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-6 14 (-7 15 , 18. Solve for X, if X³ = (). 0 2 Ans. X = -1 3

-6 14 (-7 15 , 18. Solve for X, if X³ = (). 0 2 Ans. X = -1 3

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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**Problem 18:** Solve for \( X \), if \( X^3 = \begin{pmatrix} -6 & 14 \\ -7 & 15 \end{pmatrix} \). **Answer:** \( X = \begin{pmatrix} 0 & 2 \\ -1 & 3 \end{pmatrix} \). --- **Explanation of Problem and Solution:** - We are given a matrix equation where \( X^3 \), the cube of matrix \( X \), is equal to another given matrix: \( \begin{pmatrix} -6 & 14 \\ -7 & 15 \end{pmatrix} \). - The goal is to find the matrix \( X \). - The solution provided indicates that the matrix \( X \) is \( \begin{pmatrix} 0 & 2 \\ -1 & 3 \end{pmatrix} \). In solving such problems, typically one might consider using various methods such as matrix diagonalization or calculating directly if the context allows simplification. The answer implies the matrix that, when cubed, results in the initial given matrix.
Transcribed Image Text:**Problem 18:** Solve for \( X \), if \( X^3 = \begin{pmatrix} -6 & 14 \\ -7 & 15 \end{pmatrix} \). **Answer:** \( X = \begin{pmatrix} 0 & 2 \\ -1 & 3 \end{pmatrix} \). --- **Explanation of Problem and Solution:** - We are given a matrix equation where \( X^3 \), the cube of matrix \( X \), is equal to another given matrix: \( \begin{pmatrix} -6 & 14 \\ -7 & 15 \end{pmatrix} \). - The goal is to find the matrix \( X \). - The solution provided indicates that the matrix \( X \) is \( \begin{pmatrix} 0 & 2 \\ -1 & 3 \end{pmatrix} \). In solving such problems, typically one might consider using various methods such as matrix diagonalization or calculating directly if the context allows simplification. The answer implies the matrix that, when cubed, results in the initial given matrix.
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