Example Question - determinant calculation
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Determinant Calculation for 3x3 Matrix
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 + 40 det(x) = -3x^2 + x^2 - x + 40 Combine like terms: det(x) = -2x^2 - x + 40 Now we set this determinant equal to the given value (-30) and solve for x: -2x^2 - x + 40 = -30 Move -30 to the other side: -2x^2 - x + 40 + 30 = 0 -2x^2 - x + 70 = 0 This 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.
Determinant calculation for 2x2 matrix
The given matrix M is: \[ M = \begin{pmatrix} 2 & x \\ -1 & 3 \end{pmatrix} \] And we are told that the determinant of matrix M is 9. The determinant of a 2x2 matrix \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\) is calculated by \(ad - bc\). Therefore, for the given matrix M, the determinant is calculated by: \( (2)(3) - (-1)(x) = 6 + x \) Since we know that the determinant is 9, we can set up the following equation: \( 6 + x = 9 \) Now, let's solve for \( x \): \( x = 9 - 6 \) \( x = 3 \) Hence, the value of \( x \) that satisfies the given determinant value is 3.
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