The solution set is the plane through the origin with normal vector $(1,1,1{)}^T$. The three original planes coincide, so their common intersection is this entire plane.

All three linear equations are multiples of the first, leaving the single constraint $x+y+z=0$. Choose two real numbers $s,t$ freely:A basis for this two-dimensional linear subspace is $(1,0,-1{)}^T,(0,1,-1{)}^T$.

The original PDF places complement bars on the second $A$ in each parenthesis and on the final $A$. Thus the first parenthesis is the empty set, whereas the second is the whole sample space. Their union with any further subset is still the whole space:The bars are missing from the local OCR, so using its unbarred expression would give the wrong result.

Since $A$ is a subset of the sample space $\Omega$, taking its intersection with $\Omega$ leaves it unchanged:

For a differentiable objective $f$ and regular constraint $g=r$, a constrained stationary point satisfieswhere $\lambda$ is a Lagrange multiplier. Solve these equations for both the coordinates and $\lambda$.

Geometrically, the gradient $\nabla g$ is normal to the constraint level set when $\nabla g\ne 0$. Every feasible infinitesimal displacement $\mathbf{v}$ is tangent to that level set, so $\nabla g\cdot \mathbf{v}=0$. At a constrained stationary point, the directional change $\nabla f\cdot \mathbf{v}$ must also vanish for every such tangent displacement. Thus $\nabla f$ is normal to the same tangent plane and is a scalar multiple of $\nabla g$. This is the geometric basis of the Lagrange multiplier method.

For a second constraint $h=s$, with linearly independent vectors $\nabla g$ and $\nabla h$, the feasible tangent directions are perpendicular to both. Their normal space is the linear span of these two gradients, so introduce a second Lagrange multiplier:These equations find candidates; deciding whether they are maxima, minima or saddle points requires further analysis. Constraint points where the required gradient independence fails must be examined separately.

L'Hôpital's rule applies to a quotient $f(x)/g(x)$ of differentiable functions near a limiting point, excluding that point itself, provided $g^{\prime }(x)\ne 0$ there. If the quotient has an indeterminate form $0/0$ or $\infty /\infty$, and the limit of $f^{\prime }(x)/g^{\prime }(x)$ exists as a finite or infinite value $L$, thenThe rule also applies to one-sided limits and to limits at infinity, under the corresponding hypotheses. Existence of the derivative-ratio limit and the indeterminate-form hypotheses are required; an arbitrary quotient cannot be replaced by a quotient of derivatives.

For a series $\sum u_k$ with eventually nonzero terms, supposeexists, allowing $L=+\infty$. The ratio test says that if $L<1$, the series converges absolutely; if $L>1$, it diverges; if $L=1$, the test is inconclusive. In the divergent case the terms fail to tend to zero. For example, the P-series with exponents $1$ and $2$ both have ratio limit $1$ but have different convergence behaviour.

Let $p(t)=\det (tI-A_{\alpha })$ be the real characteristic polynomial. It is a monic cubic, so $p(t)\to +\infty$ as $t\to +\infty$. At zero,The intermediate value theorem therefore gives some $t_0>0$ with $p(t_0)=0$. A root of the characteristic polynomial is an eigenvalue, hence $A_{\alpha }$ has a positive real eigenvalue for every real $\alpha$. This uses the general fact that an odd-dimensional real matrix with positive determinant has a positive eigenvalue.

Completing the square in the determinant givesfor every real $\alpha$. A square matrix is invertible exactly when its determinant is nonzero, so $A_{\alpha }$ is nonsingular for every real $\alpha$.

Expand the determinant along the first row:The cubic terms cancel in the expansion.

Substituting $\beta =1$ and $\gamma =1-\alpha$ into the matrix givesIts row sums are $4,4,6$, checking that $A_{\alpha }\mathbf{c}=\mathbf{b}$ for every real $\alpha$.

The matrix multiplication $A\mathbf{c}$ adds each row. Equating it to $\mathbf{b}$ gives the linear systemThe last equation is the sum of the first two, so it supplies no further restriction. Subtracting the first from the second gives $2\beta =2$, hence $\beta =1$, and then $\gamma =1-\alpha$. Thus all parameter choices are