Section 3.2 Determinants and Matrix Inverses

Theorem: Let $A$ be a square matrix:

(a) If a multiple of one row of $A$ is added to another row to produce a matrix $B$, then det$B=$det$A$.

 

(b) If two rows of $A$ are interchanged to produce $B$, then det$B=-$det$A$. $($det$A=-$det$B)$.

 

(c) If one row of $A$ is multiplied by$c$ to produce $B$, then det$B=c$det$A($det$A=(1/c)$det$B)$.

 

 

Fact: The above theorem could be rewrite as:

(a) If $E$ is obtained by a multiple of one row of $I_n$ is added to another row, then det$EA=$ det$E$ det$A=$ det$A$.

 

(b) If $E$ is obtained by two rows of $I_n$ are interchanged, then det$EA=$ det$E$ det$A=-$ det$A$.

 

(c) If $E$ is obtained by one row of $I_n$ is multiplied by$c$, then det$EA=$ det$E$ det$A=c$ det$A($det$A=(1/c)$det$EA)$

 

Moreover, det $E=\{\begin{array}{ll}1& \text{ if }E\text{ is the row placement of }{I}_n\\ -1& \text{ if }E\text{ is the row exchange of }{I}_n\\ c& \text{ if }E\text{ is a row scale of }{I}_n\end{array}$

 

Sketch of the Proof: 1. Show the case of $n=2$ is true. 2. Use mathematical induction on$n$. We use the fact that $n-1$ is true to show the case $n$ is true. Because most of you did not know mathematical induction, we will not prove it here.

 

 

Example 1: Compute the determinant of $A=\left[\begin{array}{cccc}1& 0& 2& 3\\ -1& 2& 4& 5\\ 0& 1& -1& 2\\ 2& 3& 0& -1\end{array}\right]$.

 

 

Exercise 1: Compute the determinant of $A=\left[\begin{array}{cccc}1& 0& 0& 3\\ 1& 4& 2& 0\\ -2& 3& -1& 2\\ 0& 1& -2& -1\end{array}\right]$.

 

Theorem: A square matrix $A$ is invertible if and only if det$A\ne 0$.

 

Fact: 1. An $n\times n$ matrix $A$ is invertible if and only if $A^T$ is invertible. When $A$ is not invertible then $A^T$ is not invertible, then $A^T$ has less than $n$ pivot positions, less than $n$ pivot columns. Hence $A^T\overrightarrow{x}=\overrightarrow{0}$ has nontrivial solution.

 

2. If we do the row inter-exchange and row replacement on an $n\times n$ matrix$A$ to obtain the Echelon form of $A$,$U$, then det $A=(-1{)}^r$ det $U$ where $r$ is the number of row exchanges. Notice that $U$ is a triangular matrix, hence det $U$ is the product of the diagonal entries. When $U$ does not have$n$ pivot positions, then the det $U=0$, i.e det $A=0$ and $A$ is not invertible.

 

Example 2: Find the determinant of $A=\left[\begin{array}{cccc}0& -1& 3& 1\\ 2& 3& 1& 2\\ 3& 5& -2& -2\\ 1& 2& 4& 6\end{array}\right]$.

 

 

Exercise 2: Find the determinant of $A=\left[\begin{array}{cccc}1& 0& 3& -1\\ 0& 2& 4& 3\\ 1& 2& 7& 2\\ 3& 4& 5& -2\end{array}\right]$.

 

Theorem: If $A$ is an $n\times n$ matrix then det $A=$det $A^T$.

 

Theorem: If $A$ and$B$ are$n\times n$ matrices then det $AB=$det $A$det $B$.

 

Example 3: Compute det $AB$ without finding$AB$, where $A=\left[\begin{array}{cc}1& 2\\ -1& 3\end{array}\right]$,$B=\left[\begin{array}{cc}2& 3\\ -2& 1\end{array}\right]$.

 

Exercise 3: Compute det $AB$ without finding $AB$, where $A=\left[\begin{array}{cc}0& 1\\ 1& 4\end{array}\right]$,$B=\left[\begin{array}{cc}1& 4\\ -2& 1\end{array}\right]$.

 

Example 4: Compute det $A^{T}B$ without finding$A{B}^T$, where $A=\left[\begin{array}{cc}1& 2\\ -2& 4\end{array}\right]$,$B=\left[\begin{array}{cc}2& -3\\ 3& -1\end{array}\right]$.

 

 

Exercise 4: Compute det $A^{T}B$ without finding $A^{T}B$, where $A=\left[\begin{array}{cc}2& 1\\ 0& 4\end{array}\right]$,$B=\left[\begin{array}{cc}-1& 3\\ 1& -1\end{array}\right]$.

 

GroupWork 1: Mark each statement True or False. Justify each answer. All matrices are $n\times n$ matrices.

a. A row replacement does not affect the determinant of a matrix.

 

b. The determinant of $A$ is the product of the diagonal in any echelon form $U$ of $A$, multiplied by $(-1{)}^r$, where $r$ is the number of row interchange made during the row operation.

 

c. det $(A+B)=$det $A+$det $B$

 

d. If two row-exchange are made in succession, then the new determinant
equals the old determinant.

 

e. The determinant of $A$ is the product of the diagonal entries.

 

f. If det $A$ is zero, then two rows or two columns are the same, or a row or a column is zero.

 

g. det $A^T=(-1)$det $A$.

 

GroupWork 2: Compute det $A^3$.

$A=\left[\begin{array}{ccc}1& 3& 0\\ 1& 2& -2\\ 0& 0& 1\end{array}\right]$

 

GroupWork 3: Show det $(A^{-1})=\frac{1}{\text{det}A}$ when$A$ is invertible.

 

GroupWork 4: In each case either prove the statement or give an example
showing that it is false. All matrices are$n\times n$ matrices.

a. det $AB=$det $B^{T}A$.

 

b. If det $A\ne 0$ and$AB=AC$, then$B=C$.

 

c. If $AB$ is invertible, then$A$ and$B$ are invertible.

 

d. det $(I+A)=1+$det $A$.

 

e. $A$ and $P$ are square matrices and $P$ is invertible then det $PA{P}^{-1}=$det $A$.

 

f. If $A^T=-A$, then det $A=-1$.