Q.Find $\frac{1}{2}(A + A')$ and $\frac{1}{2}(A - A')$, when $A = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix}$

Concept: Symmetric and Skew-Symmetric Matrices — any square matrix can be expressed as the sum of a symmetric matrix and a skew-symmetric matrix.

Step 1: Compute $A'$ (the transpose of $A$).

Since $A$ is skew-symmetric (all diagonal entries are $0$, and $a_{ij} = -a_{ji}$), its transpose is:

Step 2: Find $\frac{1}{2}(A + A')$.

Add $A$ and $A'$:

Thus $\frac{1}{2}(A + A')$ is the zero matrix.

Step 3: Find $\frac{1}{2}(A - A')$.

Subtract $A'$ from $A$: …

Why This Approach Works

Every square matrix can be uniquely split into a symmetric part and a skew-symmetric part. A symmetric matrix equals its own transpose ($S = S'$), while a skew-symmetric matrix equals the negative of its transpose ($K = -K'$). The formulas $\frac{1}{2}(A + A')$ and $\frac{1}{2}(A - A')$ are not arbitrary — they are derived from solving the system $A = S + K$ where $S$ is symmetric and $K$ is skew-symmetric. Take the transpose of both sides: $A' = S - K$, then add and subtract to isolate $S$ and $K$.

The beauty is that this decomposition works for any square matrix, but when $A$ itself has special symmetry, the result simplifies dramatically.

Step-by-Step Solution

1. Compute $A'$, the transpose of $A$.

Transpose means swapping rows and columns. For $A = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix}$, the first row $(0, a, b)$ becomes the first column, and so on:

Notice something interesting: $A'$ is exactly $-A$. Check: multiply $A$ by $-1$ and you get the same matrix. This means $A$ is skew-symmetric by definition.

2. Find the symmetric part: $\frac{1}{2}(A + A')$.

Add $A$ and $A'$:

Add entry by entry:

• $(1,1)$: $0 + 0 = 0$
• $(1,2)$: $a + (-a) = 0$
• $(1,3)$: $b + (-b) = 0$
• $(2,1)$: $(-a) + a = 0$
• $(2,2)$: $0 + 0 = 0$
• $(2,3)$: $c + (-c) = 0$
• $(3,1)$: $(-b) + b = 0$
• $(3,2)$: $(-c) + c = 0$
• $(3,3)$: $0 + 0 = 0$

Every entry cancels to zero. So $A + A' = \mathbf{0}$, the zero matrix. Therefore:

The symmetric part is the zero matrix.

3. Find the skew-symmetric part: $\frac{1}{2}(A - A')$.

Subtract $A'$ from $A$:

Subtract entry by entry (remember: subtracting a negative is adding):

• $(1,1)$: $0 - 0 = 0$ …

Method: Splitting a matrix into its symmetric and skew-symmetric parts

The matrices $\tfrac12(A+A')$ and $\tfrac12(A-A')$ are the symmetric and skew-symmetric parts of $A$; compute $A'$ first, then form each combination and halve.

Steps

Step 1: Write the transpose $A'$

Interchange rows and columns.

Step 2: Form $\tfrac12(A+A')$ and $\tfrac12(A-A')$

Add (or subtract) entry-wise, then halve every entry. …

Common Mistakes

Mistake 1: Sign errors when transposing a skew-symmetric matrix

Why it's wrong: for this $A$, $A'=-A$; miswriting a sign in $A'$ makes $A+A'$ fail to vanish. Correct approach: transpose carefully — each off-diagonal entry both moves position and, for this matrix, is the negative of its partner.

Mistake 2: Forgetting the factor $\tfrac12$ …