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Notation

To save having to write $\det$ all the time, we sometimes write the entries of a matrix inside vertical bars $|\ |$ to mean the determinant of that matrix. Using this notation (and doing a few steps in our heads), we can rewrite the previous example as:

\begin{align*}\def\vm#1{\begin{vmatrix}#1\end{vmatrix}}\vm{1&2&3\\7&8&9\\11&12&13} &= 1\vm{8&9\\12&13} -2\vm{7&9\\11&13} + 3\vm{7&8\\11&12}\\ &=-4 -2(-8)+3(-4)\\ &=0.\end{align*}

Step 4: the determinant of an $n\times n$ matrix

Definition

Example

\begin{align*} \def\vm#1{\begin{vmatrix}#1\end{vmatrix}} \vm{\color{red}1&\color{red}0&\color{red}2&\color{red}3\\0&2&1&-1\\2&0&0&1\\3&0&4&2} &= \color{red}1\vm{\color{blue}2&\color{blue}1&\color{blue}-1\\0&0&1\\0&4&2}-\color{red}0\vm{0&1&-1\\2&0&1\\3&4&2}+\color{red}2\vm{\color{orange}0&\color{orange}2&\color{orange}{-1}\\2&0&1\\3&0&2}-\color{red}3\vm{\color{purple}0&\color{purple}2&\color{purple}1\\2&0&0\\3&0&4}\\ &= 1\left(\color{blue}2\vm{0&1\\4&2}-\color{blue}1\vm{0&1\\0&2}\color{blue}{-1}\vm{0&0\\0&4}\right)-0+2\left(\color{orange}0-\color{orange}{2}\vm{2&1\\3&2}\color{orange}{-1}\vm{2&0\\3&0}\right)-3\left(\color{purple}0-\color{purple}2\vm{2&0\\3&4}+\color{purple}1\vm{2&0\\3&0}\right)\\ &=1(2(-4)-0-0)+2(-2(1)-0)-3(-2(8)+0)\\ &=-8-4+48\\ &=36. \end{align*}

Theorem: Laplace expansion along any row or column gives the determinant

1. For any fixed $i$: $\det(A)=a_{i1}C_{i1}+a_{i2}C_{i2}+\dots+a_{in}C_{in}$ (Laplace expansion along row $i$)
2. For any fixed $j$: $\det(A)=a_{1j}C_{1j}+a_{2j}C_{2j}+\dots+a_{nj}C_{nj}$ (Laplace expansion along column $j$)

Example

We can make life easier by choosing expansion rows or columns with lots of zeros, if possible. Let's redo the previous example with this in mind:

\begin{align*} \def\vm#1{\begin{vmatrix}#1\end{vmatrix}} \vm{1&\color{red}0&2&3\\0&\color{red}2&1&-1\\2&\color{red}0&0&1\\3&\color{red}0&4&2} &= -\color{red}0+\color{red}2\vm{1&2&3\\\color{purple}2&\color{purple}0&\color{purple}1\\3&4&2}-\color{red}0+\color{red}0\\ &=2\left(-\color{purple}2\vm{2&3\\4&2}+\color{purple}0-\color{purple}1\vm{1&2\\3&4}\right)\\ &=2(-2(-8)-(-2))\\ &=36. \end{align*}

Definition: upper triangular matrices

An $n\times n$ matrix $A$ is upper triangular if all the entries below the main diagonal are zero.

Definition: diagonal matrices

An $n\times n$ matrix $A$ is diagonal if the only non-zero entries are on its main diagonal.

Corollary: the determinant of upper triangular matrices and diagonal matrices

1. The determinant of an upper triangular $n\times n$ matrix is the product of its diagonal entries: $\det(A)=a_{11}a_{22}\dots a_{nn}$.
2. The determinant of an $n\times n$ diagonal matrix is the product of its diagonal entries: $\det(A)=a_{11}a_{22}\dots a_{nn}$.

Proof

1. This is true for $n=1$, trivially. For $n>1$, assume inductively that it is true for $(n-1)\times (n-1)$ matrices and use the Laplace expansion of an upper triangular $n\times n$ matrix $A$ along the first column of $A$ to see that $\det(A)=a_{11}C_{11}+0+\dots+0=a_{11}C_{11}$. Now $C_{11}$ is the determinant of the $(n-1)\times (n-1)$ matrix formed by removing the first row and and column of $A$, and this matrix is upper triangular with diagonal entries $a_{22},a_{33},\dots,a_{nn}$. By our inductive assumption, we have $C_{11}=a_{22}a_{33}\dots a_{nn}$. So $\det(A)=a_{11}C_{11}=a_{11}a_{22}a_{33}\dots a_{nn}$ as desired.
2. Any diagonal matrix is upper triangular, so this is a special case of statement 1. ■

Examples

1. For any $n$, we have $\det(I_n)=1\cdot 1\cdots 1 = 1$.
2. For any $n$, we have $\det(5I_n)=5^n$.
3. $\def\vm#1{\begin{vmatrix}#1\end{vmatrix}}\vm{1&9&43&23434&4&132\\0&3&43&2&-1423&-12\\0&0&7&19&23&132\\0&0&0&2&0&0\\0&0&0&0&-1&-903\\0&0&0&0&0&6}=1\cdot3\cdot7\cdot2\cdot(-1)\cdot6 = 252$.

Theorem: important properties of the determinant

Let $A$ be an $n\times n$ matrix.

1. $A$ is invertible if and only if $\det(A)\ne0$.
2. $\det(A^T)=\det(A)$
3. If $B$ is another $n\times n$ matrix, then $\det(AB)=\det(A)\det(B)$

Theorem: row/column operations and determinants

Let $A$ be an $n\times n$ matrix, let $c$ be a scalar and let $i\ne j$.

$A_{Ri\to x}$ means $A$ but with row $i$ replaced by $x$.

1. If $i\ne j$, then $\det(A_{Ri\leftrightarrow Rj})=-\det(A)$ (swapping two rows changes the sign of det).
2. $\det(A_{Ri\to c Ri}) = c\det(A)$ (scaling one row scales $\det(A)$ in the same way)
3. $\det(A_{Ri\to Ri + c Rj}) = \det(A)$ (adding a multiple of one row to another row doesn't change $\det(A)$)

• Also, these properties all hold if you change “row” into “column” throughout.

Corollary

If an $n\times n$ matrix $A$ has two equal rows (or columns), then $\det(A)=0$, and $A$ is not invertible.

Proof

If $A$ has two equal rows, row $i$ and row $j$, then $A=A_{Ri\leftrightarrow Rj}$ So $\det(A)=\det(A_{Ri\leftrightarrow Rj}) = -\det(A)$, so $2\det(A)=0$, so $\det(A)=0$.

If $A$ has two equal columns, then $A^T$ has two equal rows, so $\det(A)=\det(A^T)=0$.

In either case, $\det(A)=0$. So $A$ is not invertible.■

Examples

• Swapping two rows changes the sign, so $\def\vm#1{\begin{vmatrix}#1\end{vmatrix}}\vm{0&0&2\\0&3&0\\4&0&0} = -\vm{4&0&0\\0&3&0\\0&0&2}=-4\cdot 3\cdot 2 = -24$.
• Multiplying a row or a column by a constant multiplies the determinant by that constant, so \begin{align*}\vm{ 2&4&6&10\\5&0&0&-10\\9&0&81&99\\1&2&3&4} &= 2\vm{ 1&2&3&5\\5&0&0&-10\\9&0&81&99\\1&2&3&4} \\&= 2\cdot 5\vm{ 1&2&3&5\\1&0&0&-2\\9&0&81&99\\1&2&3&4}\\&= 2\cdot 5\cdot 9 \vm{ 1&2&3&5\\1&0&0&-2\\1&0&9&11\\1&2&3&4}\\&=2\cdot 5\cdot 9\cdot 2 \vm{ 1&1&3&5\\1&0&0&-2\\1&0&9&11\\1&1&3&4}\\&=2\cdot 5\cdot 9\cdot 2\cdot 3 \vm{ 1&1&1&5\\1&0&0&-2\\1&0&3&11\\1&1&1&4}.\end{align*}
• $\det(A_{R1\to R1-R4})=\det(A)$, so \begin{align*}\vm{ 1&1&1&5\\1&0&0&-2\\1&0&3&11\\1&1&1&4} &=\vm{ 0&0&0&1\\1&0&0&-2\\1&0&3&11\\1&1&1&4}=-1\vm{1&0&0\\1&0&3\\1&1&1}+0\\&=-\vm{0&3\\1&1} = -(-3)=3.\end{align*}
• Hence \begin{align*}\vm{ 2&4&6&10\\5&0&0&-10\\9&0&81&99\\1&2&3&4} &= 2\cdot 5\cdot 9\cdot 2\cdot 3 \vm{ 1&1&1&5\\1&0&0&-2\\1&0&3&11\\1&1&1&4} \\&= 2\cdot 5\cdot 9\cdot 2\cdot 3 \cdot 3 = 1620.\end{align*}

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:

1. changing the order of the rows of the matrix multiplies the determinant by $-1$;
2. multiplying one of the rows of the matrix by $c\in \mathbb{R}$ multiplies the determinant by $c$; and
3. 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*}