Question:
If \(\begin{vmatrix} 3x & x+2 \\ 3(x-1) & x-2 \end{vmatrix} = \begin{vmatrix} 1 & 5 \\ 3 & 3 \end{vmatrix}\), then the value of \(x\) is :
If \(\begin{vmatrix} 3x & x+2 \\ 3(x-1) & x-2 \end{vmatrix} = \begin{vmatrix} 1 & 5 \\ 3 & 3 \end{vmatrix}\), then the value of \(x\) is :
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Expand both determinants as \(ad - bc\). You get \(6 - 9x = -12\).
Updated On: Oct 22, 2026
- \(2\)
- \(\frac{1}{3}\)
- \(5\)
- \(3\)
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The Correct Option is A
Solution and Explanation
Step 1: Expand the left determinant:
For a 2 by 2 determinant, the value is \(ad - bc\).
\[ 3x(x-2) - (x+2)\cdot 3(x-1) \]
\[ = 3x^2 - 6x - 3(x^2 + x - 2) = 3x^2 - 6x - 3x^2 - 3x + 6 = 6 - 9x \]
Step 2: Expand the right determinant:
\[ \begin{vmatrix} 1 & 5 \\ 3 & 3 \end{vmatrix} = 1\cdot 3 - 5\cdot 3 = 3 - 15 = -12 \]
Step 3: Solve the equation:
Equate the two values.
\[ 6 - 9x = -12 \Rightarrow 9x = 18 \Rightarrow x = 2 \]
Step 4: Check the other options:
For \(x = \frac{1}{3}\), the left side is \(6 - 3 = 3\), not \(-12\).
For \(x = 5\), it is \(6 - 45 = -39\). For \(x = 3\), it is \(6 - 27 = -21\). Neither equals \(-12\).
Final Answer:
So \(x = 2\), option (A). \[ \boxed{x = 2} \]
For a 2 by 2 determinant, the value is \(ad - bc\).
\[ 3x(x-2) - (x+2)\cdot 3(x-1) \]
\[ = 3x^2 - 6x - 3(x^2 + x - 2) = 3x^2 - 6x - 3x^2 - 3x + 6 = 6 - 9x \]
Step 2: Expand the right determinant:
\[ \begin{vmatrix} 1 & 5 \\ 3 & 3 \end{vmatrix} = 1\cdot 3 - 5\cdot 3 = 3 - 15 = -12 \]
Step 3: Solve the equation:
Equate the two values.
\[ 6 - 9x = -12 \Rightarrow 9x = 18 \Rightarrow x = 2 \]
Step 4: Check the other options:
For \(x = \frac{1}{3}\), the left side is \(6 - 3 = 3\), not \(-12\).
For \(x = 5\), it is \(6 - 45 = -39\). For \(x = 3\), it is \(6 - 27 = -21\). Neither equals \(-12\).
Final Answer:
So \(x = 2\), option (A). \[ \boxed{x = 2} \]
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