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Adding the three equations for $s,t$ gives the necessary solvability condition $a+b+c=3$. Conversely, subtraction of the first two and use of the third forceIf $a+b+c=3$, these values also satisfy the first two equations: for example,The analogous calculation gives $1-s+t=b$. Thus existence and uniqueness hold exactly under the stated compatibility condition:

For the three-variable system the matrix and data vector arePerform Gaussian elimination on the augmented matrix. The operations $R_2\leftarrow 5R_2-2R_1$ and $R_3\leftarrow 5R_3+R_1$ giveEach operation is an invertible row scaling followed by a row addition, so the solution set is preserved. Next $R_3\leftarrow 7R_3+R_2$ givesBack substitution now yieldsand thenAll three pivots are nonzero. Thus the solution exists uniquely for every $s,t$, the matrix rank is $3$, and the kernel of a linear map represented by $A$ is $\{0\}$. The general rank-nullity theorem for a matrix with $m$ columns ishere $m=3$.

Finally, reverse the viewpoint and regard $x,y,z$ as prescribed. The necessary and sufficient condition already proved applies with $a=5x+2y-z$, $b=2x+5y-z$, $c=-x-y+8z$. Their sum is $6(x+y+z)$. Consequently $s,t$ can be recovered exactly whenWhen this holds, their unique values are $s=\frac{3}{2}(x-y)$ and $t=\frac{1}{2}(1+x+y-8z)$. The allowable triples form an affine plane, not all of ${\mathbb{R}}^3$.

The dual vector space is $V^{\ast }={\mathrm{Hom}}_{\mathbb{R}}(V,\mathbb{R})$, the vector space of all linear functionals on $V$. For a vector subspace $U\le V$, its annihilator of a vector subspace is

Given a basis $x_1,\dots ,x_n$, define $x_i^{\ast }$ by $x_i^{\ast }(x_j)={\delta }_{ij}$ and extend linearly. If ${\sum }_i{a}_i{x}_i^{\ast }=0$, evaluation at $x_j$ gives $a_j=0$, so the $x_i^{\ast }$ are linearly independent. Every $\varphi \in V^{\ast }$ satisfiesso they also span $V^{\ast }$ and therefore form its dual basis.

Write $\varphi =a{x}_1^{\ast }+b{x}_2^{\ast }+c{x}_3^{\ast }+d{x}_4^{\ast }$. The condition $\varphi \in U^0$ isGaussian elimination gives $b=-2c-3d$ and $a=c+2d$. Hence

Gaussian elimination without pivoting givesThus $\det A=3(\lambda -20)$ and the solution is unique exactly when $\lambda \ne 20$.

For $\lambda =20$, forward substitution in a triangular system in $Ly=b$ gives $y=(1,1,3,\mu -10{)}^T$. Consistency of $Ux=y$ requires $\boxed{\mu =10}$. backward substitution in a triangular system then gives

Gaussian elimination without row exchanges gives the LU decompositionBecause $L$ has unit diagonal and $U$ is upper triangular,

Solving the two triangular systems givesand

Gaussian elimination gives the LU decompositionThe prescribed diagonal entries of $L$ are visible, and direct multiplication gives $LU=A$.

Apply symmetric Gaussian elimination without row exchanges. At step $k$, let $d_k$ be the leading diagonal entry of the remaining symmetric Schur complement. If $d_k\le 0$, stop and report that $A$ is not positive definite. If $d_k>0$, use it to eliminate the rest of its row and column. If all steps succeed, this constructs an LDL decompositionwhere $L$ is unit lower triangular and $D=\mathrm{diag}(d_1,\dots ,d_n)$ has positive diagonal.

The test is correct from first principles. If all $d_k>0$, then for every nonzero $x$,because $L$ is invertible. Conversely, if $A$ is positive definite, its first pivot is $a_{11}>0$, and completing the square givesChoosing $s=-b^{T}y/a$ shows that the Schur complement is positive definite. Induction forces every pivot to be positive. This is also Sylvester's criterion.

At step $k$, updating the remaining matrix costs $O((n-k{)}^2)$ arithmetic operations. Hence the total iswhich proves the existence of the required algorithm.

Prepare $2^{-n/2}{\sum }_x|x⟩|0^n⟩$, query the oracle, and measure or discard the output register. For $p\ne 0$, the input register becomes the coset stateApply $H^{\otimes n}$, the quantum Fourier transform over $({\mathbb{Z}}_2{)}^n$. The two amplitudes interfere destructively unless the binary inner product satisfiesand every vector in this orthogonal subspace is sampled uniformly. Repeat until $n-1$ independent equations have been collected, then use Gaussian elimination over ${\mathbb{F}}_2$ to find their one-dimensional null space; its nonzero vector is $p$. This is Simon's algorithm. It uses $O(n)$ oracle queries with high probability and polynomial classical work. When $p=0^n$, the function is injective and the samples eventually span all of $({\mathbb{F}}_2{)}^n$, which distinguishes that case with arbitrarily high probability.

Choose finite generating sets $I=(g_1,\dots ,g_s)$ and $J=(f_1,\dots ,f_t)$, using the Hilbert basis theorem. The assumed inclusion says that each $f_j$ vanishes on $V_{\mathbb{C}}(I)$. By the Strong Hilbert Nullstellensatz, for every $j$ there is an exponent $N_j$ such that

The coefficients in an expression $f_j^{N_j}={\sum }_i{q}_{ij}{g}_i$ solve a finite system of linear equations with rational coefficients. Since it has a complex solution, Gaussian elimination gives a rational solution. Clearing the finitely many denominators produces a nonzero integer $D$ such thatfor every $j$.

For any prime $p\nmid D$, reduce these identities modulo $p$. At a common zero of ${\pi }_p(I)$ in the algebraic closure ${\overline{\mathbb{F}}}_p$, they give ${\pi }_p(f_j{)}^{N_j}=0$, hence ${\pi }_p(f_j)=0$, for all $j$. Thereforefor every prime except the finitely many divisors of $D$. This is the spreading out of an affine zero-set inclusion.

Simon's algorithm prepares hidden-subgroup coset states, applies the quantum Fourier transform over $({\mathbb{Z}}_2{)}^n$, and samples vectors $y$ satisfying $y\cdot p=0$ over ${\mathbb{F}}_2$. Repeated samples and Gaussian elimination recover $p$ with $O(n)$ oracle queries and polynomial computation.