MATH 323 Lecture 8

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Determinants by Elimination

Each elementary matrix operation corresponds to following rules:

1. Interchanging two rows/columns of matrix changes sign of determinant
2. Multiplying single row/column of matrix by scalar has effect of multiplying determinant by scalar
3. Adding multiple of row/column to another does not change value of determinant

If $U=E_{k}\,\dots \,E_{1}\,A$ is a triangular matrix, then $|U|=|E_{k}|\,\dots \,|E_{1}|=u_{11}\,\dots \,u_{nn}$. For example,

${\begin{vmatrix}2&1&3\\4&2&1\\6&-3&4\end{vmatrix}}={\begin{vmatrix}2&1&3\\0&0&-5\\0&-6&-5\end{vmatrix}}=(-1){\begin{vmatrix}2&1&3\\0&-6&-5\\0&0&-5\end{vmatrix}}=(-1)\,(2)\,(-6)\,(-5)=-60$

Cramer's Rule

Let $A=\left(a_{ij}\right)$ be a $n\times n$ nonsingular matrix.

We define the adjoint matrix of $A$ as follows:

${\begin{aligned}\mathrm {adj} A&={\begin{bmatrix}A_{11}&A_{21}&\dots &A_{n1}\\A_{12}&A_{22}&\dots &A_{n2}\\\vdots &\vdots &\ddots &\vdots \\A_{1n}&A_{2n}&\dots &A_{nn}\end{bmatrix}}\\&=\left(A^{*}\right)^{T}\end{aligned}}$

Where $A^{*}$ is the matrix created by substituting the cofactor $A_{ij}$ for element $a_{ij}$.

By Lemma 2.2.1, $a_{i1}\,A_{j1}+\dots +a_{in}\,A_{jn}={\begin{cases}\left|A\right|&i=j\\0&i\neq j\end{cases}}$, and

$A\left(\mathrm {adj} A\right)=\left|A\right|\,I={\begin{bmatrix}\left|A\right|&\dots &0\\\vdots &\ddots &\vdots \\0&\dots &\left|A\right|\end{bmatrix}}$

Assuming $A$ is nonsingular, $A^{-1}={\frac {1}{\left|A\right|}}\,\mathrm {adj} A$

Example

${\begin{aligned}A&={\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}}\\A^{*}&={\begin{bmatrix}a_{22}&-a_{21}\\-a_{12}&a_{11}\end{bmatrix}}\\\mathrm {adj} A&=\left(A^{*}\right)^{T}={\begin{bmatrix}a_{22}&-a_{12}\\-a_{21}&a_{11}\end{bmatrix}}\\A^{-1}&={\frac {1}{a_{11}\,a_{22}-a_{12}\,a_{21}}}\,{\begin{bmatrix}a_{22}&-a_{12}\\-a_{21}&a_{11}\end{bmatrix}}\end{aligned}}$

Formal Theorem

Let $A$ be a $n\times n$ nonsingular matrix, and let ${\vec {b}}\in \mathbb {R} ^{n}$. Let $A_{i}$ be the matrix obtained by replacing the $i$^th column of $A$ by ${\vec {b}}$. If ${\hat {x}}$ is the unique solution to $A\,{\vec {x}}={\vec {b}}$, then

${\hat {x}}_{i}={\frac {\left|A_{i}\right|}{\left|A\right|}}$ for $i=1,\ldots ,n$

Example

${\begin{aligned}x_{1}+2x_{2}+x_{3}&=5\\2x_{1}+2x_{2}+x_{3}&=6\\x_{1}+2x_{2}+3x_{3}=9\end{aligned}}$

${\begin{aligned}\left|A\right|&={\begin{vmatrix}1&2&1\\2&2&1\\1&2&3\end{vmatrix}}=-4\\\left|A_{1}\right|&={\begin{vmatrix}5&2&1\\6&2&1\\9&2&3\end{vmatrix}}=-4\\\left|A_{2}\right|&={\begin{vmatrix}1&5&1\\2&6&1\\1&9&3\end{vmatrix}}=-4\\\left|A_{3}\right|&={\begin{vmatrix}1&2&5\\2&2&6\\1&2&9\end{vmatrix}}=-8\end{aligned}}$

${\vec {x}}=\left\langle {\frac {-4}{-4}},{\frac {-4}{-4}},{\frac {-8}{-4}}\right\rangle =\left\langle 1,1,2\right\rangle$