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Matrix Inverses and Systems of Linear Equations
The inverse of a square matrix, and solutions to linear systems with square coefficient matrices, are intimately connected.
Matrix Inverses and Systems of Linear Equations
The inverse of a square matrix, and solutions to linear systems with square coefficient matrices, are intimately connected.
- \begin{bmatrix} -2 & 3\\ -3 & 4 \end{bmatrix}
- \begin{bmatrix} 2 & 3 & 1\\ 1 & -2 & -3\\ -2 & 4 & 6 \end{bmatrix}
- Explain why has the title it does. (Do not just state the theorem, explain the choice of the title making reference to the theorem itself.)
Solutions and Inverses
We begin with a familiar example, performed in a novel way.
Example: Solutions to Archetype B with a matrix inverse
is the system of $m=3$ linear equations in $n=3$ variables,
By we can represent this system of equations as A\vect{x}=\vect{b} where A=&&\vect{x}=\colvector{x_1\\x_2\\x_3}&&\vect{b}=\colvector{-33\\24\\5}
Now, entirely unmotivated, we define the $3\times 3$ matrix $B$, B= and note the remarkable fact that BA== \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}
Now apply this computation to the problem of solving the system of equations, \vect{x} &=I_3\vect{x}&&\\ &=(BA)\vect{x}&&\text{Substitution}\\ &=B(A\vect{x})&&\\ &=B\vect{b}&&\text{Substitution}\\
So we have \vect{x}=B\vect{b}= \colvector{-33\\24\\5}= \colvector{-3\\5\\2}
So with the help and assistance of $B$ we have been able to determine a solution to the system represented by $A\vect{x}=\vect{b}$ through judicious use of matrix multiplication. We know by that since the coefficient matrix in this example is nonsingular, there would be a unique solution, no matter what the choice of $\vect{b}$. The derivation above amplifies this result, since we were forced to conclude that $\vect{x}=B\vect{b}$ and the solution could not be anything else. You should notice that this argument would hold for any particular choice of $\vect{b}$.
The matrix $B$ of the previous example is called the inverse of $A$. When $A$ and $B$ are combined via matrix multiplication, the result is the identity matrix, which can be inserted in front of $\vect{x}$ as the first step in finding the solution. This is entirely analogous to how we might solve a single linear equation like $3x=12$. x=1x=\left(\frac{1}{3}\left(3\right)\right)x=\frac{1}{3}\left(3x\right)=\frac{1}{3}\left(12\right)=4
Here we have obtained a solution by employing the multiplicative inverse of $3$, $3^{-1}=\frac{1}{3}$. This works fine for any scalar multiple of $x$, except for zero, since zero does not have a multiplicative inverse. Consider separately the two linear equations, 0x&=12 & 0x&=0
The first has no solutions, while the second has infinitely many solutions. For matrices, it is all just a little more complicated. Some matrices have inverses, some do not. And when a matrix does have an inverse, just how would we compute it? In other words, just where did that matrix $B$ in the last example come from? Are there other matrices that might have worked just as well?
Inverse of a Matrix
Notice that if $B$ is the inverse of $A$, then we can just as easily say $A$ is the inverse of $B$, or $A$ and $B$ are inverses of each other.
Not every square matrix has an inverse. In the matrix $B$ is the inverse of the coefficient matrix of . To see this it only remains to check that $AB=I_3$. What about ? It is an example of a square matrix without an inverse.
Example: A matrix without an inverse, Archetype A
Consider the coefficient matrix from , A=
Suppose that $A$ is invertible and does have an inverse, say $B$. Choose the vector of constants \vect{b}=\colvector{1\\3\\2} and consider the system of equations $\linearsystem{A}{\vect{b}}$. Just as in , this vector equation would have the unique solution $\vect{x}=B\vect{b}$.
However, the system $\linearsystem{A}{\vect{b}}$ is inconsistent. Form the augmented matrix $\augmented{A}{\vect{b}}$ and row-reduce to \begin{bmatrix} \leading{1} & 0 & 1 & 0\\ 0 & \leading{1} & -1 & 0\\ 0 & 0 & 0 & \leading{1} \end{bmatrix} which allows us to recognize the inconsistency by .
So the assumption of $A$'s inverse leads to a logical inconsistency (the system cannot be both consistent and inconsistent), so our assumption is false. $A$ is not invertible.
It is possible this example is less than satisfying. Just where did that particular choice of the vector $\vect{b}$ come from anyway? Stay tuned for an application of the future in .
Let us look at one more matrix inverse before we embark on a more systematic study.
We will now concern ourselves less with whether or not an inverse of a matrix exists, but instead with how you can find one when it does exist. In we will have some theorems that allow us to more quickly and easily determine just when a matrix is invertible.
Condensed — the full section is in Beezer, A First Course in Linear Algebra.
Computing the Inverse of a Matrix
We have seen that the matrices from and both have inverses, but these inverse matrices have just dropped from the sky. How would we compute an inverse? And just when is a matrix invertible, and when is it not? Writing a putative inverse with $n^2$ unknowns and solving the resultant $n^2$ equations is one approach. Applying this approach to $2\times 2$ matrices can get us somewhere, so just for fun, let us do it.
There are several ways one could try to prove this theorem, but there is a continual temptation to divide by one of the eight entries involved ($a$ through $f$), but we can never be sure if these numbers are zero or not. This could lead to an analysis by cases, which is messy, messy, messy. Note how the above proof never divides, but always multiplies, and how zero/nonzero considerations are handled. Pay attention to the expression $ad-bc$, as we will see it again in a while ().
This theorem is cute, and it is nice to have a formula for the inverse, and a condition that tells us when we can use it. However, this approach becomes impractical for larger matrices, even though it is possible to demonstrate that, in theory, there is a general formula. (Think for a minute about extending this result to just $3\times 3$ matrices. For starters, we need 18 letters!) Instead, we will work column-by-column. Let us first work an example that will motivate the main theorem and remove some of the previous mystery.
Notice how the five systems of equations in the preceding example were all solved by exactly the same sequence of row operations. Would it not be nice to avoid this obvious duplication of effort? Our main theorem for this section follows, and it mimics this previous example, while also avoiding all the overhead.
We have to be just a bit careful here about both what this theorem says and what it does not say. If $A$ is a nonsingular matrix, then we are guaranteed a matrix $B$ such that $AB=I_n$, and the proof gives us a process for constructing $B$. However, the definition of the inverse of a matrix () requires that $BA=I_n$ also. So at this juncture we must compute the matrix product in the opposite order before we claim $B$ as the inverse of $A$. However, we will soon see that this is always the case, in , so the title of this theorem is not inaccurate.
What if $A$ is singular? At this point we only know that cannot be applied. The question of $A$'s inverse is still open. (But see in the next section.)
Condensed — the full section is in Beezer, A First Course in Linear Algebra.
Properties of Matrix Inverses
The inverse of a matrix enjoys some nice properties. We collect a few here. First, a matrix can have but one inverse.
When most of us dress in the morning, we put on our socks first, followed by our shoes. In the evening we must then first remove our shoes, followed by our socks. Try to connect the conclusion of the following theorem with this everyday example.
Condensed — the full section is in Beezer, A First Course in Linear Algebra.
Symbols used here
A rectangular array of numbers; a linear map.
Inequalities that allow equality; < and > exclude it.
A quantity with magnitude and direction; a column of numbers.
Scaling factor of area/volume under A; zero means singular.
The matrix that undoes A; A with rows and columns swapped.
The factor by which an eigenvector is stretched: Av = λv.
Σ u_i v_i; the length of v, √(v·v).
Questions people ask
What does a determinant mean geometrically?
It is the factor by which the matrix scales area (2×2) or volume (3×3), with a negative sign if orientation flips. Zero means the matrix flattens space and cannot be undone.
What is an eigenvector?
A direction the matrix does not turn — it only stretches it by the eigenvalue. Along eigenvectors a complicated matrix acts like multiplication by a number.
Why is matrix multiplication not commutative?
Because a matrix is a transformation and AB means "do B, then A". Rotating then reflecting is not the same as reflecting then rotating.
ჲოთრაი ჟამ.
Parts of this page are adapted from Beezer, A First Course in Linear Algebra (GFDL 1.2). Condensed and re-explained here; errors are ours.
მეტი Linear Algebra
DeterminantsMatrix inverseEigenvalues and eigenvectorsMatrix multiplicationRow reductionVector spaces, span and linear independenceOrthogonality, projections and least squaresDiagonalisation and matrix powersLinear transformations and change of basis