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:

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

and

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

### Step 1: Use the identity for \((A + B)^2\)

We know that:

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

### Step 2: Calculate \(AB\)

To find \(AB\), we multiply matrix \(A\) by matrix \(B\):

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

Calculating each element of the resulting matrix:

1. First row:
- First column: \(0 \cdot a^2 + c \cdot ab + (-b) \cdot ac = cab - bac = 0\)
- Second column: \(0 \cdot ab + c \cdot b^2 + (-b) \cdot bc = cb^2 - b^2c = 0\)
- Third column: \(0 \cdot ac + c \cdot bc + (-b) \cdot c^2 = c^2b - bc^2 = 0\)

2. Second row:
- First column: \(-c \cdot a^2 + 0 \cdot ab + a \cdot ac = -ca^2 + a^2c = 0\)
- Second column: \(-c \cdot ab + 0 \cdot b^2 + a \cdot bc = -cab + abc = 0\)
- Third column: \(-c \cdot ac + 0 \cdot bc + a \cdot c^2 = -ac^2 + ac^2 = 0\)

3. Third row:
- First column: \(b \cdot a^2 + (-a) \cdot ab + 0 \cdot ac = ba^2 - ab^2 = 0\)
- Second column: \(b \cdot ab + (-a) \cdot b^2 + 0 \cdot bc = ab^2 - ab^2 = 0\)
- Third column: \(b \cdot ac + (-a) \cdot bc + 0 \cdot c^2 = abc - abc = 0\)

Thus, we find:

\[ AB = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \]

### Step 3: Calculate \(A^2\) and \(B^2\)

Since \(AB\) is the zero matrix, we can simplify our expression for \((A + B)^2\):

\[ (A + B)^2 = A^2 + B^2 + 2 \cdot 0 = A^2 + B^2 \]

### Step 4: Conclusion

We conclude that:

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