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--- lecture_13 [2016/03/07 10:51] – 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: @@
+==== Example ====
+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 $X$.
+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 $A^{-1}=\frac1{-17}\mat{-2&-5\\-3&1}=\frac1{17}\mat{2&5\\3&-1}$.
+Hence the solution is $X=A^{-1}\mat{4&1&0\\0&2&1}=\frac1{17}\mat{2&5\\3&-1}\mat{4&1&0\\0&2&1}=\frac1{17}\mat{8&12&5\\12&1&-1}$.
 ====== The transpose of a matrix ======
@@ Line 48: / Line 57: @@
 \[ A\text{ is invertible} \iff \det(A)\ne0.\]
-  - If $A$ is a $1\times 1$ matrix, then $A=[a]$ for some number $a$, and we just define $\det[a]=a$.
+  - 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, then $A=\def\mat#1{\begin{bmatrix}#1\end{bmatrix}}\mat{a&b\\c&d}$, and we've seen that $\det\mat{a&b\\c&d}=ad-bc$.
+  - 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, then $A=\mat{a&b&c\\d&e&f\\g&h&i}$. It turns out that $\det(A)=aei-afh+bfg-bdi+cdh-ceg$.
+  - 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.
@@ Line 87: / Line 96: @@
-=== Example ===
-\begin{align*}\det\mat{1&2&3\\7&8&9\\11&12&13} &= 1\cdot C_{11} + 2 C_{12} + 3 C_{13}\\
-&= 1 \cdot (+M_{11}) + 2 \cdot (-M_{12}) + 3 \cdot(+M_{13})\\
-&= M_{11}-2M_{12}+3M_{13}\\
-&= \det\mat{8&9\\12&13} -2\det\mat{7&9\\11&13} + 3\det\mat{7&8\\11&12}\\
-&= (8\cdot 13-9\cdot 12) -2(7\cdot 13-9\cdot 11)+3(7\cdot 12-8\cdot 11)\\
-&=-4 -2(-8)+3(-4)\\
-&=-4+16-12\\
-&=0.\end{align*}
-From this, we can conclude that $\mat{1&2&3\\7&8&9\\11&12&13}$ is //not// invertible.
-=== Notation ===
-To save having to write $\det$ all the time, we sometimes write the entries of a matrix inside vertical bars $|\ |$ to mean the determinant of that matrix. Using this notation (and doing a few steps in our heads), we can rewrite the previous example as:
-\begin{align*}\def\vm#1{\begin{vmatrix}#1\end{vmatrix}}\vm{1&2&3\\7&8&9\\11&12&13} &= 1\vm{8&9\\12&13} -2\vm{7&9\\11&13} + 3\vm{7&8\\11&12}\\
-&=-4 -2(-8)+3(-4)\\
-&=0.\end{align*}
-==== Step 4: the determinant of an $n\times n$ matrix ====
-===Definition===
-{{page>determinant of an nxn matrix}}