[Linear Algebra] 5.3 Cramer's Rule, Inverses, and Volumes

Keywords

[1] $A^{-1}=\frac{C^T}{det(A)}$. Then, $(A^{-1})\mathrm{\_}ij=$ cofactor $C\mathrm{\_}ji$ divided by the determinant of $A$

[2] Cramer’s Rule computes $x=A^{-1}b$ from $x_j=det(A\text{with column}j\text{changed to}b)/det(A)$

[3] Area of parallelogram = $|ad-bc|$ if the four corners are $(0,0),(a,b),(c,d),(a+c,b+d)$

[4] Volume of box = $|det(A)|$ if the rows of $A$ (or columns of $A$) give the sides of the boc

[5] The cross product $w=u\times v$ is $\left|\begin{array}{ccc}i& j& k\\ u_1& u_2& u_3\\ v_1& v_2& v_3\end{array}\right|$. Also, $v\times u=-(u\times v)$. $w_1,w_2,w_3$ are cofactors of row 1. Notice $w^{T}u=0$ and $w^{T}v=0$.

Cramer’s Rule

Cramer’s Rule solves $Ax=b$ algebraically not by elimination. The idea is simple and makes sense.

Suppose we have a matrix $A$. First, we construct an identity matrix $I$ with its first column replaced by $x$ then its determinant is $x_1$. Then, we multiply this matrix with $A$ and the result becomes $B_1$.

$$\left[\begin{array}{ccc}& & \\ & A& \\ & & \end{array}\right]\left[\begin{array}{ccc}x\ast 1& 0& 0\\ x_2& 1& 0\\ x_3& 0& 1\end{array}\right]=\left[\begin{array}{ccc}{b}_1& a\ast 12& a\ast 13\\ b_2& a\ast 22& a\ast 23\\ b_3& a\ast 32& a\mathrm{\_}33\end{array}\right]=B_1$$ $$det(A)(x_1)=det(B_1)$$ $$x_1=\frac{det(B_1)}{det(A)}$$

Now we repeat this process for all the other $x_i$ to get the all solutions.

Cramer’s Rule

If $det(A)$ is nonzero, then $Ax=b$ is solved by determinants:

$$x_1=\frac{det(B_1)}{det(A)}{x}_2=\frac{det(B_2)}{det(A)}\cdots x_n=\frac{det(B_n)}{det(A)}$$

where the matrix $B_j$ is the matrix $A$ with $j^{th}$ column replaced by the vector $b$.

[Note] But Cramer’s Rule evalutes $n+1$ determinants and each one is the sum of $n!$ terms, resulting in $(n+1)!$ terms in total. It would be very inefficient to calculate this (factorial increase very quickly).

Explicit Formula for $x$ using Cramer’s Rule

Let’s take an example of 2 by 2 matrix $A$. We now find the columns of $A^{-1}=[xy]$ by solving

$$A{A}^{-1}=I$$

We can break down this equation for each $x$ and $y$.

$$\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}1\\ 0\end{array}\right]$$ $$\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{c}{y}_1\\ y_2\end{array}\right]=\left[\begin{array}{c}0\\ 1\end{array}\right]$$

Now, we need to calculate $|A|$ and determinants for $x_1,x_2,x_3,x_4$.

$$\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|\left|\begin{array}{cc}\mathbf{\text{1}}& b\\ \mathbf{\text{0}}& d\end{array}\right|\left|\begin{array}{cc}a& \mathbf{\text{1}}\\ c& \mathbf{\text{0}}\end{array}\right|\left|\begin{array}{cc}\mathbf{\text{0}}& b\\ \mathbf{\text{1}}& d\end{array}\right|\left|\begin{array}{cc}a& \mathbf{\text{0}}\\ c& \mathbf{\text{1}}\end{array}\right|$$

The last four determinants are $d,-c,-b,a$. Notice that they’re cofactors!.

Using the Cramer’s Rule, we get

$$x_1=\frac{d}{|A|},x_2=\frac{-c}{|A|},y_1=\frac{-b}{|A|},y_2=\frac{a}{|A|}$$

Then,

$$A^{-1}=\frac{1}{|A|}\left|\begin{array}{cc}d& -b\\ -c& a\end{array}\right|$$

The idea is that $A^{-1}$ involves the cofactors. The determinant of each $B_j$ in Cramer’s Rule is a cofactor of $A$.

Formula

The $i,j$ entry of $A^{-1}$ is the cofactor $C_{ji}$ divided by $det(A)$:

$$(A^{-1}{)}_{ij}=\frac{C_{ji}}{det(A)}$$

or

$$A\mathrm{\_}-1=\frac{C^T}{det(A)}$$

where $C$ is the cofactor matrix where the cofactors $C_{ij}$ go into $C$. The transpose of $C$ leads to $A^{-1}$

Area of a Triangle

Introduction to Linear Algebra, 5th edition

How do we find the area of a triangle with the corners $(x_1,y_1),(x_2,y_2),(x_3,y_3)$. The best way is to use the determinants.

Area by Determinant

The triangle with corners $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ has $A=\frac{determinant}{2}$.

$$A=\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$$ $$A=\frac{1}{2}\left|\begin{array}{cc}{x}_1& y_1\\ x_2& y_2\end{array}\right|,\text{if}(x_3,y_3)=(0,0)$$

Proof

For the second case, why is the area equal to $\frac{1}{2}|x_1{y}_2-x_2{y}_1|$?

Consider the case when $(x_3,y_3)=(0,0)$. Let’s remove the factor $\frac{1}{2}$ to get the parallelogram which is made of two identical triangles.

Introduction to Linear Algebra, 5th edition

This book uses an exotic proof that the area has the same properties as the determinant.

[1] When $A=I$, the parallelogram becomes the unit square. Its area is $det(I)=I$

[2] When rows are exchanged, the determinant reverses sign. The absolute value stays the same - it is the same parallelogram

[3] If row 1 is multiplied by $t$, the area is also multiplied by $t$.

Introduction to Linear Algebra, 5th edition

Hence, the area satisfies the properties of the determinant.

Cross Product

The cross product is a special application for three dimensions. Consider vectors $u=(u_1,u_2,u_3)$ and $v=(v_1,v_2,v_3)$. The cross product is a vector in three dimensions denoted by $u\times v$.

The components of cross product are 2 by 2 cofactors.

Definition

The cross product of $u=(u_1,u_2,u_3)$ and $v=(v_1,v_2,v_3)$ is a vector

$$u\times v=\left|\begin{array}{ccc}i& j& k\\ u_1& u_2& u_3\\ v_1& v_2& v_3\end{array}\right|=\left[\begin{array}{c}{u}_2{v}_3-u_3{v}_2\\ u_3{v}_1-u_1{v}_3\\ u_1{v}_2-u_2{v}_1\end{array}\right]$$

This cross product $u\times v$ is perpendicular to $u$ and $v$. The cross product $v\times u=-(u\times v)$.

Length of Cross Product

$$\Vert u\times v\Vert =\Vert u\Vert \Vert v\Vert |\sin \theta |$$

Property 1

$v\times u=-(u\times v)$ becuase it reverses two rows

Its direction is perpendicular to $u$ and $v$ pointing “up” or “down” by the right hand rule (from $u$ to $v$).

Property 2

The cross product $u\times v$ is perpendicular to $u$ and $v$.

$$u\cdot (u\times v)=u_1(u_2{v}_3-u_3{v}_2)+u_2(u_3{v}_1-u_1{v}_3)+u_3(u_1{v}_2-u_2{v}_1)=0$$

Property 3

The cross product of any vector with itself is $u\times u$ = 0. This is obvious because the determinant with two equal rows is $0$.

Cross vs Dot Product Length

$$\Vert u\times v\Vert =\Vert u\Vert \Vert v\Vert |\sin \theta |$$ $$\Vert u\cdot v\Vert =\Vert u\Vert \Vert v\Vert |\cos \theta |$$

Area of parallelogram

The length of $u\times v$ equals the area of the parallelogram with sides $u$ and $v$.

Triple Product = Determinant = Volume

Recall $u\times v$ is a vector so we can take its dot product with another vector $w$. This combination produces the triple product $(u\times v)\cdot w$ which is a scalar value which is equal to the volume of the $u,v,w$ box.

$$(u\times v)\cdot w=\left|\begin{array}{ccc}{w}_1& w_2& w_3\\ u_1& u_2& u_3\\ v_1& v_2& v_3\end{array}\right|=\left|\begin{array}{ccc}{u}_1& u_2& u_3\\ v_1& v_2& v_3\\ w_1& w_2& w_3\end{array}\right|$$

$(u\times v)\cdot w=0$ when $u,v,w$ lie in the same plane

The reasons are

[1] $u\times v$ is perpendicular to that plane so its dot product with $w$ is zero.

[2] Three vectors in a plane are dependent. The matrix is singular (det=0)

[3] Zero volume when $u,v,w$ box is squashed onto a plane

References

[1] Introduction to Linear Algebra, 5th edition