A unit vector along the line is $d=(3,4{)}^{\top }/5$. Reflection preserves the orthogonal projection $d{d}^{\top }u$ onto that line and negates its perpendicular component. Therefore $Au=d{d}^{\top }u-(u-d{d}^{\top }u)$ andAs checks, $A(3,4{)}^{\top }=(3,4{)}^{\top }$, $A(-4,3{)}^{\top }=-(-4,3{)}^{\top }$, and $A^2=I$. Its determinant is minus one, as expected for a planar reflection matrix.

Replacing row one by $R_1-R_2-R_3-R_4$ does not change the determinant: adding a multiple of another row preserves a determinant. It produces first row $(-2,0,0,0)$, whose cofactor minor is the three-dimensional identity matrix. Thus the determinant is $-2$.

The columns are homogeneous coordinate vectors $(1,v_i{)}^{\top }$ of the four vertices. Subtracting the first column from each other column preserves the determinant and reduces its magnitude to the absolute determinant of the three edge vectors based at $v_1$. That absolute determinant is the parallelepiped volume. A tetrahedron occupies one sixth of that parallelepiped, so $V=|-2|/3!=1/3$ in cubic coordinate units.

Write $U=E{E}_5{E}_4{E}_3{E}_2{E}_{1}M$ and put $F_i=E{E}_{i}E$. Inserting $E^2=I$ between adjacent elimination matrices givesEach $E_i^{-1}$ has the restricted lower triangular pattern in part (b)(v). Also $F_i^{-1}=E{E}_i^{-1}E$, as direct multiplication verifies. Hence every $F_i^{-1}$ is lower triangular. Their product is lower triangular by the allowed closure property, proving the required Pivoted LU decomposition with $L=F_1^{-1}\cdots F_5^{-1}$.

Explicitly,The multipliers in rows three and four of $L$ are permuted along with the rows of $M$; simply keeping the unpermuted elimination multipliers would give the wrong product. This is the mechanism in conjugating elimination matrices across a row permutation.

The diagonalization gives $A=SD{S}^{-1}$ and hence $A^{n-1}=S{D}^{n-1}{S}^{-1}$: the intervening factors $S^{-1}S$ cancel. The diagonal matrix power has entries ${\alpha }^{n-1},{\beta }^{n-1}$. Since $1-\beta =\alpha$ and $\alpha -1=-\beta$,Reading the second component gives the Binet formulaThe derivation includes $n=1$, and the expression gives $U_1=U_2=1$, consistent with the specified indexing.

The initial vector records both initial Fibonacci numbers: $V_1=(U_2,U_1{)}^{\top }=(1,1{)}^{\top }$. For $n\ge 2$,Conversely this vector recurrence forces the second component to be the preceding first component and forces their sum to be the next first component, recovering the scalar recurrence and its initial conditions. Iterating, or using mathematical induction, yields $V_n=A^{n-1}(1,1{)}^{\top }$ for every $n\ge 1$.

The characteristic polynomial is $\det (\lambda I-A)={\lambda }^2-\lambda -1$. Its distinct eigenvalues are $\alpha =(1+\sqrt{5})/2$ and $\beta =(1-\sqrt{5})/2$. For either eigenvalue $\lambda$, the vector $(\lambda ,1{)}^{\top }$ is an eigenvector, since $A(\lambda ,1{)}^{\top }=(\lambda +1,\lambda {)}^{\top }=\lambda (\lambda ,1{)}^{\top }$.

ChooseThe determinant of $S$ is $\alpha -\beta =\sqrt{5}\ne 0$, soIts columns being eigenvectors gives $AS=SD$, and left multiplication by $S^{-1}$ proves $S^{-1}AS=D$.