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--- lecture_13 [2016/03/08 10:57] – 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 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.