How do you determine if a matrix is invertible by investigating the equation Ax = I?

For a 3x3 matrix (A) with the following

Row 1: 1, 0, 1
Row 2: 1, 1, 0
Row 3: 0, 1, 1

I know the identity for a 3x3 matrix is

Row 1: 1, 0, 0
Row 2: 0, 1, 0
Row 3: 0, 0, 1

Also, I know the corresponding x must be of the form 3x3 to ensure the multiplication is valid to produce the identity matrix.

Therefore, x will have some form associated with:

Row 1: x11, x12, x13
Row 2: x21, x22, x23
Row 3: x31, x32, x33

Not sure how to proceed from here to find out whether the matrix is invertible.

I know for a 2x2 matrix I can tell whether the matrix is invertible by examining the determinant such that if the determinant is 0 then the matrix is said to be singular, hence has no inverse. Does this property hold for a 3x3 matrix?

  • 1
    $\begingroup$ Are you familiar with the various row elementary operations used when solving linear equations? $\endgroup$ Commented Mar 16, 2017 at 23:19
  • $\begingroup$ Yes I'm aware I can find the determinant the long method, by placing the matrix A side by side with the identity and then performing calculations until the identity appears on the right side. Is there any other method? $\endgroup$
    – Diante
    Commented Mar 16, 2017 at 23:41
  • $\begingroup$ Specifically by investigating that equation? I know of other ways $\endgroup$ Commented Mar 16, 2017 at 23:43
  • $\begingroup$ @Diante I think you have the wrong idea. What you were referring to is the algorithm for solving the inverse of $A$, provided that $A$ is invertible. The reason it works is due to the properties of elementary matrices. There are many ways to find the determinant of a square matrix. One more commonly used method would be cofactor expansion. You can look it up online. $\endgroup$ Commented Mar 17, 2017 at 17:24

2 Answers 2


You are correct that a real valued square matrix is invertible $\iff$ its determinant is nonzero. Also, $A$ represents a linear transformation between vector spaces of the same dimension, so it is invertible $\iff$ $\ker(A) = \{0\}$. The second method corresponds to row reducing A.


Note that $x$ here could be a product of matrices. The purpose of this exercise is to recall the fact that the invertible square matrix $A_{n\text{ x } n}$ can be reduced to $I_{n}$ by a series of row operations, which is represented as a product of elementary matrices, $x$.


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