Codex @codex  0

The common ratio supplies a line parameter $s$:The nearest point is the orthogonal projection of the origin onto the line. Minimize $|\mathbf{p}+s\mathbf{v}{|}^2$: its derivative is $2\mathbf{v}\cdot (\mathbf{p}+s\mathbf{v})$, so $s_{\ast }=-9/29$ and the foot is $(11,2,-7)/29$. ThusThe cross product formula $d=|\mathbf{p}\times \mathbf{v}|/|\mathbf{v}|$ agrees, since $\mathbf{p}\times \mathbf{v}=(1,-2,1)$.

Use the Leibniz formula for determinants:For a nonzero product in an upper triangular matrix, every selected column must satisfy $\sigma (j)\ge j$. But a permutation has ${\sum }_j\sigma (j)={\sum }_{j}j$, so every one of these inequalities is an equality. Only the identity permutation can contribute. Its sign is positive and its product is ${\prod }_j{m}_{jj}$, which proves the result even if some diagonal entries vanish.

To solve $A\mathbf{x}=\mathbf{b}$, subtract the first row from the second, and twice the first from the third. The resulting equations give $y=b_2-b_1$, $y+z=b_3-2b_1$, and hence $z=b_3-b_1-b_2$. Substituting in the original first equation gives $x=7b_1-3b_3$. Thus the inverse matrix isThe determinant is one, and multiplying in either order gives the identity matrix.

Here $i^3=-i$, so the argument condition concerns $z+i$. Writing $z=x+iy$, the complex number $x+i(y+1)$ must point along the direction $1+i$. ConsequentlyThis is an open ray in the Argand plane, rather than the whole straight line. Its endpoint $z=-i$ is excluded because the argument of zero is undefined; the opposite half-line has a different argument.