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32 lines
1.0 KiB
Markdown
32 lines
1.0 KiB
Markdown
#math/linear_algebra
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## theorem on product of determinants
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$det(AB)=(detA)(detB)$
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**WARNING** $det(A+B)\neq detA-detB$
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## theorem for determinant of invertible matrix
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$det(A^{-1})=\frac{1}{detA}$
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## characteristic equation
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$\operatorname{det}(A-\lambda I)=0$
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# ways of computing
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![[Pasted image 20220526165055.png]]
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Definition (Determinant of $A$ )
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![[Pasted image 20220526165329.png]]
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![[Pasted image 20220526165339.png]]
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For a $1 \times 1$ matrix $A=[a]$, the determinant of $A$, $\operatorname{denoted}$ by $\operatorname{det} A$, is defined to be
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$$
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\operatorname{det} A=a
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$$
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For $n \geq 2$, the determinant of an $n \times n$ matrix $A=\left[a_{i j}\right]$ is the sum
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$$
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\begin{aligned}
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\operatorname{det} A &=a_{11} \operatorname{det} A_{11}-a_{12} \operatorname{det} A_{12}+\ldots+(-1)^{n+1} a_{1 n} \operatorname{det} A_{1 n} \\
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&=\sum_{i=1}^{n}(-1)^{i+1} a_{1 i} \operatorname{det} A_{1 i}
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\end{aligned}
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$$
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Sometimes we use absolute value brackets to denote the determinant, i.e. we sometimes write $|A|$ to $\operatorname{denote} \operatorname{det} A$. |