Given $\mathbf { a } = \langle 1,1 \rangle$ , $\mathbf { b } = \langle - 4,2 \rangle$ , and $\mathbf { c } = \langle 5,2 \rangle$ , find s and t such that c = sa + tb.

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The Answer of Given $\mathbf { a } = \langle 1,1 \rangle$ , $\mathbf { b } = \langle - 4,2 \rangle$ , and $\mathbf { c } = \langle 5,2 \rangle$ , find s and t such that c = sa + tb.

Given a=⟨1,1⟩\mathbf { a } = \langle 1,1 \ranglea=⟨1,1⟩ b=⟨−4,2⟩\mathbf { b } = \langle - 4,2 \rangleb=⟨−4,2⟩

Given $\mathbf { a } = \langle 1,1 \rangle$ $\mathbf { b } = \langle - 4,2 \rangle$