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Let A be an invertible matrix. Thus, A A' = I where I is the identity matrix and A' the inverse of A. We can find A' two ways (that I know of; Linear Algebra I&II).

First method:

Straightforward multiplication of A by an arbitrary matrix A' (I assign variables to each element in the n by n matrix). Thus, I get simple equality and I can easily find matrix A'.

Second method:

Using the Gauss-Jordan method I can use elementary row operations to convert A into an identity matrix. Each row operation I do to A I also do to I. Thus, I transforms into the inverse of A.

I have attached an example: enter image description here

I know that there must be something fundamentally wrong with the first method, but I am not sure why. I know I am technically just finding an arbitrary matrix in the first method so it is not really the inverse. Someone, please explain.

Brian Blumberg
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    Your first method is perfectly okay, and can be used to find the inverse matrix in all situations (although this can be a little tricky for larger matrices, especially without 0s in them). In fact, your first answer is correct and your second answer is incorrect. Your method is correct, but you forgot to apply the $R_2-2R_1$ operation to the matrix on the right side. – TomKern Jul 04 '21 at 01:05
  • @TomKern true, thanks – Brian Blumberg Jul 04 '21 at 01:10
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    "I know I am technically just finding an arbitrary matrix in the first method so it is not really the inverse" - not true! If $A$ is a square matrix and you can find a matrix $B$ such that $AB = I$, then $B$ is the inverse of $A$. (The inverse is unique if it exists.) Similarly, if you can find a matrix $C$ such that $CA = I$, then $C$ is the inverse of $A$, and hence $C = B = A^{-1}$. –  Jul 04 '21 at 01:35

1 Answers1

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There’s nothing wrong with either method. In fact, they are equivalent. Unfortunately, you made an error in calculating the row operation $R_2 \rightarrow R_2 - 2R_1.$ (Edit: this error was later corrected by the question asker.)

When one is doing a row operation on an augmented matrix like

$$\left[\begin{matrix} 4 & -2 \;\\ 2 & 0 \;\\ \end{matrix}\right\rvert \left.\begin{matrix} \;1 & 0 \\ \;0 & 1 \end{matrix}\right],$$

this is the same as left-multiplying the system

$$\begin{bmatrix} 4 & -2 \\ 2 & 0 \end{bmatrix}\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $$

by the elementary matrix that the row operation represents.

The augmented matrix method is usually preferred because the symbols $a,$ $b,$ $c,$ and $d$ are unchanged if we are only multiplying by matrices on the left; so, there’s no real need to “keep track” of them. Of course, the goal is the same.

Once we reduce the matrix on the left to identity, the matrix on the right must be the inverse. There is only one matrix that could possibly satisfy that equation if $A$ is invertible. But since we’re really doing the same operations no matter which presentation we choose, we usually choose the augmented matrix because it’s more compact. The result is the same.

shoteyes
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    Yes, I realized the mistake but was too lazy to take another picture. As long as the first method is valid it's ok. Thanks! – Brian Blumberg Jul 04 '21 at 01:14