## SectionA.6Determinant

Note: 1 lecture

For square matrices we define a useful quantity called the determinant. We define the determinant of a $1 \times 1$ matrix as the value of its only entry

\begin{equation*} \det \left( \begin{bmatrix} a \end{bmatrix} \right) \overset{\text{def}}{=} a . \end{equation*}

For a $2 \times 2$ matrix we define

\begin{equation*} \det \left( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \right) \overset{\text{def}}{=} ad-bc . \end{equation*}

Before defining the determinant for larger matrices, we note the meaning of the determinant. An $n \times n$ matrix gives a mapping of the $n$-dimensional euclidean space ${\mathbb{R}}^n$ to itself. In particular, a $2 \times 2$ matrix $A$ is a mapping of the plane to itself. The determinant of $A$ is the factor by which the area of objects changes. If we take the unit square (square of side 1) in the plane, then $A$ takes the square to a parallelogram of area $\lvert\det(A)\rvert\text{.}$ The sign of $\det(A)$ denotes a change of orientation (negative if the axes get flipped). For example, let

\begin{equation*} A = \begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix} . \end{equation*}

Then $\det(A) = 1+1 = 2\text{.}$ Let us see where $A$ sends the unit square with vertices $(0,0)\text{,}$ $(1,0)\text{,}$ $(0,1)\text{,}$ and $(1,1)\text{.}$ The point $(0,0)$ gets sent to $(0,0)\text{.}$

\begin{equation*} \begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \end{bmatrix} , \qquad \begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix} , \qquad \begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 2 \\ 0 \end{bmatrix} . \end{equation*}

The image of the square is another square with vertices $(0,0)\text{,}$ $(1,-1)\text{,}$ $(1,1)\text{,}$ and $(2,0)\text{.}$ The image square has a side of length $\sqrt{2}$ and is therefore of area 2. See Figure A.7. Figure A.7. Image of the unit quare via the mapping $A\text{.}$

In general the image of a square is going to be a parallelogram. In high school geometry, you may have seen a formula for computing the area of a parallelogram with vertices $(0,0)\text{,}$ $(a,c)\text{,}$ $(b,d)$ and $(a+b,c+d)\text{.}$ The area is

\begin{equation*} \left\lvert \, \det \left( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \right) \, \right\lvert = \lvert a d - b c \rvert . \end{equation*}

The vertical lines above mean absolute value. The matrix $\left[ \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \right]$ carries the unit square to the given parallelogram.

There are a number of ways to define the determinant for an $n \times n$ matrix. Let us use the so-called cofactor expansion. We define $A_{ij}$ as the matrix $A$ with the $i^{\text{th}}$ row and the $j^{\text{th}}$ column deleted. For example, if

\begin{equation*} \text{If} \qquad A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} , \qquad \text{then} \qquad A_{12} = \begin{bmatrix} 4 & 6 \\ 7 & 9 \end{bmatrix} \qquad \text{and} \qquad A_{23} = \begin{bmatrix} 1 & 2 \\ 7 & 8 \end{bmatrix} . \end{equation*}

We now define the determinant recursively

\begin{equation*} \det (A) \overset{\text{def}}{=} \sum_{j=1}^n {(-1)}^{1+j} a_{1j} \det (A_{1j}) , \end{equation*}

or in other words

\begin{equation*} \det (A) = a_{11} \det (A_{11}) - a_{12} \det (A_{12}) + a_{13} \det (A_{13}) - \cdots \begin{cases} + a_{1n} \det (A_{1n}) & \text{if } n \text{ is odd,} \\ - a_{1n} \det (A_{1n}) & \text{if } n \text{ even.} \end{cases} \end{equation*}

For a $3 \times 3$ matrix, we get $\det (A) = a_{11} \det (A_{11}) - a_{12} \det (A_{12}) + a_{13} \det (A_{13})\text{.}$ For example,

\begin{equation*} \begin{split} \det \left( \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \right) & = 1 \cdot \det \left( \begin{bmatrix} 5 & 6 \\ 8 & 9 \end{bmatrix} \right) - 2 \cdot \det \left( \begin{bmatrix} 4 & 6 \\ 7 & 9 \end{bmatrix} \right) + 3 \cdot \det \left( \begin{bmatrix} 4 & 5 \\ 7 & 8 \end{bmatrix} \right) \\ & = 1 (5 \cdot 9 - 6 \cdot 8) - 2 (4 \cdot 9 - 6 \cdot 7) + 3 (4 \cdot 8 - 5 \cdot 7) = 0 . \end{split} \end{equation*}

It turns out that we did not have to necessarily use the first row. That is for any $i\text{,}$

\begin{equation*} \det (A) = \sum_{j=1}^n {(-1)}^{i+j} a_{ij} \det (A_{ij}) . \end{equation*}

It is sometimes useful to use a row other than the first. In the following example it is more convenient to expand along the second row. Notice that for the second row we are starting with a negative sign.

\begin{equation*} \begin{split} \det \left( \begin{bmatrix} 1 & 2 & 3 \\ 0 & 5 & 0 \\ 7 & 8 & 9 \end{bmatrix} \right) & = - 0 \cdot \det \left( \begin{bmatrix} 2 & 3 \\ 8 & 9 \end{bmatrix} \right) + 5 \cdot \det \left( \begin{bmatrix} 1 & 3 \\ 7 & 9 \end{bmatrix} \right) - 0 \cdot \det \left( \begin{bmatrix} 1 & 2 \\ 7 & 8 \end{bmatrix} \right) \\ & = 0 + 5 (1 \cdot 9 - 3 \cdot 7) + 0 = -60 . \end{split} \end{equation*}

Let us check if it is really the same as expanding along the first row,

\begin{equation*} \begin{split} \det \left( \begin{bmatrix} 1 & 2 & 3 \\ 0 & 5 & 0 \\ 7 & 8 & 9 \end{bmatrix} \right) & = 1 \cdot \det \left( \begin{bmatrix} 5 & 0 \\ 8 & 9 \end{bmatrix} \right) - 2 \cdot \det \left( \begin{bmatrix} 0 & 0 \\ 7 & 9 \end{bmatrix} \right) + 3 \cdot \det \left( \begin{bmatrix} 0 & 5 \\ 7 & 8 \end{bmatrix} \right) \\ & = 1 (5 \cdot 9 - 0 \cdot 8) - 2 (0 \cdot 9 - 0 \cdot 7) + 3 (0 \cdot 8 - 5 \cdot 7) = -60 . \end{split} \end{equation*}

In computing the determinant, we alternately add and subtract the determinants of the submatrices $A_{ij}$ multiplied by $a_{ij}$ for a fixed $i$ and all $j\text{.}$ The numbers ${(-1)}^{i+j}\det(A_{ij})$ are called cofactors of the matrix. And that is why this method of computing the determinant is called the cofactor expansion.

Similarly we do not need to expand along a row, we can expand along a column. For any $j$

\begin{equation*} \det (A) = \sum_{i=1}^n {(-1)}^{i+j} a_{ij} \det (A_{ij}) . \end{equation*}

A related fact is that

\begin{equation*} \det (A) = \det (A^T) . \end{equation*}

A matrix is upper triangular if all elements below the main diagonal are 0. For example,

\begin{equation*} \begin{bmatrix} 1 & 2 & 3 \\ 0 & 5 & 6 \\ 0 & 0 & 9 \end{bmatrix} \end{equation*}

is upper triangular. Similarly a lower triangular matrix is one where everything above the diagonal is zero. For example,

\begin{equation*} \begin{bmatrix} 1 & 0 & 0 \\ 4 & 5 & 0 \\ 7 & 8 & 9 \end{bmatrix} . \end{equation*}

The determinant for triangular matrices is very simple to compute. Consider the lower triangular matrix. If we expand along the first row, we find that the determinant is 1 times the determinant of the lower triangular matrix $\left[ \begin{smallmatrix} 5 & 0 \\ 8 & 9 \end{smallmatrix} \right]\text{.}$ So the deteriminant is just the product of the diagonal entries:

\begin{equation*} \det \left( \begin{bmatrix} 1 & 0 & 0 \\ 4 & 5 & 0 \\ 7 & 8 & 9 \end{bmatrix} \right) = 1 \cdot 5 \cdot 9 = 45 . \end{equation*}

Similarly for upper triangular matrices

\begin{equation*} \det \left( \begin{bmatrix} 1 & 2 & 3 \\ 0 & 5 & 6 \\ 0 & 0 & 9 \end{bmatrix} \right) = 1 \cdot 5 \cdot 9 = 45 . \end{equation*}

In general, if $A$ is triangular, then

\begin{equation*} \det (A) = a_{11} a_{22} \cdots a_{nn} . \end{equation*}

If $A$ is diagonal, then it is also triangular (upper and lower), so same formula applies. For example,

\begin{equation*} \det \left( \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 5 \end{bmatrix} \right) = 2 \cdot 3 \cdot 5 = 30 . \end{equation*}

In particular, the identity matrix $I$ is diagonal, and the diagonal entries are all 1. Thus,

\begin{equation*} \det(I) = 1 . \end{equation*}

The determinant is telling you how geometric objects scale. If $B$ doubles the sizes of geometric objects and $A$ triples them, then $AB$ (which applies $B$ to an object and then it applies $A$) should make size go up by a factor of $6\text{.}$ This is true in general:

This property is one of the most useful, and it is employed often to actually compute determinants. A particularly interesting consequence is to note what it means for existence of inverses. Take $A$ and $B$ to be inverses, that is $AB=I\text{.}$ Then

\begin{equation*} \det(A)\det(B) = \det(AB) = \det(I) = 1 . \end{equation*}

Neither $\det(A)$ nor $\det(B)$ can be zero. This fact is an extremely useful property of the determinant, and one which is used often in this book:

In fact, $\det(A^{-1}) \det(A) = 1$ says that

\begin{equation*} \det(A^{-1}) = \frac{1}{\det(A)}. \end{equation*}

So we know what the determinant of $A^{-1}$ is without computing $A^{-1}\text{.}$

Let us return to the formula for the inverse of a $2 \times 2$ matrix:

\begin{equation*} \begin{bmatrix} a & b \\ c & d \end{bmatrix}^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} . \end{equation*}

Notice the determinant of the matrix $[\begin{smallmatrix}a&b\\c&d\end{smallmatrix}]$ in the denominator of the fraction. The formula only works if the determinant is nonzero, otherwise we are dividing by zero.

A common notation for the determinant is a pair of vertical lines:

\begin{equation*} \begin{vmatrix} a & b \\ c & d \end{vmatrix} = \det \left( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \right) . \end{equation*}

Personally, I find this notation confusing as vertical lines usually mean a positive quantity, while determinants can be negative. Also think about how to write the absolute value of a determinant. This notation is not used in this book.

### SubsectionA.6.1Exercises

###### ExerciseA.6.1.

Compute the determinant of the following matrices:

1. $\displaystyle \begin{bmatrix} 3 \end{bmatrix}$

2. $\displaystyle \begin{bmatrix} 1 & 3 \\ 2 & 1 \end{bmatrix}$

3. $\displaystyle \begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix}$

4. $\displaystyle \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix}$

5. $\displaystyle \begin{bmatrix} 2 & 1 & 0 \\ -2 & 7 & -3 \\ 0 & 2 & 0 \end{bmatrix}$

6. $\displaystyle \begin{bmatrix} 2 & 1 & 3 \\ 8 & 6 & 3 \\ 7 & 9 & 7 \end{bmatrix}$

7. $\displaystyle \begin{bmatrix} 0 & 2 & 5 & 7 \\ 0 & 0 & 2 & -3 \\ 3 & 4 & 5 & 7 \\ 0 & 0 & 2 & 4 \end{bmatrix}$

8. $\displaystyle \begin{bmatrix} 0 & 1 & 2 & 0 \\ 1 & 1 & -1 & 2 \\ 1 & 1 & 2 & 1 \\ 2 & -1 & -2 & 3 \end{bmatrix}$

###### ExerciseA.6.2.

For which $x$ are the following matrices singular (not invertible).

1. $\displaystyle \begin{bmatrix} 2 & 3 \\ 2 & x \end{bmatrix}$

2. $\displaystyle \begin{bmatrix} 2 & x \\ 1 & 2 \end{bmatrix}$

3. $\displaystyle \begin{bmatrix} x & 1 \\ 4 & x \end{bmatrix}$

4. $\displaystyle \begin{bmatrix} x & 0 & 1 \\ 1 & 4 & 2 \\ 1 & 6 & 2 \end{bmatrix}$

###### ExerciseA.6.3.

Compute

\begin{equation*} \det \left( \begin{bmatrix} 2 & 1 & 2 & 3 \\ 0 & 8 & 6 & 5 \\ 0 & 0 & 3 & 9 \\ 0 & 0 & 0 & 1 \end{bmatrix}^{-1} \right) \end{equation*}

without computing the inverse.

###### ExerciseA.6.4.

Suppose

\begin{equation*} L = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 2 & 1 & 0 & 0 \\ 7 & \pi & 1 & 0 \\ 2^8 & 5 & -99 & 1 \end{bmatrix} \qquad \text{and} \qquad U = \begin{bmatrix} 5 & 9 & 1 & -\sin(1) \\ 0 & 1 & 88 & -1 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 1 \end{bmatrix} . \end{equation*}

Let $A = LU\text{.}$ Compute $\det(A)$ in a simple way, without computing what is $A\text{.}$ Hint: First read off $\det(L)$ and $\det(U)\text{.}$

###### ExerciseA.6.5.

Consider the linear mapping from ${\mathbb R}^2$ to ${\mathbb R}^2$ given by the matrix $A = \left[ \begin{smallmatrix} 1 & x \\ 2 & 1 \end{smallmatrix} \right]$ for some number $x\text{.}$ You wish to make $A$ such that it doubles the area of every geometric figure. What are the possibilities for $x$ (there are two answers).

###### ExerciseA.6.6.

Suppose $A$ and $S$ are $n \times n$ matrices, and $S$ is invertible. Suppose that $\det(A) = 3\text{.}$ Compute $\det(S^{-1}AS)$ and $\det(SAS^{-1})\text{.}$ Justify your answer using the theorems in this section.

###### ExerciseA.6.7.

Let $A$ be an $n \times n$ matrix such that $\det(A)=1\text{.}$ Compute $\det(x A)$ given a number $x\text{.}$ Hint: First try computing $\det(xI)\text{,}$ then note that $xA = (xI)A\text{.}$

###### ExerciseA.6.101.

Compute the determinant of the following matrices:

1. $\displaystyle \begin{bmatrix} -2 \end{bmatrix}$

2. $\displaystyle \begin{bmatrix} 2 & -2 \\ 1 & 3 \end{bmatrix}$

3. $\displaystyle \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix}$

4. $\displaystyle \begin{bmatrix} 2 & 9 & -11 \\ 0 & -1 & 5 \\ 0 & 0 & 3 \end{bmatrix}$

5. $\displaystyle \begin{bmatrix} 2 & 1 & 0 \\ -2 & 7 & 3 \\ 1 & 1 & 0 \end{bmatrix}$

6. $\displaystyle \begin{bmatrix} 5 & 1 & 3 \\ 4 & 1 & 1 \\ 4 & 5 & 1 \end{bmatrix}$

7. $\displaystyle \begin{bmatrix} 3 & 2 & 5 & 7 \\ 0 & 0 & 2 & 0 \\ 0 & 4 & 5 & 0 \\ 2 & 1 & 2 & 4 \end{bmatrix}$

8. $\displaystyle \begin{bmatrix} 0 & 2 & 1 & 0 \\ 1 & 2 & -3 & 4 \\ 5 & 6 & -7 & 8 \\ 1 & 2 & 3 & -2 \end{bmatrix}$

a) $-2$     b) $8$     c) $0$     d) $-6$     e) $-3$     f) $28$     g) $16$     h) $-24$

###### ExerciseA.6.102.

For which $x$ are the following matrices singular (not invertible).

1. $\displaystyle \begin{bmatrix} 1 & 3 \\ 1 & x \end{bmatrix}$

2. $\displaystyle \begin{bmatrix} 3 & x \\ 1 & 3 \end{bmatrix}$

3. $\displaystyle \begin{bmatrix} x & 3 \\ 3 & x \end{bmatrix}$

4. $\displaystyle \begin{bmatrix} x & 1 & 0 \\ 1 & 4 & 0 \\ 1 & 6 & 2 \end{bmatrix}$

a) $3$     b) $9$     c) $3$     d) $\nicefrac{1}{4}$

###### ExerciseA.6.103.

Compute

\begin{equation*} \det \left( \begin{bmatrix} 3 & 4 & 7 & 12 \\ 0 & -1 & 9 & -8 \\ 0 & 0 & -2 & 4 \\ 0 & 0 & 0 & 2 \end{bmatrix}^{-1} \right) \end{equation*}

without computing the inverse.

$12$

###### ExerciseA.6.104.

(challenging)   Find all the $x$ that make the matrix inverse

\begin{equation*} \begin{bmatrix} 1 & 2 \\ 1 & x \end{bmatrix}^{-1} \end{equation*}

have only integer entries (no fractions). Note that there are two answers.

$1$ and $3$