For those who don't know part 4 (Determinants)
∣ ∣ ∣ ∣ ∣ ∣ ∣ 3 3 2 2 y − 2 1 y 0 y ∣ ∣ ∣ ∣ ∣ ∣ ∣
Find the value of y other than 5 such that the determinant above is equal to 0.
The answer is 0.
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3 solutions
- Solving for the determinant, we'll find out: ( 3 y ² − 3 y ) − ( 2 y ² + 2 y ) = X = 0 = > y ² − 5 y = 0
- Factoring, we conclude: y ² − 5 y = y ( y − 5 ) = 0 .
- Finally, we have y = 0 , because it was said that y doesn't equal 5 .
we can say directly with out doing this problem.the only thing is we have to know some property .when a determinant contains ZEROS IN A ROW OR COLUMN then determinant value is equal to zero
Will a quadratic equation come and we equate it to zero.. Is my approach clear?
The problem was wrong, I modified it and I am sorry for this horrible mistake.
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Yaa ,when X = 0 then we solve for y and we get y = 0 . Done this question quite before..