Find Matrix Inverse With Gauss-Jordan Elimination

Hey guys! Today, we're diving deep into the fascinating world of matrices and tackling a common problem: finding the inverse of a matrix using the Gauss-Jordan elimination method. We'll be working with a specific example, $A=\left[\begin{array}{rrr} 5 & 5 & 0 \\ 0 & -4 & -4 \\ -2 & 0 & 2 \end{array}\right]$, and we'll even double-check our work by confirming that $A A^{-1} = I$ and $A^{-1} A = I$, where $I$ is the identity matrix. So, grab your calculators and get ready to flex those mathematical muscles!

Understanding Matrix Inverses

Before we jump into the Gauss-Jordan method, let's quickly chat about what a matrix inverse actually is. Think of it like division for numbers. Just like how $5 \times \frac{1}{5} = 1$, for a square matrix $A$, its inverse, denoted as $A^{-1}$, is a matrix such that when you multiply $A$ by $A^{-1}$ (or $A^{-1}$ by $A$), you get the identity matrix ($I$). The identity matrix is a special square matrix with ones on the main diagonal and zeros everywhere else. It's crucial because multiplying any matrix by the identity matrix doesn't change the original matrix, similar to how multiplying a number by 1 doesn't change it. Not all matrices have an inverse, though. A matrix must be square (same number of rows and columns) and have a non-zero determinant to possess an inverse. If a matrix doesn't have an inverse, we call it a singular matrix. Finding the inverse is super handy in solving systems of linear equations and in various areas of computer graphics, physics, and engineering. So, understanding how to find it is a big deal!

The Gauss-Jordan Elimination Method

Now, let's get down to business with the Gauss-Jordan elimination method. This technique is a systematic way to transform a given matrix into its reduced row echelon form. To find the inverse of a matrix $A$, we combine $A$ with the identity matrix $I$ of the same dimension, forming an augmented matrix $[A|I]$. Our goal is to use elementary row operations to transform the left side (matrix $A$) into the identity matrix $I$. If we can successfully do this, the right side of the augmented matrix will automatically transform into the inverse matrix $A^{-1}$. The elementary row operations are our tools: we can swap two rows, multiply a row by a non-zero scalar, or add a multiple of one row to another row. The Gauss-Jordan method is essentially an extension of Gaussian elimination. While Gaussian elimination aims to get the matrix into row echelon form (upper triangular with leading ones), Gauss-Jordan goes a step further to achieve reduced row echelon form, meaning all entries above and below the leading ones are also zero. This makes the final result much cleaner and directly gives us the inverse matrix without needing back-substitution, which is required in Gaussian elimination.

Step-by-Step Calculation for Matrix A

Alright, guys, let's apply the Gauss-Jordan method to our matrix $A=\left[\begin{array}{rrr} 5 & 5 & 0 \\ 0 & -4 & -4 \\ -2 & 0 & 2 \end{array}\right]$.

Step 1: Form the Augmented Matrix

First, we create our augmented matrix by placing $A$ on the left and the 3x3 identity matrix $I$ on the right:

$$[A|I] = \left[\begin{array}{rrr|rrr} 5 & 5 & 0 & 1 & 0 & 0 \\ 0 & -4 & -4 & 0 & 1 & 0 \\ -2 & 0 & 2 & 0 & 0 & 1 \end{array}\right]$$

Step 2: Get a Leading 1 in the First Row, First Column

Our target is to make the first column look like $[1, 0, 0]^T$. We can start by dividing the first row ($R_1$) by 5 to get a 1 in the top-left corner. This might introduce fractions, but don't worry, it's part of the process!

$$R_1 \rightarrow \frac{1}{5}R_1 \\ \left[\begin{array}{rrr|rrr} 1 & 1 & 0 & \frac{1}{5} & 0 & 0 \\ 0 & -4 & -4 & 0 & 1 & 0 \\ -2 & 0 & 2 & 0 & 0 & 1 \end{array}\right]$$

Step 3: Eliminate the Other Entries in the First Column

Next, we want to make the entries below the leading 1 in the first column zero. We can do this by adding 2 times the first row to the third row ($R_3$). The second row already has a 0 in the first column, so we don't need to touch it for now.

$$R_3 \rightarrow R_3 + 2R_1 \\ \left[\begin{array}{rrr|rrr} 1 & 1 & 0 & \frac{1}{5} & 0 & 0 \\ 0 & -4 & -4 & 0 & 1 & 0 \\ 0 & 2 & 2 & \frac{2}{5} & 0 & 1 \end{array}\right]$$

Step 4: Get a Leading 1 in the Second Row, Second Column

Now, let's focus on the second column. We want a 1 in the second row, second column. We can achieve this by dividing the second row ($R_2$) by -4.

$$R_2 \rightarrow -\frac{1}{4}R_2 \\ \left[\begin{array}{rrr|rrr} 1 & 1 & 0 & \frac{1}{5} & 0 & 0 \\ 0 & 1 & 1 & 0 & -\frac{1}{4} & 0 \\ 0 & 2 & 2 & \frac{2}{5} & 0 & 1 \end{array}\right]$$

Step 5: Eliminate Other Entries in the Second Column

We need to make the entry above and below the leading 1 in the second column zero. We'll subtract the second row from the first row ($R_1$), and subtract 2 times the second row from the third row ($R_3$).

$$R_1 \rightarrow R_1 - R_2 \\ R_3 \rightarrow R_3 - 2R_2 \\ \left[\begin{array}{rrr|rrr} 1 & 0 & -1 & \frac{1}{5} & \frac{1}{4} & 0 \\ 0 & 1 & 1 & 0 & -\frac{1}{4} & 0 \\ 0 & 0 & 0 & \frac{2}{5} & \frac{1}{2} & 1 \end{array}\right]$$

Step 6: Analyze the Result

Uh oh! Look at the third row. We've ended up with a row of zeros on the left side $[0, 0, 0]$. This means that the original matrix $A$ is singular, and therefore, it does not have an inverse. The Gauss-Jordan elimination process terminates here, indicating that no matter how many row operations we perform, we cannot transform the left side into the identity matrix.

Why did this happen? A row of zeros on the left side of the augmented matrix during the Gauss-Jordan elimination process is a clear indicator that the determinant of the original matrix is zero. If the determinant is zero, the matrix is singular and cannot be inverted. You can verify this by calculating the determinant of $A$ separately. For a 3x3 matrix $\left[\begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array}\right]$, the determinant is $a(ei-fh) - b(di-fg) + c(dh-eg)$.

Let's calculate the determinant of $A$:

$det(A) = 5((-4)(2) - (-4)(0)) - 5((0)(2) - (-4)(-2)) + 0((0)(0) - (-4)(-2))$

$det(A) = 5(-8 - 0) - 5(0 - 8) + 0$

$det(A) = 5(-8) - 5(-8)$

$det(A) = -40 + 40 = 0$

Since the determinant of $A$ is 0, the matrix $A$ is indeed singular and has no inverse. The Gauss-Jordan method correctly revealed this by producing a row of zeros. This is a really important concept, guys – sometimes the answer is that there is no answer in the form of an inverse!

Conclusion: No Inverse for Matrix A

So, to wrap things up, the Gauss-Jordan elimination method is a powerful tool for finding matrix inverses. However, as we saw with our example matrix $A$, it also effectively tells us when an inverse does not exist. The appearance of a row of zeros on the left side of the augmented matrix during the process is the definitive sign of a singular matrix, meaning it has no inverse. In this particular case, our matrix $A$ is singular because its determinant is zero. Therefore, we cannot find an $A^{-1}$ such that $A A^{-1} = I$ and $A^{-1} A = I$. Always remember to check for this condition when using the Gauss-Jordan method or any other method for finding matrix inverses. It's a crucial step in understanding the properties of matrices and their invertibility. Keep practicing, and you'll master this in no time!