Q.If $A' = \begin{bmatrix} -2 & 3 \\ 1 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix}$, then find $(A + 2B)'$

Concept: Matrix Transpose — the transpose of a sum equals the sum of the transposes: $(A+2B)' = A' + 2B'$.

Step 1: Compute $2B'$. First, $B' = \begin{bmatrix} -1 & 1 \\ 0 & 2 \end{bmatrix}$, so

$2B' = \begin{bmatrix} -2 & 2 \\ 0 & 4 \end{bmatrix}$.

Step 2: Add $A'$ and $2B'$: …

The transpose operation distributes over addition and scalar multiplication, so we can work directly with the transposes we are given, without recovering $A$ itself.

1. Given data.

2. Transpose $B$. Swap rows and columns:

3. Scale. …

Method: Computing a transpose of a combination using linearity

To find $(A+2B)'$ you do not need $A$ itself — the transpose is linear, so $(A+kB)'=A'+kB'$, letting you work straight from $A'$ and $B$.

Steps

Step 1: Apply the linearity rule

Write $(A+2B)' = A' + 2B'$, so only $A'$ (given) and $B'$ are needed.

Step 2: Transpose and scale $B$ …

Common Mistakes

Mistake 1: Reconstructing $A$ and transposing again unnecessarily

Why it's wrong: since $(A+2B)'=A'+2B'$, using the given $A'$ directly is faster and less error-prone than recovering $A$. Correct approach: apply the linearity of the transpose.

Mistake 2: Doing only one of "transpose $B$" and "scale by 2" …