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Note To follow this tutorial, you should know how to multiply matrices.

What is a matrix inverse?

%Q OK. Now we can add, subtract and multiply matrices. That leaves one more operation: division. Specifically, if $A$ and $B$ are, say, $n \times n$ matrices, what is $\frac{A}{B}?$
%A First ask yourself what division in the realm of real numbers actually means: it is really a form of multiplication: dividing $3$ by $7$ is the same as multiplying $3$ by $\frac{1}{7}$, the (multiplicative) inverse of $7.$ In other words, Thus, in the realm of real numbers, we could actually forget all about division and just multiply by the inverse whenever we wanted to divide.

%Q Why complicate things like this? Just tell us how to divide matrices and be done with it!
%A Patience! Writing down, say, $\frac{3}{7}$ instead of $3\cdot7^{-1}$, is fine (and also customary) for real numbers. But should $\frac{3}{7}$ really mean $3\cdot7^{-1}$ or should it mean $7^{-1}\cdot 3$? Of course, it doesn't really matter, as multiplcation of real numbers is commutative. But multiplication of matrices is not commutative:

By $\frac{A}{B}$ should we mean $A^{-1}B$ or $BA^{-1}$?

It is because the notion of "$B$ divided by $A$" is inherently ambiguous that we never ever talk about division of matrices. Instead, we content ourselves with with using multiplicative inverses of matrices.

%Q Fine. So how do we calculate the inverse, $A^{-1},$ of a given matrix $A?$
%A Not so fast! Before we try to find the inverse of a matrix $A,$ we must first know exactly what we mean by the (multiplicative) inverse. The inverse of a number $a$ is the number, often written $a^{-1},$ with the property that

$a^{-1}a = 1 \quad$ %and $\quad a^{-1}a = 1$

For example, the inverse of $76$ is the number $76^{-1} = \frac{1}{76}$, since $\frac{1}{76}\cdot 76 = 1$ and $76 \cdot \frac{1}{76} = 1.$ (By the way, not every number has an inverse: the number 0 does not–it is the only number with no multiplicative inverse, and we say that zero is "not invertible.")

Thus, the inverse of the matrix $A$ should be the matrix $A^{-1}$ with the property that

$A^{-1}A = 1 \quad$ %and $\quad A^{-1}A = I \qquad \qquad$ The identity matrix $I$ plays the role of the number 1 in the realm of matrices.

Fill in the missing entry.

Finding the inverse of a matrix

%Q How on earth do we find the inverse of a matrix?
%A If you used unknowns for all the entries of the inverse of a matrix, the condition that $AA^{-1} = I$ would give you a system of linear equations. Solving this system of equations corresponds to the following method for finding the inverse of any matrix: (See %4 for a detailed explanation). %Let $A = $%10. Determine whether or not $A$ is invertible, and its inverse if it is, as follows:

Step 1. Write down the $n\times 2n$ matrix [A,|,I]:   (Use the tab key to go from one cell to the next.)

Step 2. What matrix do you obtain if you completely row-reduce the above matrix? (It is good practice to reduce the matrix by hand, as in the %7. If you are feeling lazy and want to use technology, use the %9.)

It follows that

%Let $B = $%11. Determine whether or not $B$ is invertible, and its inverse if it is, as follows:

Step 1. Write down the $n\times 2n$ matrix [B,|,I]:   (Use the tab key to go from one cell to the next.)

Step 2. What matrix do you obtain if you completely row-reduce the above matrix? (It is good practice to reduce the matrix by hand, as in the %7. If you are feeling lazy and want to use technology, use the %9.)

It follows that

Want more practice?

Using matrix inverses to solve systems of linear equations

Look at the following system of linear equations: To solve the system, we would normally write out and row-reduce the augmented matrix for the system. However, there is another way to solve certain systems of equations like this one: Take $A$ to be the matrix of coefficients of the left-hand sides:

$A = $[-1,1,0;-4,1,1;0,-2,1],

take $X$ to be the column matrix of unknowns: $X = $[x;y;z], and take $B$ to be the column matrix of right-hand sides: $B = $[4;-1;0]. Then the product

$AX=$[-1,1,0;-4,1,1;0,-2,1][x;y;z]$=$[-x+y;-4x+y+z;-2y+z]

is the column matrix of left-hand sides. So, to say that the left-hand sides are equal to the right-hand sides is the same as saying that

$AX = B,$

and this is called the matrix form of the system of equations.

Now, notice that we can just solve this matrix equation for $X$ by "dividing both sides by $A,$ by which we mean multiplying both sides by the inverse of $A$ (assuming it exists):

$A^{-1}AX = A^{-1}B \qquad \Rightarrow \qquad IX = A^{-1}B \qquad \Rightarrow \qquad X = A^{-1}B,$

and we have solved for $X$! In this example, $A^{-1}$ can be calculated to be [3,-1,1;4,-1,1;8,-2,3] () and so

$X = A^{-1}B = $[3,-1,1;4,-1,1;8,-2,3][4;-1;0] = [13;17;34].

Thus the solution is $x = 13,\ y = 17,\ z = 34.$

Note This process works only if the matrix $A$ of coefficients is invertible, and gives a unique solution in that case as demonstrated. If the matrix is not invertible, then this method does not work, and to find the general solution, if any, you will need to reduce the augmented matrix as in %7.

Consider the following system of equations: