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Table of Contents
Corollary
If $\def\row{\text{row}}\row_j(A)=c\cdot \row_i(A)$ for some $i\ne j$ and some $c\in \mathbb{R}$, then $\det(A)=0$.
Proof
Let $B$ have the same rows as $A$, except with $\row_i(A)$ in both row $i$ and row $j$. Observe that
- row $j$ of $B$ is $c$ times row $j$ of $A$, and all the other rows are equal; and
- $A$ has two equal rows.
Hence using property 3 in the theorem and the previous corollary, we have $\det(B)=c\cdot \det(A)=c\cdot 0=0.$ ■
Corollary
If $E$ has the same rows as $A$ except in row $j$, and $\row_j(E)=\row_j(A)+c\cdot \row_i(A)$ for some $i\ne j$ and some $c\in \mathbb{R}$, then $\det(E)=\det(A)$.
Proof
Let $B$ have the same rows as $A$, except with $c\cdot \row_i(A)$ in row $j$. Observe that
- $\det(B)=0$, by the previous corollary; and
- except in row $j$, the rows of $E$, $A$ and $B$ are all equal, and $\row_j(E)=\row_j(A)+\row_j(B)$.
Hence by property 4 of the theorem, we have $\det(E)=\det(A)+\det(B)=\det(A)+0=\det(A)$. ■
We have now seen the effect of each of the three types of ERO on the determinant of a matrix:
- changing the order of the rows of the matrix multiplies the determinant by $-1$;
- multiplying one of the rows of the matrix by $c\in \mathbb{R}$ multiplies the determinant by $c$; and
- replacing row $j$ by “row $j$ ${}+{}$ $c\times {}$ (row $i$)”, where $c$ is a non-zero real number and $i\ne j$ does not change the determinant.
Moreover, since $\det(A)=\det(A^T)$, this all applies equally to columns instead of rows.
We can use EROs to put a matrix into upper triangular form, and then finding the determinant is easy: just multiply the diagonal entries together. We just have to keep track of how the determinant is changed by the EROs of types 1 and 2.
Example: using EROs to find the determinant
\begin{align*}\def\vm#1{\begin{vmatrix}#1\end{vmatrix}} \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} \\&=\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&2&2\\0&0&0&-3} \\&=-12(1)(1)(2)(-3)=72. \end{align*}
Finding the inverse of an invertible $n\times n$ matrix
Definition: the adjoint of a square matrix
Let $A$ be an $n\times n$ matrix. Recall that $C_{ij}$ is the $(i,j)$ cofactor of $A$. The matrix of cofactors of $A$ is the $n\times n$ matrix $C$ whose $(i,j)$ entry is $C_{ij}$.
The adjoint of $A$ is the $n\times n$ matrix $J=C^T$, the transpose of the matrix of cofactors.
Example: $n=2$
If $A=\def\mat#1{\begin{bmatrix}#1\end{bmatrix}}\def\vm#1{\begin{vmatrix}#1\end{vmatrix}}\mat{a&b\\c&d}$, then $C=\mat{d&-c\\-b&a}$, so the adjoint of $A$ is $J=C^T=\mat{d&-b\\-c&a}$.
Recall that $AJ=(\det A)I_2=JA$; we calculated this earlier when we looked at the inverse of a $2\times 2$ matrix. Hence if $\det A\ne0$, then $A^{-1}=\frac1{\det A}J$.
Example: $n=3$
If $A=\mat{3&1&0\\-2&-4&3\\5&4&-2}$, then the matrix of signs is $\mat{+&-&+\\-&+&-\\+&-&+}$, so \[ C=\mat{ \vm{-4&3\\4&-2}&-\vm{-2&3\\5&-2}&\vm{-2&-4\\5&4}\\ -\vm{1&0\\4&-2}&\vm{3&0\\5&-2}&-\vm{3&1\\5&4}\\ \vm{1&0\\-4&3}&-\vm{3&0\\-2&3}&\vm{3&1\\-2&-4}} = \mat{-4&11&12\\2&-6&-7\\3&-9&-10}\] so the adjoint of $A$ is \[ J=C^T=\mat{-4&2&3\\11&-6&-9\\12&-7&-10}.\]
Observe that $AJ=\mat{3&1&0\\-2&-4&3\\5&4&-2}\mat{-4&2&3\\11&-6&-9\\12&-7&-10}=\mat{-1&0&0\\0&-1&0\\0&0&-1}=-1\cdot I_3$, and $JA=\mat{-4&2&3\\11&-6&-9\\12&-7&-10}\mat{3&1&0\\-2&-4&3\\5&4&-2}=\mat{-1&0&0\\0&-1&0\\0&0&-1}=-1\cdot I_3$; and $\det(A)=-1$.
This is an illustration of the following theorem, whose proof is omitted:
Theorem: key property of the adjoint of a square matrix
If $A$ is any $n\times n$ matrix and $J$ is its adjoint, then $AJ=(\det A)I_n=JA$.
Corollary: a formula for the inverse of a square matrix
If $A$ is any $n\times n$ matrix with $\det(A)\ne 0$, then $A$ is invertible and \[A^{-1}=\frac1{\det A}J\] where $J$ is the adjoint of $A$.
Proof
Divide the equation $AJ=(\det A)I_n=JA$ by $\det A$. ■
Example
If again we take $A=\mat{3&1&0\\-2&-4&3\\5&4&-2}$, then $J=\mat{-4&2&3\\11&-6&-9\\12&-7&-10}$ and $\det(A)=-1$, so $A$ is invertible and $A^{-1}=\frac1{-1}J=-J=\mat{4&-2&-3\\-11&6&9\\-12&7&10}$.