Question

Find the value of the determinant \( \begin{vmatrix} a^2+1 & ab & ac \\ ab & b^2+1 & bc \\ ca & cb & c^2+1 \end{vmatrix} \).

Hint:

For determinants with algebraic entries, always look for simplifications through row/column operations before attempting a direct expansion. The trick of multiplying rows by variables and then taking those variables common from columns is very effective.

Answer 1

Step 1: Understanding the Concept:
The problem requires evaluating a 3x3 determinant. Instead of direct expansion, which can be cumbersome, using determinant properties and row/column operations is a more efficient approach.
Step 2: Key Formula or Approach:
The strategy is to simplify the determinant using the following properties:
1. Multiplying a row/column by a constant \(k\) multiplies the determinant's value by \(k\).
2. Taking a common factor from any one row or one column.
3. Applying operations like \(R_i \to R_i + R_j + R_k\) or \(C_i \to C_i - C_j\).
Step 3: Detailed Explanation:
Let the given determinant be \(\Delta\).
\[ \Delta = \begin{vmatrix} a^2+1 & ab & ac \\ ab & b^2+1 & bc \\ ca & cb & c^2+1 \end{vmatrix} \] Multiply Row 1 by \(a\), Row 2 by \(b\), and Row 3 by \(c\). To keep the value of the determinant unchanged, we must also divide by \(abc\).
\[ \Delta = \frac{1}{abc} \begin{vmatrix} a(a^2+1) & a^2b & a^2c \\ ab^2 & b(b^2+1) & b^2c \\ c^2a & c^2b & c(c^2+1) \end{vmatrix} \] Now, take \(a\) common from Column 1, \(b\) from Column 2, and \(c\) from Column 3.
\[ \Delta = \frac{abc}{abc} \begin{vmatrix} a^2+1 & a^2 & a^2 \\ b^2 & b^2+1 & b^2 \\ c^2 & c^2 & c^2+1 \end{vmatrix} \] \[ \Delta = \begin{vmatrix} a^2+1 & a^2 & a^2 \\ b^2 & b^2+1 & b^2 \\ c^2 & c^2 & c^2+1 \end{vmatrix} \] Apply the row operation \(R_1 \to R_1 + R_2 + R_3\).
\[ \Delta = \begin{vmatrix} 1+a^2+b^2+c^2 & 1+a^2+b^2+c^2 & 1+a^2+b^2+c^2 \\ b^2 & b^2+1 & b^2 \\ c^2 & c^2 & c^2+1 \end{vmatrix} \] Take the common factor \((1+a^2+b^2+c^2)\) from Row 1.
\[ \Delta = (1+a^2+b^2+c^2) \begin{vmatrix} 1 & 1 & 1 \\ b^2 & b^2+1 & b^2 \\ c^2 & c^2 & c^2+1 \end{vmatrix} \] Now apply the column operations \(C_2 \to C_2 - C_1\) and \(C_3 \to C_3 - C_1\).
\[ \Delta = (1+a^2+b^2+c^2) \begin{vmatrix} 1 & 0 & 0 \\ b^2 & 1 & 0 \\ c^2 & 0 & 1 \end{vmatrix} \] The determinant of a lower triangular matrix is the product of its diagonal elements.
\[ \Delta = (1+a^2+b^2+c^2) \cdot (1 \cdot 1 \cdot 1) \] \[ \Delta = 1+a^2+b^2+c^2 \] Step 4: Final Answer:
The value of the determinant is \(\boxed{1 + a^2 + b^2 + c^2}\).