Math Problem Statement
Solve for the matrix X in the equation [5 4; 3 3] · X = [2 1; 3 5].
Solution
The problem shows a matrix equation involving a matrix multiplied by a variable to produce another matrix. Here's the problem clearly written out:
5 & 4 \\ 3 & 3 \end{bmatrix} \cdot X = \begin{bmatrix} 2 & 1 \\ 3 & 5 \end{bmatrix}.$$ We need to solve for $$X$$, which is likely a $$2 \times 2$$ matrix. To do so, we'll calculate the inverse of the coefficient matrix (on the left) and apply it to the result matrix (on the right). Here's how we proceed: --- ### Step 1: Write the matrix equation in general form The equation is: $$A \cdot X = B,$$ where: $$A = \begin{bmatrix} 5 & 4 \\ 3 & 3 \end{bmatrix}, \quad B = \begin{bmatrix} 2 & 1 \\ 3 & 5 \end{bmatrix}.$$ To isolate $$X$$, multiply both sides of the equation by $$A^{-1}$$ (the inverse of $$A$$): $$X = A^{-1} \cdot B.$$ --- ### Step 2: Compute the inverse of matrix $$A$$ The formula for the inverse of a $$2 \times 2$$ matrix $$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$ is: $$A^{-1} = \frac{1}{\text{det}(A)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix},$$ where $$\text{det}(A) = ad - bc$$. For $$A = \begin{bmatrix} 5 & 4 \\ 3 & 3 \end{bmatrix}$$: - $$a = 5, \, b = 4, \, c = 3, \, d = 3$$, - $$\text{det}(A) = (5)(3) - (4)(3) = 15 - 12 = 3$$. Thus: $$A^{-1} = \frac{1}{3} \begin{bmatrix} 3 & -4 \\ -3 & 5 \end{bmatrix}.$$ Simplify: $$A^{-1} = \begin{bmatrix} 1 & -\frac{4}{3} \\ -1 & \frac{5}{3} \end{bmatrix}.$$ --- ### Step 3: Multiply $$A^{-1}$$ by $$B$$ We now compute: $$X = A^{-1} \cdot B = \begin{bmatrix} 1 & -\frac{4}{3} \\ -1 & \frac{5}{3} \end{bmatrix} \cdot \begin{bmatrix} 2 & 1 \\ 3 & 5 \end{bmatrix}.$$ Perform the matrix multiplication step-by-step. #### First row of $$X$$: 1. First element: $$(1)(2) + \left( -\frac{4}{3} \right)(3) = 2 - 4 = -2$$, 2. Second element: $$(1)(1) + \left( -\frac{4}{3} \right)(5) = 1 - \frac{20}{3} = -\frac{17}{3}$$. #### Second row of $$X$$: 1. First element: $$(-1)(2) + \left( \frac{5}{3} \right)(3) = -2 + 5 = 3$$, 2. Second element: $$(-1)(1) + \left( \frac{5}{3} \right)(5) = -1 + \frac{25}{3} = \frac{22}{3}$$. --- ### Step 4: Combine results Thus, the solution matrix $$X$$ is: $$X = \begin{bmatrix} -2 & -\frac{17}{3} \\ 3 & \frac{22}{3} \end{bmatrix}.$$ --- ### Final Answer: $$X = \begin{bmatrix} -2 & -\frac{17}{3} \\ 3 & \frac{22}{3} \end{bmatrix}.$$ --- Do you need further clarification on any step, or would you like additional examples? Let me know! Here are 5 follow-up questions to deepen your understanding: 1. How do you find the determinant of a $$2 \times 2$$ matrix? 2. What happens if the determinant of a matrix is zero? 3. How do you multiply two matrices together? 4. Can you solve the same problem using a computer or calculator tool? 5. What are the properties of an inverse matrix? **Tip**: Always check the determinant before attempting to find the inverse of a matrix; if it's zero, the inverse does not exist!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Matrix Algebra
Matrix Inversion
Matrix Multiplication
Formulas
Matrix Inverse Formula for 2x2: A^-1 = (1/det(A)) * [d -b; -c a]
Determinant of 2x2 matrix: det(A) = ad - bc
Matrix Multiplication: A · B = C
Theorems
Matrix Inversion Theorem
Suitable Grade Level
Undergraduate Level or Advanced High School