Q.If $A = \begin{bmatrix} 3 & 3 & 2 \\ 4 & 2 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & -1 & 2 \\ 1 & 2 & 4 \end{bmatrix}$, verify that

Concept: Matrix Transpose — swapping rows and columns.

(i) The transpose of $A$ is $A' = \begin{bmatrix} 3 & 4 \\ 3 & 2 \\ 2 & 0 \end{bmatrix}$. Taking its transpose again gives back $A$:

(ii) First, $A + B = \begin{bmatrix} 5 & 2 & 4 \\ 5 & 4 & 4 \end{bmatrix}$, so $(A+B)' = \begin{bmatrix} 5 & 5 \\ 2 & 4 \\ 4 & 4 \end{bmatrix}$.

Now $A' + B' = \begin{bmatrix} 3 & 4 \\ 3 & 2 \\ 2 & 0 \end{bmatrix} + \begin{bmatrix} 2 & 1 \\ -1 & 2 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} 5 & 5 \\ 2 & 4 \\ 4 & 4 \end{bmatrix}$.

Both are equal.

(iii) For any constant $k$, $kB = \begin{bmatrix} 2k & -k & 2k \\ k & 2k & 4k \end{bmatrix}$, so $(kB)' = \begin{bmatrix} 2k & k \\ -k & 2k \\ 2k & 4k \end{bmatrix}$.

And $kB' = k \begin{bmatrix} 2 & 1 \\ -1 & 2 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} 2k & k \\ -k & 2k \\ 2k & 4k \end{bmatrix}$.

They match.

The transpose of a matrix is one of those operations that feels almost too simple — just swap rows and columns — yet it obeys a clean set of algebraic rules that mirror what you'd expect from a well-behaved operation. The three properties here are the transpose analogues of what happens with addition and scalar multiplication: they're linearity properties. Let's verify each one by working directly with the given matrices.

1. Verify $(A')' = A$

First compute $A'$. Since $A$ is $2 \times 3$, its transpose will be $3 \times 2$. Take each row of $A$ and write it as a column:

Now transpose $A'$: take its rows and make them columns again.

That's exactly $A$. So $(A')' = A$ holds. This is always true — transposing twice undoes itself.

2. Verify $(A + B)' = A' + B'$

First compute $A + B$. Both matrices are $2 \times 3$, so addition is element-wise:

Now transpose this sum:

Next compute $A' + B'$ separately. We already have $A'$ from above. Find $B'$:

Now add $A'$ and $B'$:

This matches $(A + B)'$ exactly. So $(A + B)' = A' + B'$ is verified.

3. Verify $(kB)' = kB'$, where $k$ is any constant

Let $k$ be any real number. First compute $kB$: multiply every entry of $B$ by $k$.

Now transpose this:

Next compute $kB'$: first find $B'$ (already done above), then multiply by $k$:

Both results are identical. So $(kB)' = kB'$ holds for any constant $k$.

All three properties are verified by direct computation. The key insight is that transposing simply re-arranges entries without changing their values, so any operation that acts entry-wise (like addition or scalar multiplication) naturally commutes with it.

Method: Verifying transpose properties by direct computation

Use this when asked to verify identities such as $(A')' = A$, $(A+B)' = A' + B'$, and $(kB)' = kB'$.

Steps

Step 1: Recall the transpose swaps rows and columns.

$A'$ turns each row of $A$ into a column; an $m\times n$ matrix becomes $n\times m$.

Step 2: Compute both sides of each identity independently.

For $(A+B)' = A' + B'$: transpose the sum on one side; transpose separately and add on the other.

Step 3: Compare entry-by-entry.

Matching results verify the property. These identities reflect that transpose is a linear operation.

Common Mistakes

Mistake 1: Applying the reversal law to a sum.

Why it's wrong: for a sum, $(A+B)' = A' + B'$ (no order change); the reversal $B'A'$ applies only to a product $(AB)'$. Correct approach: don't swap order for sums or scalar multiples.

Mistake 2: Forgetting to transpose the shape.

Why it's wrong: $A$ is $2\times3$, so $A'$ is $3\times2$; keeping the original shape mis-places entries. Correct approach: rows become columns, so the order flips.

Mistake 3: Mishandling the scalar in $(kB)'$.

Why it's wrong: $k$ factors straight through: $(kB)' = kB'$; it is neither dropped nor changed. Correct approach: transpose $B$ and multiply by the same $k$.