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Determine whether the matrix is invertible. 95 -9 5) 4 2 -4 -3 0 3

Determine whether the matrix is invertible. 95 -9 5) 4 2 -4 -3 0 3

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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**Matrix Invertibility** --- **Problem Statement:** Determine whether the matrix is invertible. \[ \begin{pmatrix} 9 & 5 & -9 \\ 4 & 2 & -4 \\ -3 & 0 & 3 \end{pmatrix} \] To determine whether the given matrix is invertible, we need to check if its determinant is non-zero. If the determinant is non-zero, the matrix is invertible; otherwise, it is not. To find the determinant of a 3x3 matrix \(\mathbf{A}\) given as: \[ \mathbf{A} = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{pmatrix} \] The determinant (\(\det(\mathbf{A})\)) is calculated as follows: \[ \det(\mathbf{A}) = a(ei - fh) - b(di - fg) + c(dh - eg) \] For the given matrix: \[ \mathbf{A} = \begin{pmatrix} 9 & 5 & -9 \\ 4 & 2 & -4 \\ -3 & 0 & 3 \\ \end{pmatrix} \] Let \(a = 9\), \(b = 5\), \(c = -9\), \(d = 4\), \(e = 2\), \(f = -4\), \(g = -3\), \(h = 0\), \(i = 3\). Plug these values into the determinant formula: \[ \det(\mathbf{A}) = 9(2*3 - (-4)*0) - 5(4*3 - (-3)*(-4)) + (-9)(4*0 - (-3)*2) \] Simplifying each term: \[ = 9(6 - 0) - 5(12 - 12) - 9(0 + 6) \] \[ = 9 \cdot 6 - 5 \cdot 0 - 9 \cdot 6 \] \[ = 54 - 0 - 54 \] \[ = 0 \] Since \(\det(\mathbf{A}) = 0\), the matrix is **not**
Transcribed Image Text:**Matrix Invertibility** --- **Problem Statement:** Determine whether the matrix is invertible. \[ \begin{pmatrix} 9 & 5 & -9 \\ 4 & 2 & -4 \\ -3 & 0 & 3 \end{pmatrix} \] To determine whether the given matrix is invertible, we need to check if its determinant is non-zero. If the determinant is non-zero, the matrix is invertible; otherwise, it is not. To find the determinant of a 3x3 matrix \(\mathbf{A}\) given as: \[ \mathbf{A} = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{pmatrix} \] The determinant (\(\det(\mathbf{A})\)) is calculated as follows: \[ \det(\mathbf{A}) = a(ei - fh) - b(di - fg) + c(dh - eg) \] For the given matrix: \[ \mathbf{A} = \begin{pmatrix} 9 & 5 & -9 \\ 4 & 2 & -4 \\ -3 & 0 & 3 \\ \end{pmatrix} \] Let \(a = 9\), \(b = 5\), \(c = -9\), \(d = 4\), \(e = 2\), \(f = -4\), \(g = -3\), \(h = 0\), \(i = 3\). Plug these values into the determinant formula: \[ \det(\mathbf{A}) = 9(2*3 - (-4)*0) - 5(4*3 - (-3)*(-4)) + (-9)(4*0 - (-3)*2) \] Simplifying each term: \[ = 9(6 - 0) - 5(12 - 12) - 9(0 + 6) \] \[ = 9 \cdot 6 - 5 \cdot 0 - 9 \cdot 6 \] \[ = 54 - 0 - 54 \] \[ = 0 \] Since \(\det(\mathbf{A}) = 0\), the matrix is **not**
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