Math Problem Statement
Determine the values of a, b, c, d from the equality of two matrices, and find x and y given that A = B^T.
Solution
Let's break down the two problems from the image and solve them step by step.
Problem 1:
We are given two matrices:
a - 4 & -8 & 2d - 3 \\ b & 2c & -13 \end{pmatrix} = \begin{pmatrix} 3a & -8 & 11 \\ 2b - 5 & 3c + 8 & -13 \end{pmatrix}$$ We need to find the values of $$a$$, $$b$$, $$c$$, and $$d$$ by comparing corresponding elements in the two matrices. #### Step-by-step comparison: 1. $$a - 4 = 3a$$ 2. $$-8 = -8$$ (This is already satisfied) 3. $$2d - 3 = 11$$ 4. $$b = 2b - 5$$ 5. $$2c = 3c + 8$$ 6. $$-13 = -13$$ (This is already satisfied) From these equations, we can solve for $$a$$, $$b$$, $$c$$, and $$d$$. #### Solving: 1. $$a - 4 = 3a$$ $$-4 = 2a \implies a = -2$$ 2. $$2d - 3 = 11$$ $$2d = 14 \implies d = 7$$ 3. $$b = 2b - 5$$ $$b - 2b = -5 \implies -b = -5 \implies b = 5$$ 4. $$2c = 3c + 8$$ $$2c - 3c = 8 \implies -c = 8 \implies c = -8$$ Thus, the values are: $$a = -2, \quad b = 5, \quad c = -8, \quad d = 7$$ ### Problem 2: We are given the matrices: $$A = \begin{pmatrix} 4x - 2y & -4 \\ 5x - y & -3 \end{pmatrix} \quad \text{and} \quad B = \begin{pmatrix} -16 & 4 \\ -4 & -3 \end{pmatrix}$$ We are told that $$A = B^T$$, meaning $$A$$ is the transpose of $$B$$. Therefore: $$A = B^T = \begin{pmatrix} -16 & -4 \\ 4 & -3 \end{pmatrix}$$ Now we compare the corresponding elements of matrix $$A$$ and $$B^T$$: 1. $$4x - 2y = -16$$ 2. $$-4 = 4$$ (Already satisfied) 3. $$5x - y = -4$$ 4. $$-3 = -3$$ (Already satisfied) #### Solving: 1. $$4x - 2y = -16$$ $$2x - y = -8 \quad \text{(divide by 2)}$$ 2. $$5x - y = -4$$ We now have the system of equations: $$2x - y = -8 \quad \text{(1)}$$ $$5x - y = -4 \quad \text{(2)}$$ Subtract equation (1) from equation (2): $$(5x - y) - (2x - y) = -4 - (-8)$$ $$3x = 4 \implies x = \frac{4}{3}$$ Substitute $$x = \frac{4}{3}$$ into equation (1): $$2\left(\frac{4}{3}\right) - y = -8$$ $$\frac{8}{3} - y = -8$$ $$-y = -8 - \frac{8}{3} = -\frac{24}{3} - \frac{8}{3} = -\frac{32}{3}$$ $$y = \frac{32}{3}$$ Thus, the values are: $$x = \frac{4}{3}, \quad y = \frac{32}{3}$$ ### Summary of Results: 1. $$a = -2$$, $$b = 5$$, $$c = -8$$, $$d = 7$$ 2. $$x = \frac{4}{3}$$, $$y = \frac{32}{3}$$ Would you like a detailed explanation of any of the steps, or do you have any further questions? ### Follow-up Questions: 1. How do you transpose a matrix, and why is it important? 2. Can you generalize how to solve for variables when comparing matrix elements? 3. What are the conditions for matrix equality, and how do they apply here? 4. How does solving simultaneous equations using elimination work? 5. Can you explain how matrix transposition affects multiplication? #### Tip: When working with matrices, always check the dimensions to ensure they align for operations like addition, subtraction, or multiplication.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Matrix Equality
Matrix Transposition
Simultaneous Equations
Formulas
Matrix element comparison
Solving simultaneous equations
Theorems
Matrix Equality Theorem
Transpose of a Matrix
Suitable Grade Level
Grades 9-12