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• An $n\times n $ matrix $A$ is invertible if there is a matrix $C$ solving the matrix equations $AC=I_n$ and $CA=I_n$. We then say $C$ is an inverse of $A$.

• e.g. $\def\mat#1{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]}\mat{2&4\\0&1}$ is invertible, with inverse $C=\mat{0.5&-2\\0&1}$
• if $[a]$ is $1\times 1$ and $a\ne 0$, then $C=[\tfrac 1a]$ is an inverse
• $I_n$ is its own inverse for any $n$
• $0_{n\times n}$, $\mat{1&0\\0&0}$ and $\mat{1&2\\-3&-6}$ aren't invertible

Proposition: uniqueness of the inverse

If $A$ is an invertible $n\times n$ matrix, then $A$ has a unique inverse.

Proof

• $A$ invertible, so $A$ has at least one inverse.
• Suppose it has two inverses, say $C$ and $D$.
• Then $AC=I_n=CA$ and $AD=I_n=DA$.
• So $C=CI_n=C(AD)=(CA)D=I_nD=D$
• So $C=D$.
• So any two inverses of $A$ are equal.
• So $A$ has a unique inverse.■

Definition/notation: $A^{-1}$

Examples again

• $A=\mat{2&4\\0&1}$ 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, never write down $\frac AB$. It doesn't make (unambiguous) sense!

• $\frac AB$ is ambiguous, or not well defined:
  • it could mean $B^{-1}A$
  • or equally well it could mean $AB^{-1}$
  • and these are usually different because matrix multiplication isn't commutative!
• 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.■

• Easy to remember how to solve $AX=B$
• Multiply both sides (on the left) by $A^{-1}$
  • reason: want to “cancel $A$” on the left hand side
  • so we just end up with $X=\ldots$

• Example: Solve $\mat{2&4\\0&1}X=\mat{1\\3}$
• Know $\mat{2&4\\0&1}^{-1}=\mat{0.5&-2\\0&1}$
• Solution: $X=\mat{2&4\\0&1}^{-1}\mat{1\\3}=\mat{0.5&-2\\0&1}\mat{1\\3}=\mat{-5.5\\3}$
• Check: $\mat{2&4\\0&1}X = \mat{2&4\\0&1}\mat{-5.5\\3}=\mat{1\\3}$

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 $X=0_{n\times m}$
• So there is not a unique solution to $AX=B$, for $B$ the zero matrix
• If $A$ was invertible, this would contradict the uniqueness statement of the last 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 \[ AJ=(ad-bc) I_2=JA.\]

• Proof is an easy calculation!
• Note that $(ad-bc) I_2=(ad-bc)\mat{1&0\\0&1}=\mat{ad-bc&0\\0&ad-bc}$
• Now just show that $AJ$ and $JA$ both give the same matrix (exercise).

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

• Let $J=\mat{d&-b\\-c&a}$, so $AJ=\det(A) I_2=JA$ (Lemma).

If $\det(A)\ne 0$:

• Multiply by $\frac1{\det(A)}$: $\quad A(\tfrac1{\det(A)}J)=I_2=(\tfrac1{\det(A)}J) A$
• So $A$ invertible with $A^{-1}=\frac1{\det(A)}J=\frac1{\det(A)}\mat{d&-b\\-c&a}$.

If $\det(A)=0$:

• $AJ=0_{2\times 2}$
• If $J\ne 0_{2\times 2}$ then (by corollary) $A$ isn't invertible.
• If $J=0_{2\times 2}$ then $A=0_{2\times 2}$, which isn't invertible.■

Using the inverse to solve a matrix equation

• Solve $\mat{1&5\\3&-2}X=\mat{4&1&0\\0&2&1}$ for $X$
• Write $A=\mat{1&5\\3&-2}$
• $\det(A)=1(-2)-5(3)=-2-15=-17$
• So $A$ is invertible, and $A^{-1}=-\frac1{17}\mat{-2&-5\\-3&1}$
• Solution is $X=A^{-1}\mat{4&1&0\\0&2&1}=-\frac1{17}\mat{-2&-5\\-3&1}\mat{4&1&0\\0&2&1}=\frac1{17}\mat{8&12&5\\12&1&-1}$