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Table of Contents
Corollary
If $A$ is an $n\times n$ matrix and $X$ is an $n\times k$ matrix with $X\ne 0_{n\times k}$ and $AX=0_{n\times k}$, then $A$ is not invertible.
Proof
Apply the previous corollary (from the end of lecture 11) to $X$ and to the matrix $Y=0_{n\times k}$: we have $X\ne Y$ but $AY=A0_{n\times k}=0_{n\times k}$, so $AX=AY$. ■
Example
The matrix $\newcommand{\mat}[1]{\begin{bmatrix}#1\end{bmatrix}}A=\mat{1&4&5\\2&5&7\\3&6&9}$ is not invertible, since $X=\mat{1\\1\\-1}$ 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
This is a calculation (done in the lectures; you should also check it yourself). ■
Definition: the determinant of a $2\times 2$ matrix
The number $ad-bc$ is called the determinant of the $2\times 2$ matrix $A=\mat{a&b\\c&d}$. We write $\det(A)=ad-bc$ for this number.
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.
- $A$ is invertible if and only if $\det(A)\ne0$.
- If $A$ is invertible, then $A^{-1}=\frac{1}{\det(A)}\mat{d&-b\\-c&a}$.
Proof
If $A=0_{2\times 2}$, then $\det(A)=0$ and $A$ is not invertible. So the statement is true is this special case.
Now assume that $A\ne0_{2\times 2}$ and let $J=\mat{d&-b\\-c&a}$.
By the previous lemma, we have \[AJ=(\det(A))I_2=JA.\]
If $\det(A)\ne0$, then multiplying this equation through by the scalar $\frac1{\det(A)}$, we get \[ A\left(\frac1{\det(A)}J\right)=I_2=\left(\frac1{\det(A)}J\right) A,\] so if we write $B=\frac1{\det(A)}J$ to make this look simpler, then we obtain \[ AB=I_2=BA,\] so in this case $A$ is 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 $J\ne 0_{2\times2}$ (since $A\ne0_{2\times2}$, and $J$ is obtained from $A$ by swapping two entries and multiplying the others by $-1$). Hence by the previous corollary, $A$ is not invertible in this case. ■
Examples
- $\det\mat{2&3\\-4&-6}=2(-6)-3(-4)=-12-(-12)=0$, so $\mat{2&3\\-4&-6}$ is not invertible.
- $\det\mat{2&3\\-4&5}=2(5)-3(-4)=10-(-12)=22\ne0$, so $\mat{2&3\\-4&5}$ is invertible with inverse \[\mat{2&3\\-4&5}^{-1}=\frac1{22}\mat{5&-3\\4&2} = \mat{\frac 5{22}&-\frac3{22}\\\frac2{11}&\frac1{11}}.\]
The transpose of a matrix
Recall from tutorial sheet 4 that the transpose of an $n\times m$ matrix $A$ is the $m\times n$ matrix $A^T$ whose $(i,j)$ entry is the $(j,i)$ entry of $A$. In other words, to get $A^T$ from $A$, you write the rows of $A$ as columns, and vice versa; equivalently, you reflect $A$ in its main diagonal.
For example, $\mat{a&b\\c&d}^T=\mat{a&c\\b&d}$.
Exercise: simple properties of the transpose
Prove that for any matrix $A$:
- $(A^T)^T=A$; and
- $(A+B)^T=A^T+B^T$ if $A$ and $B$ are matrices of the same size; and
- $(cA)^T=c(A^T)$ for any scalar $c$.
In tutorial sheet 4, we proved:
Lemma: transposes and row-column multiplication
If $a$ is a $1\times m$ row vector and $b$ is an $m\times 1$ column vector, then \[ vw=w^Tv^T.\]