(a) left[begin{array}{llll}1 & t+2 & 3 & -6end{array}right]=3left[begin{array}{llll}r & -1 & 1 & -2end{array}right]

Question

Accepted Answer

Step:1 ：Given the equation: (left[begin{array}{llll}1 & t+2 & 3 & -6end{array}right]=3left[begin{array}{llll}r & -1 & 1 & -2end{array}right])Step:2 ：Equating the corresponding elements of the two vectors:Step:3 ：(1 = 3r)Step:4 ：(t + 2 = -3)Step:5 ：Solving for r and t:Step:6 ：(r = frac{1}{3})Step:7 ：(t = -5)Step:8 ：Final Answer: (r = boxed{frac{1}{3}}) and (t = boxed{-5})

Answer

Step: 1 ：Given the equation: (left[begin{array}{llll}1 & t+2 & 3 & -6end{array}right]=3left[begin{array}{llll}r & -1 & 1 & -2end{array}right])Step: 2 ：Equating the corresponding elements of the two vectors:Step: 3 ：(1 = 3r)Step: 4 ：(t + 2 = -3)Step: 5 ：Solving for r and t:Step: 6 ：(r = frac{1}{3})Step: 7 ：(t = -5)Step: 8 ：Final Answer: (r = boxed{frac{1}{3}}) and (t = boxed{-5})

(a) $[\begin{array}{llll} 1 & t + 2 & 3 & - 6 \end{array}] = 3 [\begin{array}{llll} r & - 1 & 1 & - 2 \end{array}]$

Answer

Popular Problems

(a) [1t+23−6]=3[r−11−2]

Answer

Popular Problems

(a) $[\begin{array}{llll} 1 & t + 2 & 3 & - 6 \end{array}] = 3 [\begin{array}{llll} r & - 1 & 1 & - 2 \end{array}]$