Question - Determinant Calculation for 3x3 Matrix
Solution:
The image contains a determinant of a 3x3 matrix equal to -30. We need to calculate the value(s) of x for which this determinant holds true. The determinant is:| 3 1 x || -1 3 4 || x 1 0 |To find the determinant, we will expand along the third column (as it seems to be the easiest due to the zero present):det = x * | -1 3 | | x 1 | - x * | 3 1 | | x 0 | + 4 * | 3 1 | | -1 3 |Solving each of these determinants:det(x) = x((-1)(1) - (3)(x)) - x((3)(0) - (1)(x)) + 4((3)(3) - (1)(-1))det(x) = x(-1 - 3x) - x(0 - x) + 4(9 + 1)det(x) = x(-1 - 3x) - x(-x) + 4(10)det(x) = -x - 3x^2 + x^2 + 40det(x) = -3x^2 + x^2 - x + 40Combine like terms:det(x) = -2x^2 - x + 40Now we set this determinant equal to the given value (-30) and solve for x:-2x^2 - x + 40 = -30Move -30 to the other side:-2x^2 - x + 40 + 30 = 0-2x^2 - x + 70 = 0This is a quadratic equation, which can often be solved by factoring, completing the square, or using the quadratic formula. To use the quadratic formula ($$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$) we identify the coefficients a = -2, b = -1, c = 70.Before using the quadratic formula, it might be helpful to check if the quadratic can be factored easily, but in this case, it cannot be factored into integers, so let's apply the quadratic formula:$$ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(-2)(70)}}{2(-2)} $$$$ x = \frac{1 \pm \sqrt{1 + 560}}{-4} $$$$ x = \frac{1 \pm \sqrt{561}}{-4} $$The two solutions for x are:$$ x = \frac{1 + \sqrt{561}}{-4} $$ and $$ x = \frac{1 - \sqrt{561}}{-4} $$These are the values of x for which the determinant of the matrix equals -30.
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