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2.3 Properties of Determinants; Cramer’s Rule: Class Notes

Properties of Determinants

For an $n\times n$ matrix $A$, $det(kA)=k^{n}det(A)$

There is one factor of $k$ for each row of $A$.

For example, let $A=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]$. Then

$$det(A)=\left|\begin{array}{cc}1& 2\\ 3& 4\end{array}\right|=-2.$$

Multiplying just the first row by 2 multiplies the determinant by 2 (we saw this in 2.2):

$$\left|\begin{array}{cc}2& 4\\ 3& 4\end{array}\right|=8-12=-4=2(-2)=2det(A).$$

But $3A$ multiplies every row by $3$:

$$3A=\left[\begin{array}{cc}3& 6\\ 9& 12\end{array}\right]\Rightarrow det(3A)=\left|\begin{array}{cc}3& 6\\ 9& 12\end{array}\right|=36-54=-18=3^2(-2).$$

Please note the difference in notation for a matrix vs. a determinant:

$$2\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]\ne 2\left|\begin{array}{cc}1& 2\\ 3& 4\end{array}\right|.$$

The left side is a matrix (every entry is doubled); the right side is a number, $2(-2)=-4$.

Lemma 2.3.2 (prelude to a later result)

Proof

Suppose $E$ results from multiplying the $i$th row of $I$ by $k$. Write the rows of $I$ as ${\overrightarrow{e}}_1,\dots ,{\overrightarrow{e}}_n$. Then

$$EB=\left[\begin{array}{c}{\overrightarrow{e}}_1\\ {\overrightarrow{e}}_2\\ ⋮\\ k{\overrightarrow{e}}_i\\ ⋮\\ {\overrightarrow{e}}_n\end{array}\right]B=\left[\begin{array}{c}{\overrightarrow{e}}_{1}B\\ {\overrightarrow{e}}_{2}B\\ ⋮\\ k{\overrightarrow{e}}_{i}B\\ ⋮\\ {\overrightarrow{e}}_{n}B\end{array}\right],$$

so the $i$th row of $EB$ is the $i$th row of $B$, multiplied by $k$. By Theorem 2.2.3(a), $det(E)=k$, so

$$det(EB)=kdet(B)=det(E)det(B).$$

Theorem 2.3.3

“If and only if” means the statement goes both ways: $A\text{ invertible}\iff det(A)\ne 0$.

Proof

$$\begin{array}{cc}\hfill {\displaystyle A\text{ invertible}}& {\displaystyle \Rightarrow A\text{ is a product of elementary matrices}}\hfill \\ \hfill {\displaystyle }& {\displaystyle \Rightarrow det(A)\ne 0\text{(using Lemma 2.3.2).}}\hfill \end{array}$$

$$\begin{array}{cc}\hfill {\displaystyle det(A)\ne 0}& {\displaystyle \Rightarrow \text{the rref of }A\text{ does not have a row of zeros}}\hfill \\ \hfill {\displaystyle }& {\displaystyle \phantom{\Rightarrow }\text{(contrapositive of Theorem 2.2.1)}}\hfill \\ \hfill {\displaystyle }& {\displaystyle \Rightarrow rref(A)=I}\hfill \\ \hfill {\displaystyle }& {\displaystyle \Rightarrow A\text{ is invertible.}}\hfill \end{array}$$

Theorem 2.3.4

But in general $det(A+B)\ne det(A)+det(B)$.

Proof

(Outline of the proof.) If $A$ and/or $B$ is not invertible, then $AB$ is not invertible. So $det(A)=0$ or $det(B)=0$, and $det(AB)=0$. Both sides of $det(AB)=det(A)det(B)$ are 0, so we have $0=0$.

If $A$ and $B$ are invertible, we write them as products of elementary matrices and use Lemma 2.3.2.

Theorem 2.3.5

Proof

$A$ invertible $\Rightarrow$ $A^{-1}$ exists and $det(A)\ne 0$.

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle A{A}^{-1}=I}& {\displaystyle \Rightarrow det\left(A{A}^{-1}\right)=det(I)=1}\hfill \\ \hfill {\displaystyle }& {\displaystyle \Rightarrow det(A)det\left(A^{-1}\right)=1\text{(Theorem 2.3.4)}}\hfill \\ \hfill {\displaystyle }& {\displaystyle \Rightarrow det\left(A^{-1}\right)=\frac{1}{det(A)}.}\hfill \end{array}}$$

Theorem 2.3.8 Equivalent Statements (extends Theorem 1.6.4)

Statement (g) is the new one added to Theorem 1.6.4.

#16

Find the values of $k$ for which the matrix $A$ is invertible.

$$A=\left[\begin{array}{cc}k& 2\\ 2& k\end{array}\right]$$

Solution

$A$ is invertible $\iff det(A)\ne 0$. Find where the determinant is zero:

$$\left|\begin{array}{cc}k& 2\\ 2& k\end{array}\right|=0\Rightarrow k^2-4=0\Rightarrow k=\pm 2.$$

#35

In each part, find the determinant given that $A$ is a $3\times 3$ matrix for which $det(A)=7$.

a. $det(3A)$

b. $det\left(A^{-1}\right)$

c. $det\left(2A^{-1}\right)$

d. $det((2A{)}^{-1})$

Solution

$A$ is $3\times 3$, so a scalar $k$ in front comes out as $k^3$.

a. $det(3A)=3^{3}det(A)=27(7)$

b. By Theorem 2.3.5, ${\displaystyle det\left(A^{-1}\right)=\frac{1}{det(A)}}$.

c. ${\displaystyle det\left(2A^{-1}\right)=2^{3}det\left(A^{-1}\right)=2^3\left(\frac{1}{7}\right)}$

d.

$${\displaystyle det((2A{)}^{-1})=\frac{1}{det(2A)}=\frac{1}{2^{3}det(A)}=\frac{1}{2^3·7}}$$

Definition

Theorem 2.3.6 Inverse of a Matrix Using Its Adjoint

#20

Decide whether the matrix is invertible, and if so, use the adjoint to find its inverse.

$$\left[\begin{array}{ccc}2& 0& 3\\ 0& 3& 2\\ -2& 0& -4\end{array}\right]$$

Solution

First find the minors. Each $M_{ij}$ is the determinant left after crossing out row $i$ and column $j$ of $A$.

$$\begin{array}{ccc}{\displaystyle M_{11}=\left|\begin{array}{cc}3& 2\\ 0& -4\end{array}\right|=-12,}\hfill & {\displaystyle M_{12}=\left|\begin{array}{cc}0& 2\\ -2& -4\end{array}\right|=4,}\hfill & {\displaystyle M_{13}=\left|\begin{array}{cc}0& 3\\ -2& 0\end{array}\right|=6,}\hfill \\ {\displaystyle M_{21}=\left|\begin{array}{cc}0& 3\\ 0& -4\end{array}\right|=0,}\hfill & {\displaystyle M_{22}=\left|\begin{array}{cc}2& 3\\ -2& -4\end{array}\right|=-2,}\hfill & {\displaystyle M_{23}=\left|\begin{array}{cc}2& 0\\ -2& 0\end{array}\right|=0,}\hfill \\ {\displaystyle M_{31}=\left|\begin{array}{cc}0& 3\\ 3& 2\end{array}\right|=-9,}\hfill & {\displaystyle M_{32}=\left|\begin{array}{cc}2& 3\\ 0& 2\end{array}\right|=4,}\hfill & {\displaystyle M_{33}=\left|\begin{array}{cc}2& 0\\ 0& 3\end{array}\right|=6.}\hfill \end{array}$$

The cofactors are $C_{ij}=(-1{)}^{i+j}{M}_{ij}$, so use the checkerboard of signs $\left[\begin{array}{ccc}+& -& +\\ -& +& -\\ +& -& +\end{array}\right]$:

$$\begin{array}{ccc}{\displaystyle C_{11}=-12,}\hfill & {\displaystyle C_{12}=-4,}\hfill & {\displaystyle C_{13}=6,}\hfill \\ {\displaystyle C_{21}=0,}\hfill & {\displaystyle C_{22}=-2,}\hfill & {\displaystyle C_{23}=0,}\hfill \\ {\displaystyle C_{31}=-9,}\hfill & {\displaystyle C_{32}=-4,}\hfill & {\displaystyle C_{33}=6.}\hfill \end{array}$$

Expand along the first row to get the determinant (using the first-row cofactors):

$$det(A)=2(-12)+0(-4)+3(6)=-6\ne 0,$$

so $A$ is invertible.

Tip: Row reduction gives the same determinant: $R_3\to R_1+R_3$ gives the upper triangular matrix $\left[\begin{array}{ccc}2& 0& 3\\ 0& 3& 2\\ 0& 0& -1\end{array}\right]$, whose determinant is the product of the diagonal entries, $2(3)(-1)=-6$.

Matrix of cofactors:

$$\left[C_{ij}\right]=\left[\begin{array}{ccc}-12& -4& 6\\ 0& -2& 0\\ -9& -4& 6\end{array}\right]$$

The transpose of $\left[C_{ij}\right]$ is the adjoint of $A$:

$${\left[C_{ij}\right]}^T=adj(A)=\left[\begin{array}{ccc}-12& 0& -9\\ -4& -2& -4\\ 6& 0& 6\end{array}\right]$$

$${\displaystyle A^{-1}=\frac{1}{det(A)}adj(A)=-\frac{1}{6}\left[\begin{array}{ccc}-12& 0& -9\\ -4& -2& -4\\ 6& 0& 6\end{array}\right]}$$

Theorem 2.3.7 Cramer’s Rule

#26

Solve by Cramer’s rule.

$$\begin{array}{cc}\hfill {\displaystyle x-4y+z}& {\displaystyle =6}\hfill \\ \hfill {\displaystyle 4x-y+2z}& {\displaystyle =-1}\hfill \\ \hfill {\displaystyle 2x+2y-3z}& {\displaystyle =-20}\hfill \end{array}$$

Solution

$$A=\left[\begin{array}{ccc}1& -4& 1\\ 4& -1& 2\\ 2& 2& -3\end{array}\right],𝐛=\left[\begin{array}{c}6\\ -1\\ -20\end{array}\right]$$

Each $3\times 3$ determinant is found by copying the first two columns to the right. Multiply along the three diagonals going down to the right and add; then subtract the products along the three diagonals going up to the right.

$$det(A):\left|\begin{array}{ccc}1& -4& 1\\ 4& -1& 2\\ 2& 2& -3\end{array}\right|\begin{array}{cc}1& -4\\ 4& -1\\ 2& 2\end{array}$$

$$\begin{array}{cc}\hfill {\displaystyle det(A)}& {\displaystyle =(3-16+8)-(-2+4+48)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =(-5)-(50)=-55.}\hfill \end{array}$$

Since $det(A)=-55\ne 0$, Cramer’s rule applies. Now replace one column at a time with the column $𝐛$.

$$det(A_1):\left|\begin{array}{ccc}6& -4& 1\\ -1& -1& 2\\ -20& 2& -3\end{array}\right|\begin{array}{cc}6& -4\\ -1& -1\\ -20& 2\end{array}$$

$$\begin{array}{cc}\hfill {\displaystyle det(A_1)}& {\displaystyle =(18+160-2)-(20+24-12)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =176-32=144.}\hfill \end{array}$$

$$det(A_2):\left|\begin{array}{ccc}1& 6& 1\\ 4& -1& 2\\ 2& -20& -3\end{array}\right|\begin{array}{cc}1& 6\\ 4& -1\\ 2& -20\end{array}$$

$$\begin{array}{cc}\hfill {\displaystyle det(A_2)}& {\displaystyle =(3+24-80)-(-2-40-72)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =(-53)-(-114)=61.}\hfill \end{array}$$

$$det(A_3):\left|\begin{array}{ccc}1& -4& 6\\ 4& -1& -1\\ 2& 2& -20\end{array}\right|\begin{array}{cc}1& -4\\ 4& -1\\ 2& 2\end{array}$$

$$\begin{array}{cc}\hfill {\displaystyle det(A_3)}& {\displaystyle =(20+8+48)-(-12-2+320)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =76-306=-230.}\hfill \end{array}$$

$${\displaystyle \begin{array}{c}{\displaystyle x=\frac{det(A_1)}{det(A)}=\frac{144}{-55}=-\frac{144}{55},}\hfill \\ {\displaystyle y=\frac{det(A_2)}{det(A)}=\frac{61}{-55}=-\frac{61}{55},}\hfill \\ {\displaystyle z=\frac{det(A_3)}{det(A)}=\frac{-230}{-55}=\frac{46}{11}.}\hfill \end{array}}$$

Proof of Cramer’s Rule (Theorem 2.3.7)

The existence of a unique solution is guaranteed by Theorem 2.3.8 (Equivalent Statements), since $det(A)\ne 0$. Then

$${\displaystyle A\overrightarrow{x}=\overrightarrow{b}\Rightarrow \overrightarrow{x}=A^{-1}\overrightarrow{b}=\frac{1}{det(A)}adj(A)\overrightarrow{b}}$$

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle }& {\displaystyle =\frac{1}{det(A)}\left[\begin{array}{cccc}{C}_{11}& C_{21}& \cdots & C_{n1}\\ C_{12}& C_{22}& \cdots & C_{n2}\\ ⋮& ⋮& & ⋮\\ C_{1n}& C_{2n}& \cdots & C_{nn}\end{array}\right]\left[\begin{array}{c}{b}_1\\ b_2\\ ⋮\\ b_n\end{array}\right]}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\frac{1}{det(A)}\left[\begin{array}{c}{b}_1{C}_{11}+b_2{C}_{21}+\cdots +b_n{C}_{n1}\\ b_1{C}_{12}+b_2{C}_{22}+\cdots +b_n{C}_{n2}\\ ⋮\\ b_1{C}_{1n}+b_2{C}_{2n}+\cdots +b_n{C}_{nn}\end{array}\right].}\hfill \end{array}}$$

Then

$${\displaystyle x_i=\frac{b_1{C}_{1i}+b_2{C}_{2i}+\cdots +b_n{C}_{ni}}{det(A)}.}$$

The numerator is the cofactor expansion along the $i$th column of $A$, with that column replaced by $\overrightarrow{b}$. That is exactly $det(A_i)$, so ${\displaystyle x_i=\frac{det(A_i)}{det(A)}}$.