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krispy kreme donuts 88 cents - Coaching Toolbox

📅 May 4, 2026 👤 scraface

May 04, 2026

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📌 Set this equal to $\mathbf{d} = egin{pmatrix} -4 \ 5 \ 1 \end{pmatrix}$, leading to the system of equations:

📌 $3v_2 + v_3 = -4$

📌 $-3v_1 + 2v_3 = 5$

📌 $-v_1 - 2v_2 = 1$

📌 From equation (3): $v_1 = -1 - 2v_2$. Substitute into equation (2):

📌 -3(-1 - 2v_2) + 2v_3 = 5 \implies 3 + 6v_2 + 2v_3 = 5 \implies 6v_2 + 2v_3 = 2 \implies 3v_2 + v_3 = 1

📌 But equation (1) states $3v_2 + v_3 = -4$. This inconsistency implies no solution exists unless $\mathbf{d}$ is orthogonal to $\mathbf{c}$. Check $\mathbf{d} \cdot \mathbf{c} = (-4)(2) + (5)(-1) + (1)(3) = -8 -5 +3 = -10

📌 eq 0$, confirming no solution exists. Thus, $oxed{ ext{No such vector } \mathbf{v} ext{ exists}}$.

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