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Using Properties of Determinants Prove that ∣ ∣ ∣ ∣ X + 4 2 X 2 X 2 X X + 4 2 X 2 X 2 X X + 4 ∣ ∣ ∣ ∣ = ( 5 X + 4 ) ( 4 − X ) 2

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प्रश्न

Using properties of determinants prove that

\[\begin{vmatrix}x + 4 & 2x & 2x \\ 2x & x + 4 & 2x \\ 2x & 2x & x + 4\end{vmatrix} = \left( 5x + 4 \right) \left( 4 - x \right)^2\]

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उत्तर

\[∆ = \begin{vmatrix}x + 4 & 2x & 2x \\ 2x & x + 4 & 2x \\ 2x & 2x & x + 4\end{vmatrix}\]

\[ = \begin{vmatrix}5x + 4 & 5x + 4 & 5x + 4 \\ 2x & x + 4 & 2x \\ 2x & 2x & x + 4\end{vmatrix} \left[\text{ Applying }R_1 \text{ to }R_1 + R_2 + R_3 \right]\]

\[ = 5x + 4\begin{vmatrix}1 & 1 & 1 \\ 2x & x + 4 & 2x \\ 2x & 2x & x + 4\end{vmatrix} \left[\text{ Take out }5x + 4\text{  common from }R_1 \right]\]

\[ = 5x + 4\begin{vmatrix}1 & 0 & 0 \\ 2x & 4 - x & 0 \\ 2x & 0 & 4 - x\end{vmatrix} \left[\text{ Applying }C_2 \text{ to }C_2 - C_1\text{ and }C_3 \text{ to }C_3 - C_1 \right]\]

\[ = 5x + 4(4 - x )^2 \left[\text{ Expanding along }R_1 \right]\]

Hence proved.

 
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अध्याय 5: Determinants - Exercise 6.2 [पृष्ठ ६०]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 5 Determinants
Exercise 6.2 | Q 38 | पृष्ठ ६०

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