Today we mostly finish Section 4.2. Read Section 4.3 and Appendix D (polynomials) for next class. Work through suggested exercises.
Homework 8 is on WeBWorK and is due Friday at 11:55pm.
Math Help Centre: M-F 12:30-5:30 in PAB48/49 and online 6pm-8pm.
My next office hour is today 2:30-3:20 in MC130.
Definition: Let $A = [a_{ij}]$ be an $n \times n$ matrix. Then the determinant of $A$ is the scalar $$ \begin{aligned} \det A = |A| &= a_{11} C_{11} + a_{12} C_{12} + \cdots + a_{1n} C_{1n} \\ &= \sum_{j=1}^{n} a_{1j} C_{1j} = \sum_{j=1}^{n} (-1)^{1+j} \, a_{1j} \det A_{1j} . \end{aligned} $$ We define the determinant of a $1 \times 1$ matrix $[a]$ to be $a$.
This is a recursive definition!
For $n = 2$: $$ \kern-6ex \begin{aligned} \det A = |A| &= \bdmat{rr} a_{11} & a_{12} \\ a_{21} & a_{22} \edmat \\ &= a_{11} |a_{22}| - a_{12} |a_{21}| = a_{11} a_{22} - a_{12} a_{21} , \end{aligned} $$ as we defined earlier.
For a $3 \times 3$ matrix $A$, we have $$ \kern-9ex \begin{aligned} \det A = \bdmat{rrr} \red{a_{11}} & \red{a_{12}} & \red{a_{13}} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \edmat &= \red{a_{11}} C_{11} + \red{a_{12}} C_{12} + \red{a_{13}} C_{13} \\ &= \red{a_{11}} \bdmat{rr} a_{22} & a_{23} \\ a_{32} & a_{33} \edmat - \red{a_{12}} \bdmat{rr} a_{21} & a_{23} \\ a_{31} & a_{33} \edmat + \red{a_{13}} \bdmat{rr} a_{21} & a_{22} \\ a_{31} & a_{32} \edmat \end{aligned} $$
The computation can be very long if there aren't many zeros! We'll learn some better methods.
The above is called the cofactor expansion along the first row. It turns out that any row or column works!
Theorem 4.1 (The Laplace Expansion Theorem): Let $A$ be any $n \times n$ matrix. Then for each $i$ we have $$ \kern-6ex \det A = a_{i1} C_{i1} + a_{i2} C_{i2} + \cdots + a_{in} C_{in} = \sum_{j=1}^{n} \, a_{ij} C_{ij} $$ (cofactor expansion along the $i$th row). And for each $j$ we have $$ \kern-6ex \det A = a_{1j} C_{1j} + a_{2j} C_{2j} + \cdots + a_{nj} C_{nj} = \sum_{i=1}^{n} \, a_{ij} C_{ij} $$ (cofactor expansion along the $j$th column).
The book proves this result at the end of this section, but we won't cover the proof.
The signs in the cofactor expansion form a checkerboard pattern: $$ \bmat{ccccc} + & - & + & - & \cdots \\ - & + & - & + & \cdots \\ + & - & + & - & \cdots \\ - & + & - & + & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots \emat $$
A triangular matrix is a square matrix that is all zero below the diagonal or above the diagonal.
Theorem 4.2: If $A$ is triangular, then $\det A$ is the product of the diagonal entries.
So how do we do better? Like always, we turn to row reduction! These properties will be what we need:
Theorem 4.3: Let $A$ be a square matrix.
a. If $A$ has a zero row, then $\det A = 0$.
b. If $B$ is obtained from $A$ by interchanging two rows,
then $\det B = - \det A$.
c. If $A$ has two identical rows, then $\det A = 0$.
d. If $B$ is obtained from $A$ by multiplying a row of $A$ by $k$,
then $\det B = k \det A$.
e. If $A$, $B$ and $C$ are identical in all rows except the $i$th row,
and the $i$th row of $C$ is the sum of the $i$th rows of $A$ and $B$,
then $\det C = \det A + \det B$.
f. If $B$ is obtained from $A$ by adding a multiple of one row to another,
then $\det B = \det A$.
All of the above statements are true with rows replaced by columns.
The bold statements are the ones that are useful for understanding how row operations change the determinant.
Example: Use row operations to compute $\det A$ for $A$ shown below by reducing to triangular form:
Solution: $$ \kern-8ex \begin{aligned} A = \bmat{rrrr} 2 & \phm 4 & \phm 6 & \phm 8 \\ 1 & 2 & 1 & 2 \\ 2 & 2 & 12 & 8 \\ 1 & 2 & 3 & 9 \emat &\lra{(1/2)R_1} \bmat{rrrr} 1 & \phm 2 & \phm 3 & \phm 4 \\ 1 & 2 & 1 & 2 \\ 2 & 2 & 12 & 8 \\ 1 & 2 & 3 & 9 \emat \\ \lra{\mystackthree{\scriptstyle R_2 - R_1}{\scriptstyle R_3 - 2R_1}{\scriptstyle R_4 - R_1}} \bmat{rrrr} 1 & 2 & 3 & 4 \\ 0 & 0 & -2 & -2 \\ 0 & -2 & 6 & 0 \\ 0 & 0 & 0 & 5 \emat &\lra{\,R_2 \leftrightarrow R_3\,} \bmat{rrrr} 1 & \phm 2 & 3 & 4 \\ 0 & -2 & 6 & 0 \\ 0 & 0 & -2 & -2 \\ 0 & 0 & 0 & 5 \emat \end{aligned} $$ The determinant of the last matrix is $(1)(-2)(-2)(5) = 20$, so the determinant of the original matrix is $\query{-40}$
Example 27-1: Same for $A = \bmat{rrr} 2 & 3 & -1 \\ -4 & -6 & 2 \\ 2 & 5 & 3 \emat$. On board.
Note that you can even mix and match row and column operations, if it simplifies the work.
Question: What happens if you do something like "replace $R_3$ with $R_1 + 3 R_3$"?
Questions: Let $E$ be an elementary matrix, formed by performing
one row operation to the identity matrix $I$.
(a) If two rows were exchanged, then $\det E = \query{-1}$
(b) If a row was scaled by $k$, then $\det E = \query{k}$
(c) If a multiple of one row was added to another, then $\det E = \query{1}$
(The above is Theorem 4.4 in the book.)
Row reduction of an $n \times n$ matrix requires roughly $n^3$ operations in general, which is much less than $n$ factorial. E.g. 3 hours vs. 1 year to do a $10 \times 10$ matrix by hand (rough estimate).
Use Laplace expansion for matrices with lots of zeros, or matrices with variables like $\lambda$. Use row reduction for $4 \times 4$ or larger matrices with not many zeros. For a $3 \times 3$ matrix, which is better depends on the matrix.
Theorem 4.6: A square matrix $A$ is invertible if and only if $\det A \neq 0$.
The book proves this using elementary matrices, but here is a simpler proof.
Proof: If $A$ is invertible, then by the Fundamental Theorem, the reduced row echelon form of $A$ is $I$. Each elementary row operation either leaves the determinant the same or multiplies by a non-zero number. Since $\det I = 1 \neq 0$, we must also have $\det A \neq 0$.
On the other hand, if $A$ is not invertible, then the reduced row echelon form $R$ has a zero row, so $\det R = 0.$ Again, $\det A = k \det R$ for some $k$, so $\det A = 0$ too. $\quad\Box$
Example: The $4 \times 4$ matrix above is invertible, but the $3 \times 3$ is not. The computations illustrate the proof of the theorem.
Question: Is the matrix $\bmat{rrr} 1 & 0 & 0 & 0 \\ 2 & 5 & 0 & 0 \\ 3 & 6 & 0 & 0 \\ 4 & 7 & 8 & 9 \\ \emat$ invertible?
Theorems 4.7 to 4.10: Let $A$ be an $n \times n$ matrix. Then:
4.7: $\det(k A) = \query{k^n \det A}$
4.8: $\det(A B) = \query{(\det A) (\det B)}$
4.9: $\det(A^{-1}) = \query{\frac{1}{\det A}}$, if $A$ is invertible.
4.10: $\det(A^T) = \query{\det A}$
Note: There is no formula for $\det(A+B)$.
Proofs:
4.7: Follows from Theorem 4.3(d), since each row is multiplied by $k$.
4.8: The book uses elementary matrices to prove this, and I won't repeat the argument.
I'll do an example below.
4.9: This follows from 4.8. Since $\det(A A^{-1}) = \det(I) = 1$
we have $(\det A)(\det A^{-1}) = 1$, so $\det A^{-1} = 1/\det A$.
4.10: Computing $\det A$ using expansion along the first row
produces the same thing as computing $\det (A^T)$ by expanding along
the first column. (Proof by induction on the size of the matrix.)
$\quad\Box$
Example 27-2: Illustrate all four statements with $2 \times 2$ matrices, on board.
True/false: If $A$ and $B$ are invertible, then so is $A + B$.
True/false: If $A$ is not invertible, then $AB$ is not invertible.
Notation: If $A$ is an $n \times n$ matrix and $\vb \in \R^n$, we write $A_i(\vb)$ for the matrix obtained from $A$ by replacing the $i$th column with the vector $\vb$: $$ A_i(\vb) = [ \va_1 \cdots \va_{i-1} \, \vb \,\, \va_{i+1} \cdots \va_n \,] $$
Theorem 4.11: Let $A$ be an invertible $n \times n$ matrix and let $\vb$ be in $\R^n$. Then the unique solution $\vx$ of the system $A \vx = \vb$ has components $$ x_i = \frac{\det(A_i(\vb))}{\det A},\quad \text{for } i = 1, \ldots, n $$
Example 4.16: On board: Use Cramer's rule to solve $$ \begin{aligned} x_1 + 2 x_2 &= 2 \\ -x_1 + 4 x_2 &= 1 \end{aligned} $$
Proof: Suppose $A \vx = \vb$. Consider $I_i(\vx) = [ \ve_1 \cdots \vx \cdots \ve_n\,]$. Then $$ \kern-6ex \begin{aligned} A\, I_i(\vx) &= A \, [ \ve_1 \cdots \vx \cdots \ve_n\,] = [ A \ve_1 \cdots A \vx \cdots A \ve_n\,]\\ &= [ \va_1 \cdots \vb \cdots \va_n\,] = A_i(\vb) . \end{aligned} $$ So $$ \kern-6ex (\det A)(\det I_i(\vx)) = \det (A \, I_i(\vx)) = \det(A_i(\vb)). $$ But $$ \det I_i(\vx) = \bdmat{ccccc} 1 & \cdots & x_1 & \cdots & 0 \\ \vdots & \ddots & \vdots & & \vdots \\ 0 & \cdots & x_i & \cdots & 0 \\ \vdots & & \vdots & \ddots & \vdots \\ 0 & \cdots & x_n & \cdots & 1 \\ \edmat = x_i $$ by expanding along $i$th row. So the claim follows.$\quad\Box$
Note: This is not an efficient method. For an $n \times n$ system, you have to compute $n+1$ determinants. But the work in computing $1$ determinant is enough to solve the system by our usual method.