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What an inverse is and why it is unique, the two by two formula, Gauss-Jordan on the augmented identity, the reversal rule for products, and why elimination usually beats inversion.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to state what makes a matrix invertible and prove that an inverse is unique when it exists, apply the two by two formula in the right order and check it by multiplying back, invert a larger matrix by reducing the wide array $[A \mid I]$ and recognise from a zero row that no inverse exists, find the entry that makes a matrix singular, apply $(AB)^{-1} = B^{-1}A^{-1}$, and say why solving $Ax = b$ by elimination is usually preferable to computing an inverse.
You can multiply matrices, and you can reduce one to its reduced row echelon form. An inverse is the meeting point: it is defined by multiplication and it is computed by reduction, and neither half makes sense without the other.
A square matrix $A$ is invertible — equivalently non-singular — when there is a matrix $A^{-1}$ with $AA^{-1} = A^{-1}A = I$. A square matrix with no such partner is singular. The determinant of $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is $ad - bc$; a $2 \times 2$ matrix is invertible exactly when it is non-zero. Only square matrices are candidates: the word does not apply to a rectangular one, which can have a one-sided inverse at best.
$A^{-1}$ is the matrix for which $AA^{-1} = A^{-1}A = I$. Two facts come almost free.
It is unique. If $BA = I$ and $AC = I$ then $B = BI = B(AC) = (BA)C = IC = C$, so a left inverse and a right inverse of the same matrix are the same matrix, and there is only one of it. This is why we may write the inverse.
Not every matrix has one. $\begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}$ sends $(2, -1)$ to $(0, 0)$; no matrix can bring $(0,0)$ back to $(2,-1)$, because it also has to leave $(0,0)$ alone. Any matrix that sends a non-zero vector to zero is singular, and that is the whole of the reason.
For $2 \times 2$ there is a formula worth knowing by heart:
$$\begin{pmatrix} a & b \\ c & d \end{pmatrix}^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}, \qquad ad - bc \ne 0.$$
Swap the diagonal, negate the other two, divide by the determinant. There is no comparable formula worth using for larger matrices; from $3 \times 3$ upwards the method is reduction.
Gauss-Jordan. Write $[A \mid I]$ and row-reduce until the left block is $I$. The right block is then $A^{-1}$. The reason is the next lesson's, in one sentence: every row operation is multiplication on the left by some matrix, so reducing $A$ to $I$ means finding $E$ with $EA = I$, and the same operations applied to $I$ produce $EI = E$. If the left block cannot be reduced to $I$ — a row of zeros appears — then $A$ is singular and the algorithm has proved it.
Products reverse. $(AB)^{-1} = B^{-1}A^{-1}$, and $(A^{T})^{-1} = (A^{-1})^{T}$.
Another way: picture
A matrix is an action on the plane or on space. Invertible means the action can be run backwards: nothing was lost. Singular means it flattened something — a plane onto a line, space onto a plane — and once two different inputs have arrived at the same output, no rule can decide which one to send back. That is why the test does $A$ send some non-zero vector to zero? is the same question as is $A$ singular? rather than a separate one.
Another way: steps
Having met $x = A^{-1}b$, it is tempting to solve every system that way. Do not.
Inverting an $n \times n$ matrix is roughly three times the work of eliminating on $[A \mid b]$ — reducing one wide array of $n$ extra columns rather than one. So for a single right-hand side, inversion costs three times as much and then still requires a multiplication. It is also less accurate in floating point, because the entries of $A^{-1}$ can be enormous when $A$ is nearly singular, and the errors in them are then multiplied into the answer.
The honest uses of an inverse are these. It is a formula: $x = A^{-1}b$ says what the solution is as a function of $b$, which elimination does not. It is an object of algebra: $(AB)^{-1} = B^{-1}A^{-1}$ and $A^{-1}$ appearing in $P^{-1}AP$ are statements about matrices, not requests for arithmetic. And where many systems share one $A$, the work done once is reused — though even then, the factorisation of the next lesson is the better way to reuse it.
The rule of thumb professionals use: if you have written $A^{-1}b$ in a program, you meant solve $Ax = b$, and you should say that instead.
For a square matrix $A$, the following say the same thing, and the course will keep adding to the list:
This is the invertible matrix theorem, and it is the reason the subject hangs together: a question about solutions, a question about columns and a question about undoing all turn out to be one question.
Every word of it needs "square". For a $3 \times 2$ matrix the columns can perfectly well be independent while $Ax = b$ is unsolvable for most $b$; the two conditions come apart immediately. Keeping track of which claims need squareness is most of the care this subject asks for.
There is no division. $\dfrac{b}{A}$ means nothing. $A^{-1}b$ and $bA^{-1}$ are different expressions and usually only one of them is even defined, so the side an inverse is applied on is part of the answer.
Negate the off-diagonal, not the diagonal. In the $2 \times 2$ formula, $a$ and $d$ swap places and keep their signs; $b$ and $c$ stay put and change sign. Multiplying back catches the error at once, which is why it is worth the ten seconds.
$(AB)^{-1}$ is $B^{-1}A^{-1}$. The other order is wrong for exactly the same reason it is wrong for transposes, and for the reason you take a coat off before a jumper.
"Singular" is not "nearly zero". A matrix of enormous entries can be singular and a matrix of tiny ones invertible. The test is the determinant, or equivalently whether the reduction reaches the identity.
A rectangular matrix has no inverse at all. Not a hard one to find — there is no such thing. $A^{T}A$ may be invertible, and that is a different and very useful statement, but it is not an inverse of $A$.
$A = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{pmatrix}$. Write $[A \mid I]$; the left block is already upper triangular with pivots $1$.
No downward pass is needed here.
Row $2$ minus $3$ times row $3$ clears the $3$; on the right, row $2$ becomes $(0, 1, -3)$.
Every operation is done to the whole wide row.
Row $1$ minus $2$ times the new row $2$ clears the $2$: the right block is $\begin{pmatrix} 1 & -2 & 6 \\ 0 & 1 & -3 \\ 0 & 0 & 1 \end{pmatrix}$, and multiplying back gives $I$. Note the $6$ — it is the product of the two multipliers, not either of them.
Corrections accumulate; that is why reduction, not guesswork.
$B = \begin{pmatrix} 3 & 6 \\ 1 & 2 \end{pmatrix}$: the determinant is $6 - 6 = 0$, so the formula's denominator vanishes and there is no inverse.
The arithmetic test.
Reducing $[B \mid I]$: row $1$ minus $3$ times row $2$ gives a zero row on the left. The reduction can go no further, so no sequence of operations turns $B$ into $I$.
The algorithmic test, reaching the same verdict.
And $B\begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$: a non-zero vector sent to zero. Three descriptions, one fact.
The structural test — and the one that explains the other two.
Determinant: $2 \times 3 - 1 \times 4 = 2$, which is not zero, so yes.
Always the determinant first.
Swap the diagonal and negate the rest: $\begin{pmatrix} 3 & -1 \\ -4 & 2 \end{pmatrix}$, then divide by $2$.
The formula, in the order it is stated.
$A^{-1} = \begin{pmatrix} 3/2 & -1/2 \\ -2 & 1 \end{pmatrix}$, and multiplying back gives $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$.
Write the inverse of $A = \begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix}$.
This task has no paper form; do it on a device.
Write the inverse of $L = \begin{pmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 2 & 5 & 1 \end{pmatrix}$.
This task has no paper form; do it on a device.
For which value of $k$ does $\begin{pmatrix} 3 & 8 \\ 3 & k \end{pmatrix}$ fail to have an inverse?
Answer:
Put the stages of inverting $\begin{pmatrix} 5 & 2 \\ 2 & 1 \end{pmatrix}$ by row reduction into the order they are carried out.
Number the steps in order (write the number in the box):
$A$ and $B$ are invertible matrices of the same size. What is $(AB)^{-1}$?
Solve $Ax = b$ for the column $x$, where $A = \begin{pmatrix} 1 & 2 \\ 2 & 5 \end{pmatrix}$ and $b = \begin{pmatrix} 5 \\ 3 \end{pmatrix}$.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Write the inverse of $A = \begin{pmatrix} 1 & 1 \\ 3 & 4 \end{pmatrix}$.
This task has no paper form; do it on a device.
You can invert a matrix by formula and by reduction, detect a singular one, and use the reversal rule. Say in your own words why a matrix that sends some non-zero vector to zero can have no inverse. Next: the matrices that carry out row operations, and the factorisation they produce.
10. Your turn: is $\begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix}$ invertible, and if so what is its inverse?, step 3
Never skip the check.