2.2 - Singular systems and three planes

2.2 - Singular systems and three planes

A zero determinant does not tell you whether equations contradict one another or leave a choice of solutions. You will reduce singular systems, distinguish these outcomes, and express a whole line or plane of common solutions using parameters.

Reducing a system

If an inverse does not exist, can the equations still have solutions? Compare x+y+z=4x+y+z=4, x−y+z=2x-y+z=2, 2x+2z=62x+2z=6. The third equation is the sum of the first two, including its constant. It supplies no new restriction.

Elimination makes this dependence visible. An augmented matrix places the constant column beside the coefficient matrix, separated by a vertical line. Each row represents one complete equation.

Equivalent row operations

You may swap two rows, multiply one row by a nonzero scalar, or add a multiple of one row to another. Apply the operation to the constants as well as the coefficients. Each operation is reversible, so it preserves the complete solution set.

The notation R2←R2−R1R_2\leftarrow R_2-R_1 means replace row 2 by row 2 minus row 1. Not in the formula booklet: learn it.

A zero row must be read as an equation. Its right-hand entry makes the decisive difference.

Worked example 1: a redundant equation

Question. Reduce x+y+z=4x+y+z=4, x−y+z=2x-y+z=2, 2x+2z=62x+2z=6, and find its solutions.

Plan. Subtract row 1 from row 2, then subtract twice row 1 from row 3.

(11141−1122026)⟶(11140−20−20−20−2).\left(\begin{array}{ccc|c}1&1&1&4\\1&-1&1&2\\2&0&2&6\end{array}\right) \longrightarrow \left(\begin{array}{ccc|c}1&1&1&4\\0&-2&0&-2\\0&-2&0&-2\end{array}\right).

Now R3←R3−R2R_3\leftarrow R_3-R_2 gives 0=00=0. Row 2 gives y=1y=1, and row 1 gives x+z=3x+z=3. Choose z=tz=t.

Answer. (x,y,z)=(3−t,1,t)(x,y,z)=(3-t,1,t), for every real tt; there are infinitely many solutions.

Check. Substitution gives (3−t)+1+t=4(3-t)+1+t=4, (3−t)−1+t=2(3-t)-1+t=2, and 2(3−t)+2t=62(3-t)+2t=6 for every tt. All three original equations hold.

Change only the third constant from 66 to 77. The coefficient matrix remains singular, but agreement of the constants can change.

Worked example 2: a contradiction after elimination

Question. Reduce x+y+z=4x+y+z=4, x−y+z=2x-y+z=2, 2x+2z=72x+2z=7, and state the number of solutions.

Plan. Apply the same reversible operations as in Worked example 1.

(11141−1122027)⟶(11140−20−20−20−1)⟶(11140−20−20001).\left(\begin{array}{ccc|c}1&1&1&4\\1&-1&1&2\\2&0&2&7\end{array}\right) \longrightarrow \left(\begin{array}{ccc|c}1&1&1&4\\0&-2&0&-2\\0&-2&0&-1\end{array}\right) \longrightarrow \left(\begin{array}{ccc|c}1&1&1&4\\0&-2&0&-2\\0&0&0&1\end{array}\right).

The last row says 0=10=1, which no coordinates can satisfy.

Answer. There is no solution.

Check. Adding the first two original equations requires 2x+2z=62x+2z=6. The third requires the same expression to equal 77, independently confirming the contradiction.

You need not always reduce to a diagonal array. Stop once the remaining equations show a contradiction or show which coordinates are free.

Watch out

The row (0,0,0∣0)(0,0,0\mid0) adds no condition. The row (0,0,0∣c)(0,0,0\mid c), with c≠0c\ne0, makes the system impossible.

Track the constants through this elimination.

Consistency and inconsistency

A system is consistent if it has at least one solution and inconsistent if it has none. A nonzero determinant guarantees consistency and uniqueness. A zero determinant requires you to inspect the constants as well as the coefficient rows.

For three unknowns with a singular coefficient matrix, a consistent system has infinitely many solutions: elimination leaves at least one free variable. A singular system cannot have exactly one solution.

[DIAGRAM: asset_name: Three different intersections of planes; asset_slug: cam9231-three-plane-cases; caption: Three planes can meet at one point, share a line, or have no common point; description: Three exact coordinate constructions compare x equals 0 and y equals 0 with a third plane. With z equals 0 the common set is the origin. With x plus y equals 0 the common set is the whole z-axis. With x plus y equals 1 the three planes have no common point although each pair intersects in a line. The planes are pale blue, sage and lavender, with the common point or line highlighted where present.]
Three planes can meet at one point, share a line, or have no common point

Each panel concerns the intersection of all three planes. Three pairwise intersections need not share a point.

Worked example 3: a singular case that stays consistent

Question. Investigate the consistency of x+y=3x+y=3, y+z=4y+z=4, x+kz=−1x+kz=-1 for all real kk.

Plan. Locate the singular value, then inspect it without dividing by a possibly zero expression.

The coefficient determinant is

det⁡(11001110k)=k+1.\det\begin{pmatrix}1&1&0\\0&1&1\\1&0&k\end{pmatrix}=k+1.

For k≠−1k\ne-1, there is a unique solution. Eliminate using y=4−zy=4-z and x=z−1x=z-1. The third equation gives

(z−1)+kz=−1⇒(k+1)z=0.(z-1)+kz=-1\quad\Rightarrow\quad(k+1)z=0.

For k≠−1k\ne-1, this gives z=0z=0. For k=−1k=-1, it is 0=00=0, leaving zz free.

Answer. The system is consistent for every kk. Its unique solution is (−1,4,0)(-1,4,0) for k≠−1k\ne-1; at k=−1k=-1, its solutions are (t−1,4−t,t)(t-1,4-t,t) for real tt.

Check. The point (−1,4,0)(-1,4,0) satisfies the original equations for every kk, immediately confirming consistency. Substituting the whole family at k=−1k=-1 verifies the free parameter.

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