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• $\def\mat#1{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]}A=\mat{2&4\\0&1}$ is invertible, and $A^{-1}=\mat{\tfrac12&-2\\0&1}$.
  • i.e. $\mat{2&4\\0&1}^{-1}=\mat{\tfrac12&-2\\0&1}$
• if $a$ is a non-zero scalar, then $[a]^{-1}=[\tfrac 1a]$
• $I_n^{-1}=I_n$ for any $n$
• $0_{n\times n}^{-1}$, $\mat{1&0\\0&0}^{-1}$ and $\mat{1&2\\-3&-6}^{-1}$ do not exist
  • because these matrices aren't invertible

Warning

If $A$, $B$ are matrices, then $\frac AB$ doesn't make sense!

• Never write this down as it will almost always lead to mistakes.
• In particular, $A^{-1}$ definitely doesn't mean $\frac 1A$. Never write this either!

Proposition: solving $AX=B$ when $A$ is invertible

If $A$ is an invertible $n\times n$ matrix and $B$ is an $n\times k$ matrix, then $ AX=B$ has a unique solution: $X=A^{-1}B$.

Proof

• First check that $X=A^{-1}B$ really is a solution:
  • $AX=A(A^{-1}B)=(AA^{-1})B=I_nB=B$
• Uniqueness: suppose $X$ and $Y$ are both solutions
• Then $AX=B$ and $AY=B$, so $AX=AY$.
• Multiply both sides on the left by $A^{-1}$:
  • $A^{-1}AX=A^{-1}AY$, or $I_nX=I_n Y$, or $X=Y$.
• So any two solutions are equal.■

Corollary

If $A$ is an $n\times n$ matrix and there is a non-zero $n\times m$ matrix $K$ so that $AK=0_{n\times m}$, then $A$ is not invertible.

Proof

• $AX=0_{n\times m}$ has (at least) two solutions:
  • $X=K$,
  • and $X=0_{n\times m}$
• If $A$ was invertible, this would contradict the Proposition.
• So $A$ cannot be invertible. ■

Example

• Let $A=\mat{1&2\\-3&-6}$. $A$ isn't invertible… why?
• One column of $A$ is $2$ times the other… exploit this.
• Let $K=\mat{-2\\1}$
• $K$ is non-zero but $AK=\mat{1&2\\-3&-6}\mat{-2\\1}=\mat{0\\0}=0_{2\times 1}$
• So $A$ is not invertible, by the Corollary.

Example

• $A=\mat{1&4&5\\2&5&7\\3&6&9}$ is not invertible
  • $X=\mat{1\\1\\-1}$ is non-zero
  • and $AX=0_{3\times 1}$.

$2\times 2$ matrices: determinants and invertibility

Question

Which $2\times 2$ matrices are invertible? For the invertible matrices, can we find their inverse?

Lemma

If $A=\mat{a&b\\c&d}$ and $J=\mat{d&-b\\-c&a}$, then we have \[ AJ=\delta I_2=JA\] where $\delta=ad-bc$.

• Proof is a calculation!

Definition: the determinant of a $2\times 2$ matrix

Theorem: the determinant determines the invertibility (and inverse) of a $2\times 2$ matrix

Let $A=\mat{a&b\\c&d}$ be a $2\times 2$ matrix.

1. $A$ is invertible if and only if $\det(A)\ne0$.
2. If $A$ is invertible, then $A^{-1}=\frac{1}{\det(A)}\mat{d&-b\\-c&a}$.

Proof

• true for $A=0_{2\times 2}$!
• Now suppose $A\ne0_{2\times 2}$. Let $J=\mat{d&-b\\-c&a}$.
• By the previous lemma, $AJ=(\det(A))I_2=JA$.
• If $\det(A)\ne0$, multiply by $\frac1{\det(A)}$ and write $B=\tfrac1{\det(A)}J$: \[ AB= A(\tfrac1{\det(A)}J)=I_2=(\tfrac1{\det(A)}J) A=BA\]
• So $ AB=I_2=BA$, so $A$ invertible with inverse $B=\frac1{\det(A)}J=\frac1{\det(A)}\mat{d&-b\\-c&a}$.

• If $\det(A)=0$, then $AJ=0_{2\times 2}$ and $J\ne 0_{2\times2}$ [why?]
• Hence by the previous corollary, $A$ is not invertible in this case. ■