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\( \left[ \begin{array} { c c } { x - 2 } & { 4 - x } \\ { 3 } & { x } \end{array} \right] = 0 \)

Ask by Gordon Love. in Kenya
Feb 24,2025

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Tutor-Verified Answer

Answer

The matrix equation has no solution.

Solution

Solve the system of equations \( \begin{cases} x-2=0,\\ 3=0 \end{cases} \). Solve the system of equations by following steps: - step0: Solution: \(\left\{ \begin{array}{l}x-2=0\\3=0\end{array}\right.\) - step1: Calculate: \(\left\{ \begin{array}{l}x=2\\x \in \varnothing \end{array}\right.\) - step2: Find the intersection: \(x \in \varnothing \) The system of equations has no solution. Therefore, the matrix equation \( \left[ \begin{array} { c c } { x - 2 } & { 4 - x } \\ { 3 } & { x } \end{array} \right] = 0 \) has no solution.

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Beyond the Answer

To solve the matrix equation \( \left[ \begin{array} { c c } { x - 2 } & { 4 - x } \\ { 3 } & { x } \end{array} \right] = 0 \), we need to set each element of the matrix to zero. This gives us the following equations: 1. \( x - 2 = 0 \) 2. \( 4 - x = 0 \) 3. \( 3 = 0 \) 4. \( x = 0 \) Looking at these equations, we can quickly identify that the third equation, \( 3 = 0 \), is not valid, meaning there is no solution for this matrix equation that satisfies all elements being zero. The inconsistencies in the equations suggest that the matrix does not represent a valid system that can be solved; hence, the system is inconsistent. This little exercise provides a peek into matrix theory where it’s vital to check not just the variables, but also the logical coherence of the equations derived from them! In real-world applications, matrices are used in everything from computer graphics transformations to solving systems of equations in engineering. For example, in robotics, matrices help to determine the position and movement of parts in a 3D space, making them essential for designing and controlling robots effectively.

Related Questions

ISCELÁNEA cribir, por simple inspección, el resultado de: \( \begin{array}{lll}(x+2)^{2} & \text { 14. }(x+y+1)(x-y-1) & \text { 27. }\left(2 a^{3}-5 b^{4}\right)^{2} \\ (x+2)(x+3) & \text { 15. }(1-a)(a+1) & \text { 28. }\left(a^{3}+12\right)\left(a^{3}-15\right) \\ (x+1)(x-1) & \text { 16. }(m-8)(m+12) & \text { 29. }\left(m^{2}-m+n\right)\left(n+m+m^{2}\right) \\ (x-1)^{2} & \text { 17. }\left(x^{2}-1\right)\left(x^{2}+3\right) & \text { 30. }\left(x^{4}+7\right)\left(x^{4}-11\right) \\ (n+3)(n+5) & \text { 18. }\left(x^{3}+6\right)\left(x^{3}-8\right) & \text { 31. }(11-a b)^{2} \\ (m-3)(m+3) & \text { 19. }\left(5 x^{3}+6 m^{4}\right)^{2} & \text { 32. }\left(x^{2} y^{3}-8\right)\left(x^{2} y^{3}+6\right) \\ (a+b-1)(a+b+1) & \text { 20. }\left(x^{4}-2\right)\left(x^{4}+5\right) & \text { 33. }(a+b)(a-b)\left(a^{2}-b^{2}\right) \\ (1+b)^{3} & \text { 21. }(1-a+b)(b-a-1) & \text { 34. }(x+1)(x-1)\left(x^{2}-2\right) \\ \left(a^{2}+4\right)\left(a^{2}-4\right) & \text { 22. }\left(a^{x}+b^{n}\right)\left(a^{x}-b^{n}\right) & \text { 35. }(a+3)\left(a^{2}+9\right)(a-3) \\ \left(3 a b-5 x^{2}\right)^{2} & \text { 23. }\left(x^{a+1}-8\right)\left(x^{a+1}+9\right) & \text { 36. }(x+5)(x-5)\left(x^{2}+1\right) \\ (a b+3)(3-a b) & \text { 24. }\left(a^{2} b^{2}+c^{2}\right)\left(a^{2} b^{2}-c^{2}\right) & \text { 37. }(a+1)(a-1)(a+2)(a-2) \\ (1-4 a x)^{2} & \text { 25. }(2 a+x)^{3} & \text { 36. }\left(x^{2}-11\right)\left(x^{2}-2\right)\end{array} \) \( \left(a^{2}+8\right)\left(a^{2}-7\right) \)
Algebra Mexico Feb 26, 2025
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