For those who don't know part 2 (Determinants)

Algebra Level 2

X = ∣ 0 1 2 3 5 5 5 7 5 ∣ X=\left| \begin{matrix} 0 & 1 & 2 \\ 3 & 5 & 5 \\ 5 & 7 & 5 \end{matrix} \right|

Find the value of determinant ( X X )

If you don't know how to expand determinants, look at this WIKI


The answer is 2.

Abdulrahman El Shafei by Abdulrahman El Shafei , 23, Egypt

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5 solutions

If you don't know how to expand determinants, look at this WIKI . X = ∣ 0 1 2 3 5 5 5 7 5 ∣ X = 0 × ∣ 5 5 7 5 ∣ − 1 × ∣ 3 5 5 5 ∣ + 2 × ∣ 3 5 5 7 ∣ X = 0 × ( ( 5 × 5 ) − ( 5 × 7 ) ) − 1 × ( ( 3 × 5 ) − ( 5 × 5 ) ) + 2 × ( ( 3 × 7 ) − ( 5 × 5 ) ) X = 0 + 10 − 8 X = 2 X=\left| \begin{matrix} 0 & 1 & 2 \\ 3 & 5 & 5 \\ 5 & 7 & 5 \end{matrix} \right| \\ X=0\times \begin{vmatrix} 5 & 5 \\ 7 & 5 \end{vmatrix}-1\times \begin{vmatrix} 3 & 5 \\ 5 & 5 \end{vmatrix}+2\times \begin{vmatrix} 3 & 5 \\ 5 & 7 \end{vmatrix}\\ X=0\times ((5\times 5)-(5\times 7))-1\times ((3\times 5)-(5\times 5))+2\times ((3\times 7)-(5\times 5))\\ X=0+10-8\\ X=\boxed { 2 }

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Or you can use the Sarrus' rule too. But, yeah, the general method is much more efficient and reliable. :D

Prasun Biswas - 6 years, 5 months ago
Hobart Pao
Jan 14, 2015

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Chew-Seong Cheong
Dec 31, 2014

We know that in calculating for determinant, the row operation r o w i − C ˙ r o w j row_i - C\dot{}row_j will not change the value of the determinant. Therefore, we should do row operations to create more 0 0 's to ease the final calculation.

The following should that by making two zeros in a column calculation is much reduced. The row operation is r o w 2 − 3 5 r o w 3 → r o w 3 row_2 - \frac {3}{5}row_3 \rightarrow row_3 .

X = ∣ 0 1 2 3 5 5 5 7 5 ∣ = ∣ 0 1 2 0 0.8 2 5 7 5 ∣ = ( 1 ) ( 2 ) ( 5 ) − ( 2 ) ( 0.8 ) ( 5 ) = 10 − 8 = 2 X = \begin{vmatrix} 0&1&2\\ 3&5&5\\ 5&7&5 \end{vmatrix} = \begin{vmatrix} 0&1&2\\ 0&0.8&2\\ 5&7&5 \end{vmatrix} = (1)(2)(5)-(2)(0.8)(5)=10-8 = \boxed{2}

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If the elements were large, then this would've been a much more elegant way. But since the elements aren't much large values, we can simply expand along R 1 R_1 (or any other row/column one prefers) without using any properties of determinants.

Prasun Biswas - 6 years, 5 months ago
Fox To-ong
Jan 10, 2015

by SARRUS RULE:67 -65 = 2

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Mihir Gokhale
Jan 3, 2015

sarrus rule can be used .so,

67-65=2

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