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

Linear Algebra: A Modern Introduction

4th Edition

ISBN:9781285463247

Author:David Poole

Publisher:David Poole

Chapter2: Systems Of Linear Equations

Section2.2: Direct Methods For Solving Linear Systems

Problem 3CEXP

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Question

**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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