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--- lecture_13 [2015/03/03 10:28] – [Theorem: the determinant determines the invertibility (and inverse) of a $2\times 2$ matrix] rupert
+++ lecture_13 [2017/03/07 15:08] (current) – [Step 3: the determinant of a $3\times 3$ matrix using Laplace expansion along the first row] rupert
@@ Line 1: / Line 1: @@
-==== Corollary ====
+==== Example ====
-If $A$ is an $n\times n$ matrix and $X$ is a non-zero $n\times k$ matrix with $AX=0_{n\times k}$, then $A$ is not invertible.
-=== Proof ===
+Let's solve the matrix equation $\def\mat#1{\begin{bmatrix}#1\end{bmatrix}}\mat{1&5\\3&-2}X=\mat{4&1&0\\0&2&1}$ for $X$.
-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 ====
+Write $A=\mat{1&5\\3&-2}$. Then $\det(A)=1(-2)-5(3)=-2-15=-17$ which isn't zero, so $A$ is invertible. And $A^{-1}=\frac1{-17}\mat{-2&-5\\-3&1}=\frac1{17}\mat{2&5\\3&-1}$.
-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 =====
+Hence the solution is $X=A^{-1}\mat{4&1&0\\0&2&1}=\frac1{17}\mat{2&5\\3&-1}\mat{4&1&0\\0&2&1}=\frac1{17}\mat{8&12&5\\12&1&-1}$.
-==== Question ====
-Which $2\times 2$ matrices are invertible? For the invertible matrices, can we find their inverse?
+====== The transpose of a matrix ======
-==== Lemma ====
+We defined this in tutorial sheet 4:
-If $A=\mat{a&b\\c&d}$ and $J=\mat{d&-b\\-c&a}$, then we have
+{{page>transpose}}
-\[ AJ=\delta I_2=JA\]
+==== Exercise: simple properties of the transpose ====
-where $\delta=ad-bc$.
+Prove that for any matrix $A$:
-=== Proof ===
+  * $(A^T)^T=A$; 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:
-This is a calculation (done in the lectures; you should also check it yourself). ■
+==== Lemma: transposes and row-column multiplication ====
-==== Definition: the determinant of a $2\times 2$ matrix ====
+If $a$ is a $1\times m$ row vector and $b$ is an $m\times 1$ column vector, then
+\[ ab=b^Ta^T.\]
-{{page>determinant of a 2x2 matrix}}
+==== Observation: the transpose swaps rows with columns ====
-==== Theorem: the determinant determines the invertibility (and inverse) of a $2\times 2$ matrix ====
+Formally, for any matrix $A$ and any $i,j$, we have
+\begin{align*}\def\col#1{\text{col}_{#1}}\def\row#1{\text{row}_{#1}}
+\row i(A^T)&=\col i(A)^T\\\col j(A^T)&=\row j(A)^T
+.\end{align*}
-Let $A=\mat{a&b\\c&d}$ be a $2\times 2$ matrix.
+==== Theorem: the transpose reverses the order of matrix multiplication ====
-  - $A$ is invertible if and only if $\det(A)\ne0$.
+If $A$ and $B$ are matrices and the [[matrix product]] $AB$ is defined, then $B^TA^T$ is also defined. Moreover, in this case we have
-  - If $A$ is invertible, then $A^{-1}=\frac{1}{\det(A)}\mat{d&-b\\-c&a}$.
+\[ (AB)^T=B^TA^T.\]
 === 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.
+If $AB$ is defined, then $A$ is $n\times m$ and $B$ is $m\times k$ for some $n,m,k$, so $B^T$ is $k\times m$ and $A^T$ is $m\times n$, so $B^TA^T$ is defined. Moreover, in this case $B^TA^T$ is an $k\times n$ matrix, and $AB$ is an $n\times k$ matrix, so $(AB)^T$ is a $k\times n$ matrix. Hence $B^TA^T$ has the same size as $(AB)^T$. To show that they are equal, we calculate, using the fact that the transpose swaps rows with columns:
+\begin{align*}
+\text{the }(i,j)\text{ entry of }(AB)^T&= \text{the }(j,i)\text{ entry of }AB
+\\&= \row j(A)\cdot\col i(B)
+\\&=\col i(B)^T\cdot \row j(A)^T \quad\text{by the previous Lemma}
+\\&=\row i(B^T)\cdot \col j(A^T) \quad\text{by the Observation}
+\\&=\text{the }(i,j)\text{ entry of }B^TA^T
+\end{align*}
+Hence $(AB)^T=B^TA^T$. ■
+====== Determinants of $n\times n$ matrices ======
+Given any $n\times n$ matrix $A$, it is possible to define a number $\det(A)$  (as a formula using the entries of $A$) so that
+\[ A\text{ is invertible} \iff \det(A)\ne0.\]
+  - If $A$ is a $1\times 1$ matrix, say $A=[a]$, then we just define $\det[a]=a$.
+  - If $A$ is a $2\times 2$ matrix, say $A=\def\mat#1{\begin{bmatrix}#1\end{bmatrix}}\mat{a&b\\c&d}$, then we've seen that $\det(A)=ad-bc$.
+  - If $A$ is a $3\times 3$ matrix, say $A=\mat{a&b&c\\d&e&f\\g&h&i}$, then it turns out that $\det(A)=aei-afh+bfg-bdi+cdh-ceg$.
+  - If $A$ is a $4\times 4$ matrix, then the formula for $\det(A)$ is more complicated still, with $24$ terms.
+  - If $A$ is a $5\times 5$ matrix, then the formula for $\det(A)$ has $120$ terms.
+Trying to memorise a formula in every case (or even in the $3\times 3$ case!) isn't convenient unless we understand it somehow. We will approach this is several steps.
+==== Step 1: minors ====
+=== Definition ===
+{{page>minor}}
+=== Examples ===
+  * If $A=\mat{3&5\\-4&7}$, then $M_{11}=\det[7]=7$, $M_{12}=\det[-4]=-4$, $M_{21}=5$, and $M_{22}=3$.
+  * If $A=\mat{1&2&3\\7&8&9\\11&12&13}$, then $M_{23}=\det\mat{1&2\\11&12}=1\cdot 12-2\cdot 11=-10$ and $M_{32}=\det\mat{1&3\\7&9}=-12$.
+==== Step 2: cofactors ====
+===Definition===
+{{page>cofactor}}
+Note that $(-1)^{i+j}$ is $+1$ or $-1$, and can looked up in the matrix of signs:
+$\mat{+&-&+&-&\dots\\-&+&-&+&\dots\\+&-&+&-&\dots\\\vdots&\vdots&\vdots&\vdots&\ddots}$.
+This matrix starts with a $+$ in the $(1,1)$ entry (corresponding to $(-1)^{1+1}=(-1)^2=+1$) and the signs then alternate.
+===Examples===
+  * If $A=\mat{3&5\\-4&7}$, then $C_{11}=+M_{11}=\det[7]=7$, $C_{12}=-M_{12}=-\det[-4]=4$, $C_{21}=-5$, and $C_{22}=3$.
+  * If $A=\mat{1&2&3\\7&8&9\\11&12&13}$, then $C_{23}=-M_{23}=-(-10)=10$ and $C_{33}=+M_{33}=\det\mat{1&2\\7&8}=-6$.
+==== Step 3: the determinant of a $3\times 3$ matrix using Laplace expansion along the first row ====
+===Definition===
-Now assume that $A\ne0_{2\times 2}$ and let $J=\mat{d&-b\\-c&a}$.
+{{page>determinant of a 3x3 matrix}}
-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 \[\mat{2&3\\-4&5}^{-1}=\frac1{22}\mat{5&-3\\4&2} = \mat{\frac 5{22}&-\frac3{22}\\\frac2{11}&\frac1{11}}.\]