Given the matrix A=[x-squareroot2009 , -1] [1  , x+squareroot2009] and I2, what is x if det(A)=0? [x-squareroot2009 , -1]-first row of matrix

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

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The matrix A is [x - sqrt 2009 , -1] [1, x + sqrt 2009].

The determinant of this 2x2 square matrix is: a1*b2 - a2*b1

=> (x - sqrt 2009)(x + sqrt 2009) - (-1)*1

As the determinant is equal to 0

(x - sqrt 2009)(x + sqrt 2009) - (-1)*1 = 0

=> x^2 - 2009 + 1 = 0

=> x^2 - 2008 = 0

=> x^2 = 2008

This gives the values of x as sqrt 2008 and -sqrt 2008

The values of x are sqrt 2008 and -sqrt 2008

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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To calculate the determinant of the matrix A, we'll have to recall that the determinant of the square matrix (2x2) is the following:

det(A) = a11*a22 - a12*a21

a11*a22 - the elements that belong to the main diagonal of the matrix

a12*a21 - the elemnents of the secondary diagonal of the matrix

Therefore, det(A) = (x-sqrt2009)(x+sqrt2009) - (-1)*1

We notice that the first product returns the difference of two squares:

det(A) = x^2 - 2009 + 1

det(A) = x^2 - 2008

We'll cancel det (A):

det(A) = 0 <=> x^2 - 2008 = 0 => x^2 = 2008

x1 = +sqrt2008

x2 = -sqrt2008

The solutions of the equation det(A)=0 are x1 = +sqrt2008 and x2 = -sqrt2008.

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