Question

(7.1)Compute the product AB for

0 4 0 1 0 3

A=2 3 1 , B= 1 1 5

3 0 1 2 3 -1

(7.2) Use your answer in (7.1) to evaluate det(AB) and compare it to det(A) det(B).

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

Answer 1

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

(7.1) We know that dimension of matrix A=3X3 and dimension of matrix B = 3X3

Then, dimension of matrix AB will be 3X3

$AB=\begin{bmatrix} 0+4+0&0+4+0&0+20+0\\2+3+2&0+3+3&6+15-1\\3+0+2&0+0+3&9+0-1 \end{bmatrix}$

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

(7.2) We know that if A and B are n x n matrices, then

det(AB) = (det A)(det B)

Proof:-

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

$det(AB)=4(48-60)-4(56-100)+20(21-30)\\det(AB)=-52\ \ ......(1)$

and we know that

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

$det(A)=0(3-0)-4(2-3)+0(0-9)\\det(A)=4$

$det(B)=1(-1-15)-0(-1-10)+3(3-2)\\det(B)=-13$

and

$det(A)\times det(B)=4\times (-13)=-52\ \ .....(2)$

So, from equation (1) and (2) , It is clear that

$$\boxed{det(AB)=det(A)\times det(B)}$$

(7.3)

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

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

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

$det(A+B)=1(0-15)-4(0-25)+3(9-20)\\det(A+B)=52$

and

$det(A)=0(3-0)-4(2-3)+0(0-9)\\det(A)=4$

$det(B)=1(-1-15)-0(-1-10)+3(3-2)\\det(B)=-13$

So,

$det(A)+det(B)=-9$

So, we can say that $det(A+B)\neq det(A)+det(B)$