how we can solve a determinant

A=[1-(x-1)^2,1-x^2,1-(x+1^)2; 2^2-(x-1)^2, 2^2-x^2, 2^2-(x+1)^2; 3^2-(x-1)^2, 3^2-x^2, 3^2-(x+1)^2];
det(A)=0;

det means determinant; please give general solution so that rows and colons can also be made more than 3.

Answer 1

[Torsten]

fun=@(x)det([1-(x-1)^2,1-x^2,1-(x+1^)2; 2^2-(x-1)^2, 2^2-x^2, 2^2-(x+1)^2; 3^2-(x-1)^2, 3^2-x^2, 3^2-(x+1)^2]);

sol=fzero(fun,1);

Best wishes

Torsten.

Answer 2

[Roger Stafford]

@Arjun. As you undoubtedly are aware, any row of a determinant can be subtracted from another row without changing the value of the determinant. Suppose you subtract the first row of your determinant from the second, and then subtract that first row from the third row. The result would be:

   det([[1-(x-1)^2,1-x^2,1-(x+1)^2;
             3    ,   3 ,   3     ;
             8    ,   8 ,   8     ]);

Since both the second and third rows are multiples of all ones, the determinant must be identically zero.

That means for all x, the determinant of A is zero. You can never solve for x from det(A) = 0 since it is zero for all possible values of x. If you doubt this reasoning, try using random values for x and evaluating the determinant. You will get only tiny round-off errors as a result.

The same reasoning would apply to the more general case of n rows and columns that you asked about.

[Arjun Kumar]

thank you for your answer. i agree with this