Answer to Question #201446 in Linear Algebra for Rebecca

Question

a. LCompute the product AB for

A = 0 4 0 2 3 1 3 0 1 and B = 1 0 3 1 1 5 2 3 -1

b.) Use your answer in a.) to evaluate det(AB) and compare it to det(A) det(B)

c.) Determine whether or not if det(A+B) is related to det(A) + det(B).

Answer 1

(a)

$$AB=\begin{pmatrix} 0 & 4 & 0 \\ 2 & 3 & 1 \\ 3 & 0 &1 \end{pmatrix}\cdot\begin{pmatrix} 1 & 0 & 3 \\ 1 & 1 & 5 \\ 2 & 3 & -1 \end{pmatrix}$$

$$=\begin{pmatrix} 4 & 4 & 20 \\ 7 & 6 & 20 \\ 5 & 3 &8 \end{pmatrix}$$

(b)

$$\det(AB)=\begin{vmatrix} 4 & 4 & 20 \\ 7 & 6 & 20 \\ 5 & 3 &8 \end{vmatrix}$$

$$=4\begin{vmatrix} 6 & 20 \\ 3 & 8 \end{vmatrix}-4\begin{vmatrix} 7 & 20 \\ 5 & 8 \end{vmatrix}+20\begin{vmatrix} 7 & 6 \\ 5 & 3 \end{vmatrix}$$

$$=4(48-60)-4(56-100)+20(21-30)=-52$$

$$\det(A)=\begin{vmatrix} 0 & 4 & 0 \\ 2 & 3 & 1 \\ 3 & 0 &1 \end{vmatrix}$$

$$=0\begin{vmatrix} 3 & 1 \\ 0 & 1 \end{vmatrix}-4\begin{vmatrix} 2 & 1 \\ 3 & 1 \end{vmatrix}+0\begin{vmatrix} 2 & 3 \\ 3 & 0 \end{vmatrix}$$

$$=0-4(2-3)+0=4$$

$$\det(B)=\begin{vmatrix} 1 & 0 & 3 \\ 1 & 1 & 5 \\ 2 & 3 & -1 \end{vmatrix}$$

$$=1\begin{vmatrix} 1 & 5 \\ 3 & -1 \end{vmatrix}-0\begin{vmatrix} 1 & 5 \\ 2 & -1 \end{vmatrix}+3\begin{vmatrix} 1 & 1 \\ 2 & 3 \end{vmatrix}$$

$$=1(-1-15)-0+3(3-2)=-13$$

$$\det(AB)=-52=\det(A)\det(B)$$

(c)

$$A+B=\begin{pmatrix} 0 & 4 & 0 \\ 2 & 3 & 1 \\ 3 & 0 &1 \end{pmatrix}+\begin{pmatrix} 1 & 0 & 3 \\ 1 & 1 & 5 \\ 2 & 3 & -1 \end{pmatrix}$$

$$=\begin{pmatrix} 1 & 4 & 3 \\ 3 & 4 & 6 \\ 5 & 3 & 0 \end{pmatrix}$$

$$\det(A+B)=\begin{vmatrix} 1 & 4 & 3 \\ 3 & 4 & 6 \\ 5 & 3 & 0 \end{vmatrix}$$

$$=1\begin{vmatrix} 4 & 6 \\ 3 & 0 \end{vmatrix}-4\begin{vmatrix} 3 & 6 \\ 5 & 0 \end{vmatrix}+3\begin{vmatrix} 3 & 4 \\ 5 & 3 \end{vmatrix}$$

$$=-18-4(-30)+3(9-20)=69$$

$$\det(A+B)\not=\det(A)+\det(B)$$