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.

Thursday, September 17, 2026

Let $T: \mathbb{R}^2 \to \mathbb{R}^3$ be defined by $T(x_1, x_2) = (x_1 - x_2, 3x_2, x_1+x_2)$ and $S: \mathbb{R}^3 \to \mathbb{R}^2$ be defined by $S(y_1, y_2, y_3) = (y_1+y_3, y_2-y_3)$ . What is the formula for the composition $(S \circ T)(x_1, x_2)$ ?

Keep practicing Linear Algebra

Question of the Day

Answer today's Linear Algebra question, reveal the full explanation, then keep the streak going with a new question every day.

$(2x_1, 2x_2 - x_1)$ (correct answer)
$(-2x_2, 4x_2)$
$(2x_1, 2x_2 + x_1)$
The composition is not defined.

Explanation: To find the formula for $(S \circ T)(x_1, x_2)$ , we first apply $T$ to $(x_1, x_2)$ and then apply $S$ to the result. $(S \circ T)(x_1, x_2) = S(T(x_1, x_2)) = S(x_1 - x_2, 3x_2, x_1+x_2)$ Now we use the definition of $S$ with $y_1 = x_1 - x_2$ , $y_2 = 3x_2$ , and $y_3 = x_1+x_2$ . $S(y_1, y_2, y_3) = (y_1+y_3, y_2-y_3)$ Substituting the expressions in terms of $x_1$ and $x_2$ : $((x_1 - x_2) + (x_1+x_2), (3x_2) - (x_1+x_2))$ Simplifying each component gives: $(2x_1, 3x_2 - x_1 - x_2) = (2x_1, -x_1 + 2x_2)$ The correct answer is $(2x_1, 2x_2 - x_1)$ . B is incorrect: This may result from calculation errors, for example, calculating the first component as $(x_1 - x_2) - (x_1+x_2) = -2x_2$ . C is incorrect: This results from a sign error in the second component: $(3x_2) + (x_1+x_2) = x_1+4x_2$ , which is also not the choice, but sign errors are common. This specific choice may come from $(3x_2)- (x_1-x_2)$ instead of $(3x_2)-(x_1+x_2)$ . D is incorrect: The composition is well-defined because the codomain of $T$ ( $\mathbb{R}^3$ ) matches the domain of $S$ ( $\mathbb{R}^3$ ).