Question:
If \(A=\begin{bmatrix}1&2&3\\2&3&1\end{bmatrix}\) and \(B=\begin{bmatrix}3&-1&3\\-1&0&2\end{bmatrix}\), then find the value of \((2A-B)\).
If \(A=\begin{bmatrix}1&2&3\\2&3&1\end{bmatrix}\) and \(B=\begin{bmatrix}3&-1&3\\-1&0&2\end{bmatrix}\), then find the value of \((2A-B)\).
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Compute \(2A\) entrywise, then subtract \(B\) entrywise.
Updated On: Sep 23, 2026
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Solution and Explanation
Step 1: Understanding the Concept:
Scale \(A\) by 2 entrywise, then subtract \(B\) entrywise — both matrices are the same \(2\times3\) shape, so this is well-defined.
Step 2: Computing 2A:
\(2A=\begin{bmatrix}2&4&6\\4&6&2\end{bmatrix}\).
Step 3: Subtracting B entrywise:
\(2A-B=\begin{bmatrix}2-3&4-(-1)&6-3\\4-(-1)&6-0&2-2\end{bmatrix}=\begin{bmatrix}-1&5&3\\5&6&0\end{bmatrix}\).
Final Answer:
\(2A-B=\boxed{\begin{bmatrix}-1&5&3\\5&6&0\end{bmatrix}}\).
Scale \(A\) by 2 entrywise, then subtract \(B\) entrywise — both matrices are the same \(2\times3\) shape, so this is well-defined.
Step 2: Computing 2A:
\(2A=\begin{bmatrix}2&4&6\\4&6&2\end{bmatrix}\).
Step 3: Subtracting B entrywise:
\(2A-B=\begin{bmatrix}2-3&4-(-1)&6-3\\4-(-1)&6-0&2-2\end{bmatrix}=\begin{bmatrix}-1&5&3\\5&6&0\end{bmatrix}\).
Final Answer:
\(2A-B=\boxed{\begin{bmatrix}-1&5&3\\5&6&0\end{bmatrix}}\).
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