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A triangular matrix has determinant equal to the product of its diagonal entries. Since $L$ is a unit lower triangular matrix, $\det L=1$. ThusEquivalently, the elimination used only row additions, which leave the determinant unchanged; there were no row swaps or row scalings to account for.

Define the characteristic polynomial using $p(\lambda )=\det (\lambda I-P)$, where $I$ is the identity matrix. Expanding the determinant along the first row givesThus the characteristic equation isThe convention $\det (P-\lambda I)$ changes the sign of this cubic polynomial but gives the same roots.

The elementary row operations used above add multiples of other rows, so preserve the determinant. The determinant of an upper triangular matrix is the product of its diagonal entries. HenceEquivalently, $\det L=1$ and multiplicativity of the determinant gives $\det A=\det L\det U=6$.

Source gap / TODO: the original printed page 4, confirmed visually, omits the unknown vector. It literally equates a $3\times 3$ matrix with a $3\times 1$ vector, so it does not define a well-formed system of linear equations. The following answer is conditional on the usual intended interpretation $A\mathbf{x}=(1,k^2,k{)}^T$; the missing factor is not present in the source.

Under that interpretation, the rows satisfy $r_1-2r_2+r_3=0$. Applying this linear relation to the right-hand side gives the necessary consistency conditionIt is sufficient too: the upper-left $2\times 2$ minor has determinant $4-9=-5\ne 0$, so the first two rows are linearly independent. The third row is $2r_2-r_1$, and imposes precisely the condition already found. ThereforeFor completeness, setting $x_3=t$ givesand direct substitution satisfies all three linear equations exactly for those two values of $k$. An unconditional answer to the malformed printed equality awaits a source correction.

For $1=(1,1,1{)}^T$, the entries of $P^{T}1$ are the column sums of $P$. Henceso $1$ is an eigenvalue of $P^T$, with eigenvector $1$.

The determinant is unchanged by matrix transposition, soThus $P-I$ has a nontrivial null space, givingThe same argument with $\lambda I$ shows that a matrix and its transpose have the same characteristic polynomial. It does not say that they have the same eigenvectors: here $P1=(0.9,1.2,0.9{)}^T\ne 1$.

Let $c_1,\dots ,c_4$ be the original columns. The first three are linearly independent: restricting them to rows 2, 3 and 4 gives the triangular matrixwhose determinant is one. Also $c_4=c_1-c_2+c_3$, so a basis for the column space isThese are columns of the original matrix. Taking columns of its reduced row echelon form would generally give the wrong column space.

The determinant of the Jacobian matrix is $s^2+t^2$. Wherever $(s,t)\ne (0,0)$, the inverse function theorem permits a local inverse of the parabolic coordinates transformation, andReading off its entries givesThese are local formulas on a chosen branch: $(s,t)$ and $(-s,-t)$ have the same image, and at the origin the Jacobian matrix is singular, so this inverse-derivative formula does not apply there.

Denote the columns by $c_1,\dots ,c_4$. The first two are linearly independent, because the top two coordinates have determinant $3\cdot 2-(-1)\cdot 1=7$. Direct calculation givesThus a column-space basis is $\{(3,1,9{)}^T,(-1,2,4{)}^T\}$, andFor the last description, the third row equals twice the first row plus three times the second, so every column lies in that plane. Since the two independent columns already span a two-dimensional vector subspace, they fill the plane.

The target vector satisfies $5=2(-2)+3(3)$, so it is in the column space. An especially simple linear combination isIn terms of the displayed basis, the same vector is $-c_1/7+11c_2/7$. The column space itself is unique; the freedom is in choosing its basis.

For square matrices of the same size,For an upper or lower triangular matrix $T$, its determinant is the product of its diagonal entries:These formulas include singular matrices and zero diagonal entries.

Let $e_i$ be standard coordinate columns. The required elementary matrix isIt is the identity matrix with the extra entry $\alpha$ in position $(j,i)$. Left matrix multiplication implements the row operation: all other rows are unchanged. Since $j>i$, $E$ is a lower triangular matrix with every diagonal entry one. Part (a) therefore gives $\det (E)=1$ andHence adding a multiple of one row to another leaves the determinant unchanged, even if $A$ is singular.

Write the determinant in its Leibniz formula for determinants, where $S_n$ is the set of permutations of $\{1,\dots ,n\}$:The permutation sign is justified by the given row-interchange property: a permutation matrix is obtained from the identity matrix by row interchanges, and every interchange changes its determinant by a factor $-1$. Reversing a sequence of interchanges constructs the inverse permutation with the same number of interchanges. Consequently $\mathrm{sgn}({\sigma }^{-1})=\mathrm{sgn}(\sigma )$. The Leibniz formula for determinants can itself be obtained by expanding the rows multilinearly in the standard basis: repeated selected coordinate rows contribute zero, and the remaining terms are precisely these signed permutation matrices.

Since $(A^T{)}_{ij}=a_{ji}$, the same formula givesReindex the product with $j=\sigma (i)$ and reindex the sum with $\tau ={\sigma }^{-1}$. Both reindexings are bijections, soThus $\det (A)=\det (A^T)$ for every real square matrix, including singular matrices. This proof does not assume the analogous column-operation rules in order to establish the transpose identity.

For a skew-symmetric matrix, $A^T=-A$. By parts (c) and (d),If $n$ is odd, $\det (A)=-\det (A)$, hence $2\det (A)=0$ over $\mathbb{R}$. Therefore $\det (A)=0$, so $A$ is a singular matrix. The conclusion depends on odd dimension: in dimension two, $\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$ is skew-symmetric and has determinant one.

For $A$, the triangular matrix formula gives the characteristic polynomialThe three eigenvalues are distinct, so part (d) already proves the existence of an eigenvector basis. Explicit choices, ordered by eigenvalues $1,3,2$, areTheir column matrix has determinant one, and direct multiplication gives respectively one, three and two times the indicated vector.

For $B$, expansion along the last row givesAgain the three eigenvalues are distinct. An explicit eigenvector basis, ordered by eigenvalues $4,2,3$, isIts column matrix has determinant $-2\ne 0$. In particular, $B$'s diagonal entries are not its eigenvalues: it is not a triangular matrix.

Finally $C=I_3$, so every nonzero vector is an eigenvector with eigenvalue one. Choose the standard basis $e_1,e_2,e_3$. These are linearly independent even though their eigenvalues coincide. Distinct eigenvalues are sufficient for an eigenvector basis, and are not necessary.

The determinant of the unit lower triangular matrix $L$ is 1. There is one row swap, so $\det P=-1$, and the upper triangular matrix has determinant $1\cdot (-2)\cdot (3/2)\cdot 2=-6$. Therefore

Accumulate the product of the $2\times 2$ matrices from left to right, which matches forward linked list traversal.The four-value assignment is simultaneous, as permitted by the source. With $M(x)=\left(\begin{array}{cc}x& 1\\ 1& 0\end{array}\right)$, the loop invariant after reading the prefix $x_0,\dots ,x_{t-1}$ iswith the empty product initialized to the identity matrix. Multiplication on the right by $M(x)$ produces exactly the four assignments above. The associativity of matrix multiplication allows this left-to-right accumulation even though the recursive conversion groups operations from the right.

At completion, the first column is the numerator/denominator pair for the whole continued fraction: the last factor sends $(1,0{)}^T$ to $(x_n,1{)}^T$, and each earlier factor applies the recurrence from part (d). The vector $(1,0{)}^T$ is an algebraic device, not a finite input ratio requiring division by zero. For a nonempty list the resulting $c$ is positive; for a singleton it is $1$. The product has determinant $(-1{)}^{n+1}$, so its first-column entries are coprime. Return $(a,c)$ in linear node-processing time and $O(1)$ auxiliary space, without changing the list.

Use multiplicativity of the determinant and the determinant criterion for an invertible matrix:Neither factor can therefore be zero, so both matrices are nonsingular. The determinant of a triangular matrix is the product of its diagonal entries:Each product is nonzero, forcing every diagonal factor to be nonzero. These explicitly stated determinant properties justify both conclusions.

Let column $j$ of $A^{-1}$ be $x$, so $Ax=e_j$. Every diagonal entry of the invertible lower triangular matrix $A$ is nonzero, since its determinant is the product of those entries. In row one through row $j-1$, the right-hand side is zero. Forward substitution first gives $x_1=0$, then $x_2=0$, and successively $x_i=0$ for all $i<j$. These are precisely the entries above the diagonal in column $j$ of $A^{-1}$. Every column satisfies this, proving that $A^{-1}$ is lower triangular. This is triangularity of an inverse matrix.

A triangular matrix has determinant equal to the product of its diagonal entries, so $\det L=1$ and $\det U=3\cdot 4\cdot (-11)=-132$. A permutation matrix exchanging two rows has determinant $-1$. Using multiplicativity of the determinant in $PA=LU$ yields $(-1)\det A=-132$. Hence $\det A=132$.

Subtract the first vertex from the other two, giving side vectors $(2,1)$ and $(3,4)$. Their determinant is $2\cdot 4-3\cdot 1=5$, the signed area of the corresponding parallelogram. The triangle occupies half that area, hence its area is $5/2$. Taking the absolute value would remove any sign change caused by reversing the vertex order. This is the determinant formula for triangle area.

Let $c_1=(1,1,1{)}^T$ and $c_2=(2,-2,6{)}^T$. The remaining columns are $3c_1$ and $2c_2$, so these two vectors span the column space. They are linearly independent: the determinant of their first two coordinates is $-4\ne 0$. Therefore $\{(1,1,1{)}^T,(2,-2,6{)}^T\}$ is a basis of the range, and the matrix rank is two.

Equivalently the range is the plane $y_3=2y_1-y_2$: both displayed basis vectors satisfy this equation, and their two-dimensional span fills that plane.