Consider the matrices A, B, and C, where A is defined as in Prob. 1 and

$$B = \begin{bmatrix} 2 & 3 & 1 \\ 2 & 1 & 1 \\ 1 & 0 & 3 \end{bmatrix},$$

$$C = \begin{bmatrix} 3 & 2 & 1 \\ 2 & 1 & 1 \\ 1 & 0 & 3 \end{bmatrix}$$.

Note that A, B, and C are identical except for the first row, and for that row the relationship *c*_*ij* = *a*_*ij* + *b*_*ij*; *j* = 1, 2, 3; holds. Show that det (A) + det (B) = det (C), and hence demonstrate Theorem 1.5.12.