Q.For the matrix $A = \begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix}$, verify that

The key idea is that for any square matrix $A$, the sum $A + A'$ is always symmetric, and the difference $A - A'$ is always skew-symmetric. We verify this by direct computation.

First, find the transpose $A' = \begin{bmatrix} 1 & 6 \\ 5 & 7 \end{bmatrix}$.

(i) Compute $A + A' = \begin{bmatrix} 1+1 & 5+6 \\ 6+5 & 7+7 \end{bmatrix} = \begin{bmatrix} 2 & 11 \\ 11 & 14 \end{bmatrix}$.

Its transpose is itself, so it is symmetric. …

The Core Idea

This problem is not about doing random matrix arithmetic — it's about seeing a beautiful structural fact. Every square matrix can be split into two special parts: a symmetric part and a skew-symmetric part. The symmetric part is $(A + A')/2$ and the skew-symmetric part is $(A - A')/2$. Here, we're just checking the numerators before halving.

Why does this work? Because transposition flips the roles. When you add $A$ and $A'$, the off-diagonal entries add up to the same number in symmetric positions — that's exactly what symmetry means. When you subtract, the off-diagonals become negatives of each other, which is the definition of skew-symmetry.

Step-by-Step Verification

1. Write down $A$ and find $A'$

We have:

The transpose swaps rows and columns:

Notice how the off-diagonal entries $5$ and $6$ have swapped places.

2. Compute $A + A'$

Add entry by entry:

Now check symmetry: a matrix is symmetric if it equals its own transpose. Look at the off-diagonals — both are $11$. The transpose of this matrix is:

It's identical to the original. So $A + A'$ is symmetric. Done.

3. Compute $A - A'$

Subtract entry by entry:

Now check skew-symmetry: a matrix is skew-symmetric if its transpose equals its negative. Take the transpose: …

Method: Verifying $A+A'$ is symmetric and $A-A'$ is skew symmetric

For any square matrix, $A+A'$ is always symmetric and $A-A'$ is always skew symmetric; verify by forming each combination and comparing it with its own transpose.

Steps

Step 1: Write $A'$ and form the combinations

Compute $A'$, then $A+A'$ and $A-A'$ entry-wise.

Step 2: Transpose each combination …

Common Mistakes

Mistake 1: Expecting the diagonal of $A-A'$ to be non-zero

Why it's wrong: $A-A'$ is skew symmetric, so every diagonal entry must be 0 (here both are); a non-zero diagonal signals an arithmetic slip. Correct approach: verify the diagonal entries cancel to 0.

Mistake 2: A sign error when transposing $A-A'$ …