Question:
The determinant
\[
\det \begin{bmatrix} \frac{a^2 + b^2}{c} & c & c a & \frac{b^2 + c^2}{a} & a b & b & \frac{c^2 + a^2}{b} \end{bmatrix}
\]
is equal to:
The determinant
\[
\det \begin{bmatrix} \frac{a^2 + b^2}{c} & c & c a & \frac{b^2 + c^2}{a} & a b & b & \frac{c^2 + a^2}{b} \end{bmatrix}
\]
is equal to:
Show Hint
For symmetric polynomial or rational matrix determinants, substitution saves immense time! Let \(a = 1, b = 1, c = 1\). The determinant becomes \(\begin{vmatrix} 2 & 1 & 1 1 & 2 & 1 1 & 1 & 2 \end{vmatrix} = 2(4-1) - 1(2-1) + 1(1-2) = 6 - 1 - 1 = 4\). Comparing with options: (a) 0, (b) 8, (c) 2, (d) 4. Option (D) matches instantly!
Updated On: May 20, 2026
- \( (a - b)(b - c)(c - a) \)
- \( (a + b)(b + c)(c + a) \)
- \( 2abc \)
- \( 4abc \)
Show Solution
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The Correct Option is D
Solution and Explanation
Concept:
The value of a determinant can be simplified by applying elementary operations or factored out by clearing denominators. When a determinant expression is homogeneous and symmetric with respect to its variables (\(a, b, c\)), choosing strategic numeric values is a powerful method to quickly evaluate the algebraic expression without full expansion.
Step 1: Clearing denominators by taking out common scalar factors from the rows.
Let the given determinant be \(\Delta\). Take out \(\frac{1}{c}\) from the 1st row (\(R_1\)), \(\frac{1}{a}\) from the 2nd row (\(R_2\)), and \(\frac{1}{b}\) from the 3rd row (\(R_3\)): \[ \Delta = \frac{1}{abc} \begin{vmatrix} a^2 + b^2 & c^2 & c^2 a^2 & b^2 + c^2 & a^2 b^2 & b^2 & c^2 + a^2 \end{vmatrix} \]
Step 2: Applying row operations to reduce the cyclic expressions.
Apply the row operations \(R_1 \rightarrow R_1 - R_2 - R_3\): \[ \Delta = \frac{1}{abc} \begin{vmatrix} (a^2+b^2) - a^2 - b^2 & c^2 - (b^2+c^2) - b^2 & c^2 - a^2 - (c^2+a^2) a^2 & b^2 + c^2 & a^2 b^2 & b^2 & c^2 + a^2 \end{vmatrix} \] \[ \Delta = \frac{1}{abc} \begin{vmatrix} 0 & -2b^2 & -2a^2 a^2 & b^2 + c^2 & a^2 b^2 & b^2 & c^2 + a^2 \end{vmatrix} \] Taking out the common factor \(-2\) from the first row: \[ \Delta = \frac{-2}{abc} \begin{vmatrix} 0 & b^2 & a^2 a^2 & b^2 + c^2 & a^2 b^2 & b^2 & c^2 + a^2 \end{vmatrix} \]
Step 3: Creating more zeros using row differences and expanding.
Apply the operations \(R_2 \rightarrow R_2 - R_1\) and \(R_3 \rightarrow R_3 - R_1\): \[ \Delta = \frac{-2}{abc} \begin{vmatrix} 0 & b^2 & a^2 a^2 & c^2 & 0 b^2 & 0 & c^2 \end{vmatrix} \] Now, expand the simplified determinant along the first row (\(R_1\)): \[ \Delta = \frac{-2}{abc} \left[ 0 - b^2(a^2c^2 - 0) + a^2(0 - b^2c^2) \right] \] \[ \Delta = \frac{-2}{abc} \left[ -a^2b^2c^2 - a^2b^2c^2 \right] = \frac{-2}{abc} \left[ -2a^2b^2c^2 \right] \] \[ \Delta = \frac{4a^2b^2c^2}{abc} = 4abc \]
Step 1: Clearing denominators by taking out common scalar factors from the rows.
Let the given determinant be \(\Delta\). Take out \(\frac{1}{c}\) from the 1st row (\(R_1\)), \(\frac{1}{a}\) from the 2nd row (\(R_2\)), and \(\frac{1}{b}\) from the 3rd row (\(R_3\)): \[ \Delta = \frac{1}{abc} \begin{vmatrix} a^2 + b^2 & c^2 & c^2 a^2 & b^2 + c^2 & a^2 b^2 & b^2 & c^2 + a^2 \end{vmatrix} \]
Step 2: Applying row operations to reduce the cyclic expressions.
Apply the row operations \(R_1 \rightarrow R_1 - R_2 - R_3\): \[ \Delta = \frac{1}{abc} \begin{vmatrix} (a^2+b^2) - a^2 - b^2 & c^2 - (b^2+c^2) - b^2 & c^2 - a^2 - (c^2+a^2) a^2 & b^2 + c^2 & a^2 b^2 & b^2 & c^2 + a^2 \end{vmatrix} \] \[ \Delta = \frac{1}{abc} \begin{vmatrix} 0 & -2b^2 & -2a^2 a^2 & b^2 + c^2 & a^2 b^2 & b^2 & c^2 + a^2 \end{vmatrix} \] Taking out the common factor \(-2\) from the first row: \[ \Delta = \frac{-2}{abc} \begin{vmatrix} 0 & b^2 & a^2 a^2 & b^2 + c^2 & a^2 b^2 & b^2 & c^2 + a^2 \end{vmatrix} \]
Step 3: Creating more zeros using row differences and expanding.
Apply the operations \(R_2 \rightarrow R_2 - R_1\) and \(R_3 \rightarrow R_3 - R_1\): \[ \Delta = \frac{-2}{abc} \begin{vmatrix} 0 & b^2 & a^2 a^2 & c^2 & 0 b^2 & 0 & c^2 \end{vmatrix} \] Now, expand the simplified determinant along the first row (\(R_1\)): \[ \Delta = \frac{-2}{abc} \left[ 0 - b^2(a^2c^2 - 0) + a^2(0 - b^2c^2) \right] \] \[ \Delta = \frac{-2}{abc} \left[ -a^2b^2c^2 - a^2b^2c^2 \right] = \frac{-2}{abc} \left[ -2a^2b^2c^2 \right] \] \[ \Delta = \frac{4a^2b^2c^2}{abc} = 4abc \]
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