Understanding Distributive Property of Matrices
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Beginner
Start here! Easy to understand
Beginner Explanation
Step 1: Write the expression $A(B + C)$. Example: let A = [[1,2],[0,-1]], B = [[3,0],[1,4]], C = [[-1,2],[2,1]]. Step 2: Compute B + C = [[2,2],[3,5]]. Step 3: Multiply A by (B+C): A(B+C) = [[1*2+2*3, 1*2+2*5],[0*2+(-1)*3, 0*2+(-1)*5]] = [[8,12],[-3,-5]].
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Practice Problems
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1
Quick Quiz
Single Choice Quiz
Beginner
If $A = \begin{bmatrix} 1 & 2 \\ 0 & -1 \end{bmatrix}$, $B = \begin{bmatrix} 0 & -1 \\ 1 & 1 \end{bmatrix}$, and $C = \begin{bmatrix} -2 & 0 \\ 0 & 1 \end{bmatrix}$, what is $A(B + C)$?
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2
Real-World Problem
Question Exercise
Intermediate
Teenager Scenario
You have two activity matrices: $A = \begin{bmatrix}2 & 3\\1 & 0\end{bmatrix}$, $B = \begin{bmatrix}1 & 2\\3 & 4\end{bmatrix}$, and $C = \begin{bmatrix}0 & 1\\2 & 1\end{bmatrix}$. Use the distributive property to compute $A(B + C)$. Ensure dimensions match (all are 2×2).
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3
Thinking Challenge
Thinking Exercise
Intermediate
Think About This
Given matrices $A (2×3) = \begin{bmatrix}1 & 0 & 2\\0 & 3 & -1\end{bmatrix}$, $B (3×2) = \begin{bmatrix}2 & 1\\0 & -1\\4 & 3\end{bmatrix}$, and $C (3×2) = \begin{bmatrix}-1 & 2\\3 & 0\\1 & 4\end{bmatrix}$, verify the distributive property by computing both $A(B + C)$ and $AB + AC$.
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4
Challenge Quiz
Single Choice Quiz
Advanced
Given matrices $A$, $B$, and $C$, which expression correctly uses the distributive property: $(B + C)A$?
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