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Evaluate ∣ ∣ ∣ ∣ 2 3 − 5 7 1 − 2 − 3 4 1 ∣ ∣ ∣ ∣ by Two Methods.

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प्रश्न

Evaluate

\[\begin{vmatrix}2 & 3 & - 5 \\ 7 & 1 & - 2 \\ - 3 & 4 & 1\end{vmatrix}\] by two methods.

 
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उत्तर

Let 

\[∆ = \begin{vmatrix}2 & 3 & - 5 \\ 7 & 1 & - 2 \\ - 3 & 4 & 1\end{vmatrix}\]
First method
\[∆ = \left( - 1 \right)^{1 + 1} 2\left( 1 + 8 \right) + \left( - 1 \right)^{1 + 2} 3\left( 7 - 6 \right) + \left( - 1 \right)^{1 + 3} \left( - 5 \right)\left( 28 + 3 \right)\]
\[ = 2\left( 1 + 8 \right) - 3\left( 7 - 6 \right) - 5\left( 28 + 3 \right)\]
\[ = 18 - 3 - 155\]
\[ = - 140\]
Second method is the Sarus Method, where we adjoin the first two columns to the right to get
`|[2,3,-5,2,3],[7,1,-2,7,1],[-3,4,1,-3,4]|`

=(2 x 1 x 1 + 3 x -2 x -3 -5 x 7 x 4)-(-5 x 1 x -3 + 2 x -2 x 4 + 3 x 7 x 1)

=(2 + 18 - 140) - (15 - 16 + 21)

= -120 - 20

= -140

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अध्याय 5: Determinants - Exercise 6.1 [पृष्ठ १०]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 5 Determinants
Exercise 6.1 | Q 5 | पृष्ठ १०

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