If `A=[{:(0,c,-b),(-c,0,a),(b,-a,0):}]`and `B=[{:(a^(2),ab,ac),(ab,b^(2),bc),(ac,bc,c^(2)):}]`, then `(A+B)^(2)=`

Options

A. `A`
B. `B`
C. `I`
D. `A^(2)+B^(2)`

Explanation

To solve the problem, we need to find \((A + B)^2\) where:

\[ A = \begin{pmatrix} 0 & c & -b \\ -c & 0 & a \\ b & -a & 0 \end{pmatrix} \]

\[ B = \begin{pmatrix} a^2 & ab & ac \\ ab & b^2 & bc \\ ac & bc & c^2 \end{pmatrix} \]

### Step 1: Calculate \(A + B\)

First, we will add the matrices \(A\) and \(B\):

\[ A + B = \begin{pmatrix} 0 + a^2 & c + ab & -b + ac \\ -c + ab & 0 + b^2 & a + bc \\ b + ac & -a + bc & 0 + c^2 \end{pmatrix} \]

This simplifies to:

\[ A + B = \begin{pmatrix} a^2 & c + ab & ac - b \\ ab - c & b^2 & a + bc \\ b + ac & bc - a & c^2 \end{pmatrix} \]

### Step 2: Calculate \((A + B)^2\)

Now, we need to compute \((A + B)^2 = (A + B)(A + B)\). We will denote \(C = A + B\):

\[ C = \begin{pmatrix} a^2 & c + ab & ac - b \\ ab - c & b^2 & a + bc \\ b + ac & bc - a & c^2 \end{pmatrix} \]

To compute \(C^2\), we perform matrix multiplication:

\[ C^2 = C \cdot C \]

The element in the \(i^{th}\) row and \(j^{th}\) column of \(C^2\) is given by:

\[ (C^2)_{ij} = \sum_{k=1}^{3} C_{ik} C_{kj} \]

We will calculate each element of the resulting matrix \(C^2\).

### Step 3: Calculate each element of \(C^2\)

1. **First Row:**
- \( (C^2)_{11} = a^2 \cdot a^2 + (c + ab)(ab - c) + (ac - b)(b + ac) \)
- \( (C^2)_{12} = a^2(c + ab) + (c + ab)b^2 + (ac - b)(bc - a) \)
- \( (C^2)_{13} = a^2(ac - b) + (c + ab)(a + bc) + (ac - b)c^2 \)

2. **Second Row:**
- \( (C^2)_{21} = (ab - c)a^2 + b^2(ab - c) + (a + bc)(b + ac) \)
- \( (C^2)_{22} = (ab - c)(c + ab) + b^2b^2 + (a + bc)(bc - a) \)
- \( (C^2)_{23} = (ab - c)(ac - b) + b^2(a + bc) + (a + bc)c^2 \)

3. **Third Row:**
- \( (C^2)_{31} = (b + ac)a^2 + (bc - a)(ab - c) + c^2(b + ac) \)
- \( (C^2)_{32} = (b + ac)(c + ab) + (bc - a)b^2 + c^2(a + bc) \)
- \( (C^2)_{33} = (b + ac)(ac - b) + (bc - a)(a + bc) + c^2c^2 \)

### Step 4: Simplifying the Result

After calculating each element, we will notice that due to the properties of matrices \(A\) and \(B\) being skew-symmetric and symmetric respectively, the terms involving \(AB\) and \(BA\) will yield zero when multiplied. Therefore, the final result simplifies to:

\[ (A + B)^2 = A^2 + B^2 \]

### Conclusion

Thus, the final answer is:

\[ (A + B)^2 = A^2 + B^2 \]