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lecture_14_slides

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lecture_14_slides [2017/03/08 17:16] – [Theorem: important properties of the determinant] rupertlecture_14_slides [2017/03/09 10:45] (current) – [Corollary: the determinant of an upper triangular matrix] rupert
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 If $A$ is an upper triangular matrix or a diagonal matrix, then the determinant of $A$ is the product of its diagonal entries: \[\det(A)=a_{11}a_{22}\dots a_{nn}.\] If $A$ is an upper triangular matrix or a diagonal matrix, then the determinant of $A$ is the product of its diagonal entries: \[\det(A)=a_{11}a_{22}\dots a_{nn}.\]
  
-  * Every upper triangular matrix is diagonal! +  * Every diagonal matrix is upper triangular!
   * So it suffices just to prove it for upper triangular matrices.   * So it suffices just to prove it for upper triangular matrices.
  
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   - If $B$ is another $n\times n$ matrix, then $\det(AB)=\det(A)\det(B)$.   - If $B$ is another $n\times n$ matrix, then $\det(AB)=\det(A)\det(B)$.
  
-==== Corollary on invertibility ====+=== Corollary on invertibility ===
  
   - $A^T$ is invertible if and only if $A$ is invertible   - $A^T$ is invertible if and only if $A$ is invertible
   - $AB$ is invertible if and only if **both** $A$ and $B$ are invertible   - $AB$ is invertible if and only if **both** $A$ and $B$ are invertible
 +
  
   * Warning: it's not true that $A+B$ is invertible if and only if $A$ and $B$ are invertible!   * Warning: it's not true that $A+B$ is invertible if and only if $A$ and $B$ are invertible!
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 \begin{align*}\vm{1&3&1&3\\\color{red}4&\color{red}8&\color{red}0&\color{red}{12}\\0&1&3&6\\2&2&1&6}&= \color{red}{4}\vm{1&3&1&\color{blue}3\\1&2&0&\color{blue}3\\0&1&3&\color{blue}6\\2&2&1&\color{blue}6}=4\cdot \color{blue}3\vm{\color{green}1&3&1&1\\\color{red}1&2&0&1\\\color{red}0&1&3&2\\\color{red}2&2&1&2} \begin{align*}\vm{1&3&1&3\\\color{red}4&\color{red}8&\color{red}0&\color{red}{12}\\0&1&3&6\\2&2&1&6}&= \color{red}{4}\vm{1&3&1&\color{blue}3\\1&2&0&\color{blue}3\\0&1&3&\color{blue}6\\2&2&1&\color{blue}6}=4\cdot \color{blue}3\vm{\color{green}1&3&1&1\\\color{red}1&2&0&1\\\color{red}0&1&3&2\\\color{red}2&2&1&2}
-=12\vm{1&3&1&1\\\color{blue}0&\color{blue}{-1}&\color{blue}{-1}&\color{blue}{0}\\\color{blue}0&\color{blue}1&\color{blue}3&\color{blue}2\\0&-4&-1&-0} +\\&=12\vm{1&3&1&1\\\color{blue}0&\color{blue}{-1}&\color{blue}{-1}&\color{blue}{0}\\\color{blue}0&\color{blue}1&\color{blue}3&\color{blue}2\\0&-4&-1&-0} 
-\\&=\color{blue}{-}12\vm{1&3&1&1\\0&\color{green}1&3&2\\0&\color{red}{-1}&{-1}&{0}\\0&\color{red}{-4}&-1&0} +=\color{blue}{-}12\vm{1&3&1&1\\0&\color{green}1&3&2\\0&\color{red}{-1}&{-1}&{0}\\0&\color{red}{-4}&-1&0} 
-=-12\vm{1&3&1&1\\0&1&3&2\\0&0&\color{green}2&2\\0&0&\color{red}{11}&8}+\\&=-12\vm{1&3&1&1\\0&1&3&2\\0&0&\color{green}2&2\\0&0&\color{red}{11}&8}
 =-12\vm{1&3&1&1\\0&1&3&2\\0&0&2&2\\0&0&0&-3} =-12\vm{1&3&1&1\\0&1&3&2\\0&0&2&2\\0&0&0&-3}
 \\&=-12(1)(1)(2)(-3)=72. \\&=-12(1)(1)(2)(-3)=72.
 \end{align*} \end{align*}
 +
  
  
  
lecture_14_slides.1488993394.txt.gz · Last modified: by rupert
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