbb{R}$ such that $$ A = a M_1 + b M_2 + c M_3. $$ This means $$ \begin{pmatrix} 3 & -2 \\ 2 & 2 \end{pmatrix} = a \begin{pmatrix} 2 & 0 \\ 2 & 0 \end{pmatrix} + b \begin{pmatrix} 1 & k \\ 0 & -k \end{pmatrix} + c \begin{pmatrix} k & 6 \\ 1 & -6 \end{pmatrix}. $$ 3. **Set up the system of equations:** Equate each entry: - Top-left: $3 = 2a + b + ck$ - Top-right: $-2 = 0a + bk + 6c = bk + 6c$ - Bottom-left: $2 = 2a + 0b + c = 2a + c$ - Bottom-right: $2 = 0a - bk - 6c = -bk - 6c$ 4. **Simplify the system:** From top-right and bottom-right equations: $$ -2 = bk + 6c \\ 2 = -bk - 6c $$ Add these two equations: $$ -2 + 2 = (bk - bk) + (6c - 6c) = 0 $$ which is true for all $b,c$. So these two equations are dependent. Rewrite bottom-right: $$ 2 = -bk - 6c \implies bk + 6c = -2 $$ But top-right says $bk + 6c = -2$ as well, so consistent. 5. **From bottom-left:** $$ 2 = 2a + c \implies c = 2 - 2a $$ 6. **Substitute $c$ into top-left:** $$ 3 = 2a + b + k(2 - 2a) = 2a + b + 2k - 2ak = b + 2a(1 - k) + 2k $$ Rearranged: $$ b = 3 - 2a(1 - k) - 2k $$ 7. **Substitute $b$ and $c$ into $bk + 6c = -2$:** $$ (3 - 2a(1 - k) - 2k)k + 6(2 - 2a) = -2 $$ Expand: $$ 3k - 2a k (1 - k) - 2k^2 + 12 - 12a = -2 $$ Group $a$ terms: $$ -2a k (1 - k) - 12a + 3k - 2k^2 + 12 = -2 $$ Move constants: $$ -2a k (1 - k) - 12a = -2 - 3k + 2k^2 - 12 = -14 - 3k + 2k^2 $$ Factor $a$: $$ a(-2k(1-k) - 12) = -14 - 3k + 2k^2 $$ Simplify inside parentheses: $$ -2k + 2k^2 - 12 = 2k^2 - 2k - 12 $$ So: $$ a(2k^2 - 2k - 12) = -14 - 3k + 2k^2 $$ 8. **Solve for $a$ when denominator nonzero:** $$ a = \frac{-14 - 3k + 2k^2}{2k^2 - 2k - 12} $$ 9. **Check denominator zero cases:** $$ 2k^2 - 2k - 12 = 0 \implies k^2 - k - 6 = 0 $$ Solve quadratic: $$ k = \frac{1 \pm \sqrt{1 + 24}}{2} = \frac{1 \pm 5}{2} $$ So $k=3$ or $k=-2$. 10. **For $k=3$:** Denominator zero, check if numerator zero: $$ -14 - 3(3) + 2(9) = -14 - 9 + 18 = -5 \neq 0 $$ No solution for $a$, so no $a,b,c$ exist. $A \notin W$. 11. **For $k=-2$:** Numerator: $$ -14 - 3(-2) + 2(4) = -14 + 6 + 8 = 0 $$ So $a$ can be any real number. Infinite solutions. 12. **Summary:** - For $k \neq 3, -2$, $a,b,c$ exist uniquely and $A \in W$. - For $k=3$, no solution. - For $k=-2$, infinite solutions. --- **Final answer:** $$ \boxed{ \begin{cases} A \in W & \text{if } k \neq 3 \\ A \notin W & \text{if } k = 3 \end{cases} } $$ --- **Slug:** matrix span **Subject:** linear algebra **Desmos:** {"latex":"","features":{"intercepts":true,"extrema":true}} **q_count:** 1