Question:
If \(\begin{vmatrix}x-4&2x&2x\\2x&x-4&2x\\2x&2x&x-4\end{vmatrix}=(A+Bx)(x-A)^2\), then the ordered pair \((A,B)\) is equal to :
If \(\begin{vmatrix}x-4&2x&2x\\2x&x-4&2x\\2x&2x&x-4\end{vmatrix}=(A+Bx)(x-A)^2\), then the ordered pair \((A,B)\) is equal to :
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Add all columns to the first column to factor out \(5x-4\), then simplify with row operations.
Updated On: Nov 11, 2026
- \((4,5)\)
- \((-4,-5)\)
- \((-4,3)\)
- \((-4,5)\)
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The Correct Option is D
Solution and Explanation
Apply \(C_1\to C_1+C_2+C_3\): each entry of the first column becomes \(x-4+4x=5x-4\).
Determinant \(=(5x-4)\begin{vmatrix}1&2x&2x\\1&x-4&2x\\1&2x&x-4\end{vmatrix}\).
Apply \(R_2\to R_2-R_1,\ R_3\to R_3-R_1\): the determinant becomes \(\begin{vmatrix}1&2x&2x\\0&-x-4&0\\0&0&-x-4\end{vmatrix}=(x+4)^2\).
So the given determinant \(=(5x-4)(x+4)^2\).
Comparing with \((A+Bx)(x-A)^2\): \(A=-4\), \(B=5\).
Answer: (4) \((-4,5)\)
Determinant \(=(5x-4)\begin{vmatrix}1&2x&2x\\1&x-4&2x\\1&2x&x-4\end{vmatrix}\).
Apply \(R_2\to R_2-R_1,\ R_3\to R_3-R_1\): the determinant becomes \(\begin{vmatrix}1&2x&2x\\0&-x-4&0\\0&0&-x-4\end{vmatrix}=(x+4)^2\).
So the given determinant \(=(5x-4)(x+4)^2\).
Comparing with \((A+Bx)(x-A)^2\): \(A=-4\), \(B=5\).
Answer: (4) \((-4,5)\)
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