Inverse matrices

Lecture 7

Author

Affiliation

Minjae Park

Auburn University
MATH 2660 - Spring 2026

Published

January 23, 2026

Recap

A linear equation has the form \(A\vec{x}=\vec{b}\), where \(A\) is an \(n\times m\) matrix, \(\vec{x}\in\mathbb{R}^m\) is the vector of unknowns, and \(\vec{b}\in\mathbb{R}^n\) is the given vector.
The corresponding augmented matrix is written as \([A\mid\vec{b}]\).
To solve a system of linear equations, we apply Gauss–Jordan elimination to the augmented matrix using elementary row operations to obtain the reduced row-echelon form (RREF).
Depending on the final RREF, the system may have a unique solution, infinitely many solutions, or no solution.

Consider the system of linear equations \[ \left\{\begin{aligned} x+2y+3z &= 9\\ 2x-y+z &= 8\\ 3x-z &= 3 \end{aligned}\right. \]

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Consider the augmented matrix \[ \left[\begin{array}{ccc|c} 1 & 3 & 1 & 0\\ 0 & 0 & 1 & 2\\ 0 & 0 & 0 & 0 \end{array}\right]. \]

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Recall that a linear transformation \(A:\mathbb{R}^2\to\mathbb{R}^2\) describes how points in the plane are deformed linearly.
If \(\vec{v}_1=A\vec{e}_1\) and \(\vec{v}_2=A\vec{e}_2\), then a vector \[\vec{x}=\langle x_1,x_2 \rangle=x_1\vec{e}_1+x_2\vec{e}_2\] is mapped to \[\vec{y}=A\vec{x}=x_1\vec{v}_1+x_2\vec{v}_2.\]
Geometrically, \(A\) is completely determined by how the unit square spanned by \(\vec{e}_1,\vec{e}_2\) is transformed into the parallelogram spanned by \(\vec{v}_1,\vec{v}_2\).

Each coordinate vector defines an axis, and together they generate a grid on the plane.
The deformed picture can be understood by tracking how this grid changes. For example, Aubie’s head is located at \((0,1)\) in every coordinate system.

Since a linear transformation is determined by how it transforms the grid of unit squares, we may ask whether there exists a linear transformation \[B:\mathbb{R}^2\to\mathbb{R}^2\] that maps the parallelogram spanned by \(\vec{v}_1,\vec{v}_2\) back to the unit square spanned by \(\vec{e}_1,\vec{e}_2\).
Such a transformation \(B\) would undo the effect of \(A\), and vice versa.
In other words, for any vector \(\vec{x}, \vec{y}\in \mathbb{R}^2\), \[\vec{x}=B(A(\vec{x})) = B\circ A (\vec{x}),\] \[\vec{y}=A(B(\vec{y})) = A\circ B (\vec{y}).\]

Consider two linear transformations \(A,B:\mathbb{R}^n\to\mathbb{R}^n\).
They are represented by matrices \[A=[\vec{a}_1\ \vec{a}_2\ \dots \ \vec{a}_n],\qquad B=[\vec{b}_1\ \vec{b}_2\ \dots \ \vec{b}_n].\]
Fact: the composition \(C = A\circ B\), read as “apply \(B\) first, then apply \(A\)”, is also a linear transformation.
How can we represent the matrix of the linear transformation \(C\)?

The matrix \(C = [\vec{c}_1\ \vec{c}_2\ \dots \ \vec{c}_n]\) is determined by how it acts on the coordinate vectors \(\vec{e}_i\).
Observe that \[\vec{c}_i = C\vec{e}_i = A(B\vec{e}_i)=A \vec{b}_i.\]
This is exactly how we defined the matrix multiplication \(AB\).
Hence, the matrix representing \(A\circ B\) is the matrix product \(AB\).
Similarly, the composition \(B\circ A\) is represented by the matrix \(BA\).

Let \(A:\mathbb{R}^n \to \mathbb{R}^n\) be a linear transformation.
If another linear transformation \(B:\mathbb{R}^n \to \mathbb{R}^n\) undoes what \(A\) does (and vice versa), then \[AB=BA=I_n,\] where \(I_n\) is the \(n \times n\) identity matrix.
Fact: if such a matrix \(B\) exists, it is unique, and \(A\) is called invertible.
This matrix \(B\) is called the inverse matrix of \(A\), denoted by \(A^{-1}\).

Suppose we want to solve the linear equation \[A\vec{x}=\vec{b}.\]
If \(A\) has an inverse matrix, we can “undo” \(A\) by multiplying both sides by \(A^{-1}\): \[\vec{x}=A^{-1}A\vec{x}=A^{-1}\vec{b}.\]
This produces a single vector solution, so the solution is unique.
Therefore, a matrix \(A\) has an inverse if and only if it can be reduced to the identity matrix \(I_n\) using elementary row operations.

Consider the case where \(A=[\vec{v}_1\ \vec{v}_2]\) is a \(2\times2\) matrix.
Geometrically, \(A\) transforms the square grid spanned by \(\vec{e}_1\) and \(\vec{e}_2\) into the parallelogram grid spanned by \(\vec{v}_1\) and \(\vec{v}_2\).
If this parallelogram forms a genuine grid (that is, it has nonzero area), then it can be “unsqueezed” back to the unit square, and the matrix is invertible.
However, if \(\vec{v}_1\) and \(\vec{v}_2\) collapse into a single line and fail to form a grid, information is lost and the transformation cannot be undone.
For example, this happens when \(\vec{v}_1=c \vec{v}_2\).
This geometric viewpoint motivates the concept of the determinant, which we will study later.

Computing inverse matrices is an important task in numerical analysis.
In general, finding inverses can be computationally expensive and numerically sensitive.
There are several algorithms and approximation methods designed for this purpose.
In this course, we will not focus on these methods in detail; you may use computer software when needed.
However, for a \(2\times2\) matrix, there is a simple formula you can verify by hand: \[A^{-1}=\frac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}, \qquad \text{when } A=\begin{bmatrix} a & b \\ c & d \end{bmatrix} \text{ and } ad-bc\neq0.\]

In the next class, we will interpret elementary row operations as special linear transformations represented by matrices called elementary matrices.
In fact, we already learned those—scaling, transposing, and shearing.
Since Gauss–Jordan elimination is a sequence of such operations, it can be written as a product of elementary matrices.
If a matrix is invertible, Gauss–Jordan elimination reduces it to the identity matrix.
In this process, we explicitly construct a matrix that multiplies the given matrix to give the identity—that matrix is the inverse.