If a,b,c are real number, and D=`|{:(,a,1+2i,3-5i),(,1-2i,b,-7-3i),(,3+5i,-7+3i,c):}|` then show that D is purely real.

Answer

To show that the determinant \( D \) of the given matrix is purely real, we will follow these steps:

### Step 1: Write down the determinant
The determinant \( D \) is given by the matrix:

\[ D = \begin{vmatrix} a & 1 + 2i & 3 - 5i \\ 1 - 2i & b & -7 - 3i \\ 3 + 5i & -7 + 3i & c \end{vmatrix} \]

### Step 2: Find the conjugate of the determinant
To find the conjugate \( D^* \) of the determinant, we replace \( i \) with \( -i \):

\[ D^* = \begin{vmatrix} a & 1 - 2i & 3 + 5i \\ 1 + 2i & b & -7 + 3i \\ 3 - 5i & -7 - 3i & c \end{vmatrix} \]

### Step 3: Use the property of determinants
We know that the determinant of a matrix is equal to the determinant of its transpose. Therefore, we can write:

\[ D^* = \begin{vmatrix} a & 1 - 2i & 3 + 5i \\ 1 + 2i & b & -7 + 3i \\ 3 - 5i & -7 - 3i & c \end{vmatrix} \]

### Step 4: Compare \( D \) and \( D^* \)
Now, we can see that \( D \) and \( D^* \) have the same structure. If we denote \( D \) as \( D = x + iy \) where \( x \) and \( y \) are real numbers, we can express the conjugate as:

\[ D^* = x - iy \]

### Step 5: Set \( D = D^* \)
Since \( D = D^* \), we have:

\[ x + iy = x - iy \]

This implies that the imaginary part \( y \) must be zero:

\[ y = 0 \]

### Step 6: Conclude that \( D \) is purely real
Since the imaginary part \( y \) is zero, we conclude that \( D \) is purely real:

\[ D = x \]

Thus, we have shown that the determinant \( D \) is purely real.