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Linear Algebra · Question of the Day

Linear Algebra Question of the Day

A fresh daily question to build accuracy, reinforce recall, and turn practice into a steady habit.
Friday, October 9, 2026

Let B1={(1,0),(0,1)}B_1 = \{(1, 0), (0, 1)\} and B2={(3,1),(2,1)}B_2 = \{(3, 1), (2, 1)\} be two bases for R2\mathbb{R}^2. If the coordinate vector of v⃗\vec{v} with respect to B2B_2 is [v⃗]B2=(−25)[\vec{v}]_{B_2} = \begin{pmatrix} -2 \\ 5 \end{pmatrix} , what is [v⃗]B1[\vec{v}]_{B_1}?

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Question of the Day

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Let B1={(1,0),(0,1)}B_1 = \{(1, 0), (0, 1)\} and B2={(3,1),(2,1)}B_2 = \{(3, 1), (2, 1)\} be two bases for R2\mathbb{R}^2. If the coordinate vector of v⃗\vec{v} with respect to B2B_2 is [v⃗]B2=(−25)[\vec{v}]_{B_2} = \begin{pmatrix} -2 \\ 5 \end{pmatrix} , what is [v⃗]B1[\vec{v}]_{B_1}?

  1. (43)\begin{pmatrix} 4 \\ 3 \end{pmatrix} (correct answer)
  2. (−6−2)\begin{pmatrix} -6 \\ -2 \end{pmatrix}
  3. (4−3)\begin{pmatrix} 4 \\ -3 \end{pmatrix}
  4. (62)\begin{pmatrix} 6 \\ 2 \end{pmatrix}

Explanation: To find [v⃗]B1[\vec{v}]_{B_1}, we first express v⃗\vec{v} in standard coordinates: v⃗=−2(3,1)+5(2,1)=(−6,2)+(10,5)=(4,3)\vec{v} = -2(3,1) + 5(2,1) = (-6,2) + (10,5) = (4,3). Since B1B_1 is the standard basis, [v⃗]B1=(4,3)[\vec{v}]_{B_1} = (4,3). Choice B results from incorrectly computing v⃗=−2(3,1)−5(2,1)\vec{v} = -2(3,1) - 5(2,1). Choice C comes from sign error in the second component. Choice D results from computing 2(3,1)+5(2,1)−(4,1)2(3,1) + 5(2,1) - (4,1) through misreading the coefficients.