Math Problem Statement
Find the value of x such that the determinant of the given 3x3 matrix is zero:
\[\begin{bmatrix}
x+4 & 3 & 3 \\
3 & x+4 & 5 \\
5 & 5 & x+4
\end{bmatrix} = 0.\]
Solution
The provided image is a determinant problem where you need to solve for such that the determinant of the given matrix is equal to zero.
The matrix is:
x+4 & 3 & 3 \\ 3 & x+4 & 5 \\ 5 & 5 & x+4 \end{bmatrix}$$ ### Steps to Solve: 1. **Write down the determinant of the matrix:** $$\text{Determinant} = (x+4)\begin{vmatrix} x+4 & 5 \\ 5 & x+4 \end{vmatrix} - 3\begin{vmatrix} 3 & 5 \\ 5 & x+4 \end{vmatrix} + 3\begin{vmatrix} 3 & x+4 \\ 5 & 5 \end{vmatrix}.$$ 2. **Expand each $$2 \times 2$$ determinant:** - For $$\begin{vmatrix} x+4 & 5 \\ 5 & x+4 \end{vmatrix}$$: $$(x+4)(x+4) - (5)(5) = (x+4)^2 - 25$$ - For $$\begin{vmatrix} 3 & 5 \\ 5 & x+4 \end{vmatrix}$$: $$(3)(x+4) - (5)(5) = 3(x+4) - 25 = 3x + 12 - 25 = 3x - 13$$ - For $$\begin{vmatrix} 3 & x+4 \\ 5 & 5 \end{vmatrix}$$: $$(3)(5) - (x+4)(5) = 15 - 5(x+4) = 15 - 5x - 20 = -5x - 5$$ 3. **Substitute back into the determinant equation:** $$\text{Determinant} = (x+4)\left[(x+4)^2 - 25\right] - 3(3x - 13) + 3(-5x - 5).$$ 4. **Expand and simplify:** - First term: $$(x+4)\left[(x+4)^2 - 25\right] = (x+4)\left[x^2 + 8x + 16 - 25\right] = (x+4)(x^2 + 8x - 9)$$ $$= x^3 + 8x^2 - 9x + 4x^2 + 32x - 36 = x^3 + 12x^2 + 23x - 36.$$ - Second term: $$-3(3x - 13) = -9x + 39.$$ - Third term: $$3(-5x - 5) = -15x - 15.$$ Combine everything: $$x^3 + 12x^2 + 23x - 36 - 9x + 39 - 15x - 15 = x^3 + 12x^2 - x - 12.$$ 5. **Set the determinant equal to zero:** $$x^3 + 12x^2 - x - 12 = 0.$$ 6. **Solve the cubic equation.** You can factorize the cubic equation or use numerical methods. The factorization process shows: $$x^3 + 12x^2 - x - 12 = (x + 4)(x^2 + 8x - 3) = 0.$$ This gives two possible solutions: - $$x + 4 = 0 \Rightarrow x = -4$$ - Solve $$x^2 + 8x - 3 = 0$$ using the quadratic formula: $$x = \frac{-8 \pm \sqrt{8^2 - 4(1)(-3)}}{2(1)} = \frac{-8 \pm \sqrt{64 + 12}}{2} = \frac{-8 \pm \sqrt{76}}{2} = \frac{-8 \pm 2\sqrt{19}}{2} = -4 \pm \sqrt{19}.$$ ### Final Solution: $$x = -4, \quad x = -4 + \sqrt{19}, \quad x = -4 - \sqrt{19}.$$ Let me know if you'd like further clarification or a detailed explanation of any step!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Matrix Determinant
Cubic Equations
Algebra
Formulas
Determinant of a 3x3 matrix formula
Quadratic formula: x = (-b ± √(b² - 4ac)) / 2a
Theorems
Properties of determinants
Suitable Grade Level
Grades 11-12