Answer to Question #130471 in Linear Algebra for Evans Malepe

Question

Suppose A and B are 3 × 3 matrices. Show that If B is obtained from A by adding 2 times the first row of A to the last row of A, then

det (A) = det (B)

Answer 1

A: $\begin{bmatrix} a & b & c\\ d & e & f\\ g & h & i \end{bmatrix}$

B: $\begin{bmatrix} a & b & c\\ d & e & f\\ g+2a & h+2b & i+2c \end{bmatrix}$

$det(A)=det(B)$

$a*det(\begin{bmatrix} e & f \\ h & i \end{bmatrix})-b*det(\begin{bmatrix} d & f \\ g & i \end{bmatrix})+c*det(\begin{bmatrix} d & e \\ g & h \end{bmatrix})=a*det(\begin{bmatrix} e & f \\ h+2b & i+2c \end{bmatrix})-b*det(\begin{bmatrix} d & f \\ g+2a & i+2c \end{bmatrix})+c*det(\begin{bmatrix} d & e \\ g+2a & h+2b \end{bmatrix})$

$a(ei-fh)-b(di-fg)+c(dh-eg)=a(e(i+2c)-f(h+2b))-b(d(i+2c)-f(g+2a))+c(d(h+2b)-e(g+2a))$

$aei-afh-bdi+bfg+cdh-ecg=bfg-afh+eai-bdi-ecg+cdh$

$0=0$

Proved.