5.1 Determinate

Section 5.1: The Determinant Function Classwork

1. Compute the determinants of the following $2\times 2$ matrices.

a. $A=\left[\begin{array}{cc}2& -1\\ 3& 1\end{array}\right]$

$$\det (A)=(2\times 1)-(-1\times 3)=2+3=5$$

b. $A=\left[\begin{array}{cc}6& 2\\ -9& 3\end{array}\right]$

$$\det (A)=(6\times 3)-(2\times -9)=18+18=36$$

2. Compute the determinant for the indicated minors for the following matrices.

a. $A=\left[\begin{array}{ccc}1& 0& 4\\ 2& 1& 3\\ 0& 4& -1\end{array}\right]$, $\det (M_{11})$

Minor matrix by removing first row and first column:

$$M_{11}=\left[\begin{array}{cc}1& 3\\ 4& -1\end{array}\right]$$ $$\det (M_{11})=(1\times -1)-(3\times 4)=-1-12=-13$$

b. $A=\left[\begin{array}{ccc}1& 3& -2\\ 6& 2& 1\\ 3& 1& 0\end{array}\right]$, $\det (M_{23})$

Minor matrix by removing second row and third column:

$$M_{23}=\left[\begin{array}{cc}1& 3\\ 3& 1\end{array}\right]$$ $$\det (M_{23})=(1\times 1)-(3\times 3)=1-9=-8$$

c. $A=\left[\begin{array}{ccc}4& -1& 0\\ 2& 1& 1\\ 1& 3& 3\end{array}\right]$, $\det (M_{32})$

Minor matrix by removing third row and second column:

$$M_{32}=\left[\begin{array}{cc}4& 0\\ 2& 1\end{array}\right]$$ $$\det (M_{32})=(4\times 1)-(0\times 2)=4$$

3. Compute the determinants of the following matrices.

a. $A=\left[\begin{array}{ccc}3& -1& 0\\ 2& 2& 1\\ 0& 4& 1\end{array}\right]$

$$\det (A)=3\left|\begin{array}{cc}2& 1\\ 4& 1\end{array}\right|-(-1)\left|\begin{array}{cc}2& 1\\ 0& 1\end{array}\right|+0\left|\begin{array}{cc}2& 2\\ 0& 4\end{array}\right|$$ $$=3(2-4)+1(2-0)=3(-2)+2=-6+2=-4$$

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

$$\det (A)=1\left|\begin{array}{cc}0& 1\\ 2& -1\end{array}\right|-1\left|\begin{array}{cc}3& 1\\ 0& -1\end{array}\right|+(-2)\left|\begin{array}{cc}3& 0\\ 0& 2\end{array}\right|$$ $$=1(0-2)-1(-3-0)-2(6-0)=-2+3-12=-11$$

4. Verify that $\det (AB)=\det (A)\det (B)$ for the given matrices:

Given:

$$A=\left[\begin{array}{cc}4& 1\\ 2& 1\end{array}\right],B=\left[\begin{array}{cc}-1& 0\\ 4& 1\end{array}\right]$$

Compute $\det (A)$:

$$\det (A)=(4\times 1)-(1\times 2)=4-2=2$$

Compute $\det (B)$:

$$\det (B)=(-1\times 1)-(0\times 4)=-1$$

Compute $AB$:

$$AB=\left[\begin{array}{cc}(4\times -1)+(1\times 4)& (4\times 0)+(1\times 1)\\ (2\times -1)+(1\times 4)& (2\times 0)+(1\times 1)\end{array}\right]=\left[\begin{array}{cc}-4+4& 0+1\\ -2+4& 0+1\end{array}\right]=\left[\begin{array}{cc}0& 1\\ 2& 1\end{array}\right]$$ $$\det (AB)=(0\times 1)-(1\times 2)=-2$$ $$\det (A)\det (B)=2\times (-1)=-2$$

Thus, verified: $\det (AB)=\det (A)\det (B)$.

5. Prove that the determinant of the following matrix gives the equation of a line passing through two points:

$$\left|\begin{array}{ccc}x& y& 1\\ x_1& y_1& 1\\ x_2& y_2& 1\end{array}\right|=0$$

Expanding along the first row:

$$\det (A)=x\left|\begin{array}{cc}{y}_1& 1\\ y_2& 1\end{array}\right|-y\left|\begin{array}{cc}{x}_1& 1\\ x_2& 1\end{array}\right|+1\left|\begin{array}{cc}{x}_1& y_1\\ x_2& y_2\end{array}\right|$$

Computing minors:

$$\left|\begin{array}{cc}{y}_1& 1\\ y_2& 1\end{array}\right|=y_1-y_2,\left|\begin{array}{cc}{x}_1& 1\\ x_2& 1\end{array}\right|=x_1-x_2,\left|\begin{array}{cc}{x}_1& y_1\\ x_2& y_2\end{array}\right|=x_1{y}_2-x_2{y}_1$$

Thus, the equation simplifies to:

$$(x_1-x_2)y-(y_1-y_2)x+(x_1{y}_2-x_2{y}_1)=0$$

which is the equation of a line passing through $(x_1,y_1)$ and $(x_2,y_2)$.