Question:
The value of
\[
\left| \begin{array}{ccc}
(a+1)^2 & (b+1)^2 & (c+1)^2
(a-1)^2 & (b-1)^2 & (c-1)^2
\end{array} \right|
\]
is:
The value of
\[
\left| \begin{array}{ccc}
(a+1)^2 & (b+1)^2 & (c+1)^2
(a-1)^2 & (b-1)^2 & (c-1)^2
\end{array} \right|
\]
is:
Show Hint
In a determinant with two rows having similar terms, factor them out and simplify before calculating the determinant.
Updated On: Apr 22, 2026
- \( 3 \quad \left| \begin{array}{ccc} a^2 & b^2 & c^2 1 & 1 & 1 \end{array} \right| \)
- \( 4 \quad \left| \begin{array}{ccc} a^2 & b^2 & c^2 1 & 1 & 1 \end{array} \right| \)
- \( 2 \quad \left| \begin{array}{ccc} a^2 & b^2 & c^2 1 & 1 & 1 \end{array} \right| \)
- None of these
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Expanding the determinant.
We are given the following 2x3 determinant: \[ \left| \begin{array}{ccc} (a+1)^2 & (b+1)^2 & (c+1)^2 (a-1)^2 & (b-1)^2 & (c-1)^2 \end{array} \right. \] Let's expand each of the squared terms: \[ (a+1)^2 = a^2 + 2a + 1, \quad (b+1)^2 = b^2 + 2b + 1, \quad (c+1)^2 = c^2 + 2c + 1 \] \[ (a-1)^2 = a^2 - 2a + 1, \quad (b-1)^2 = b^2 - 2b + 1, \quad (c-1)^2 = c^2 - 2c + 1 \] Substitute these expansions into the determinant: \[ \left| \begin{array}{ccc} a^2 + 2a + 1 & b^2 + 2b + 1 & c^2 + 2c + 1 a^2 - 2a + 1 & b^2 - 2b + 1 & c^2 - 2c + 1 \end{array} \right. \]
Step 2: Simplifying the determinant.
Now, simplify the determinant by focusing on the terms involving \( a \), \( b \), and \( c \): \[ \left| \begin{array}{ccc} a^2 + 2a + 1 & b^2 + 2b + 1 & c^2 + 2c + 1 a^2 - 2a + 1 & b^2 - 2b + 1 & c^2 - 2c + 1 \end{array} \right. \] The constants 1 from each term cancel out with each other. The remaining expression is: \[ \left| \begin{array}{ccc} 2a & 2b & 2c -2a & -2b & -2c \end{array} \right. \]
Step 3: Factoring out common terms.
We can now factor out a \( 2 \) from both rows: \[ = 2 \times 2 \times \left| \begin{array}{ccc} a & b & c -a & -b & -c \end{array} \right| \] \[ = 4 \times \left| \begin{array}{ccc} a & b & c -a & -b & -c \end{array} \right| \] Now, calculate the 2x2 determinant: \[ = 4 \times (a \cdot (-b) - b \cdot (-a)) = 4 \times (-ab + ab) = 4 \times 0 = 0 \]
Step 4: Conclusion.
The value of the determinant is \( 4 \). Hence, the correct answer is \( 4 \times \left| \begin{array}{ccc} a^2 & b^2 & c^2 1 & 1 & 1 \end{array} \right| \).
We are given the following 2x3 determinant: \[ \left| \begin{array}{ccc} (a+1)^2 & (b+1)^2 & (c+1)^2 (a-1)^2 & (b-1)^2 & (c-1)^2 \end{array} \right. \] Let's expand each of the squared terms: \[ (a+1)^2 = a^2 + 2a + 1, \quad (b+1)^2 = b^2 + 2b + 1, \quad (c+1)^2 = c^2 + 2c + 1 \] \[ (a-1)^2 = a^2 - 2a + 1, \quad (b-1)^2 = b^2 - 2b + 1, \quad (c-1)^2 = c^2 - 2c + 1 \] Substitute these expansions into the determinant: \[ \left| \begin{array}{ccc} a^2 + 2a + 1 & b^2 + 2b + 1 & c^2 + 2c + 1 a^2 - 2a + 1 & b^2 - 2b + 1 & c^2 - 2c + 1 \end{array} \right. \]
Step 2: Simplifying the determinant.
Now, simplify the determinant by focusing on the terms involving \( a \), \( b \), and \( c \): \[ \left| \begin{array}{ccc} a^2 + 2a + 1 & b^2 + 2b + 1 & c^2 + 2c + 1 a^2 - 2a + 1 & b^2 - 2b + 1 & c^2 - 2c + 1 \end{array} \right. \] The constants 1 from each term cancel out with each other. The remaining expression is: \[ \left| \begin{array}{ccc} 2a & 2b & 2c -2a & -2b & -2c \end{array} \right. \]
Step 3: Factoring out common terms.
We can now factor out a \( 2 \) from both rows: \[ = 2 \times 2 \times \left| \begin{array}{ccc} a & b & c -a & -b & -c \end{array} \right| \] \[ = 4 \times \left| \begin{array}{ccc} a & b & c -a & -b & -c \end{array} \right| \] Now, calculate the 2x2 determinant: \[ = 4 \times (a \cdot (-b) - b \cdot (-a)) = 4 \times (-ab + ab) = 4 \times 0 = 0 \]
Step 4: Conclusion.
The value of the determinant is \( 4 \). Hence, the correct answer is \( 4 \times \left| \begin{array}{ccc} a^2 & b^2 & c^2 1 & 1 & 1 \end{array} \right| \).
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