Encode and verify a matrix solution

Try

Three balances give x+y=4, y+z=6 and x+z=6. Adding the first and third and subtracting the second leaves 2x=4, so x=2. Then y=2 and z=4. Check: the three sums are 4, 6 and 6. This small calculation is the reference case; it is separate from the live tool’s starting values.

A matrix records the same equations compactly. Choose the unknown order (x,y,z). The coefficient rows are (1,1,0), (0,1,1), (1,0,1), and the right-side column is (4,6,6). A zero records an absent variable; it must not be omitted and shift the remaining columns.

Predict

Write this as Ax=b. Here A is a 3×3 coefficient matrix, x denotes the three-entry unknown column in this notation, and b is the right-side column. The scalar x in the first equation is the first entry of that column; keep the surrounding notation clear.

An inverse B=A⁻¹ must satisfy AB=BA=I, where I has ones on its main diagonal and zeros elsewhere. Predict the consequence of multiplying Ax=b on the left by B. It gives x=Bb. Matrix multiplication follows row-by-column rules; changing the variable order without changing the coefficient columns changes the problem.

Experiment

Use the exact-fraction tool. Its first preset has a different right side from the worked example, and editable entries have bounded integer menus. Predict whether the current system has one solution before calculating. Then inspect the inverse, determinant, solution and equation residuals. Changes are temporary exploration; retain the reasoning in the practice answers if you want a reviewable record.

Encode, invert, then verify a three-variable system

Write the unknowns in the order x, y, z. In Ax=b, the x in the matrix notation denotes the column vector of all three unknowns. Each row below is one equation, with its right side in the last column. Predict whether the equations determine one solution before calculating. The tool uses exact fractions for these small integer inputs.

Starting systemUnique solution with fractional inverseA zero first entry is not singularityDependent equations, consistent dataDependent equations, conflicting dataLoad starting systemReset matrix

Ax=b: A has three coefficient columns; b is the right-side column.

Equation	x coefficient	y coefficient	z coefficient	Right side b
1	-3-2-10123	-3-2-10123	-3-2-10123	-6-5-4-3-2-10123456
2	-3-2-10123	-3-2-10123	-3-2-10123	-6-5-4-3-2-10123456
3	-3-2-10123	-3-2-10123	-3-2-10123	-6-5-4-3-2-10123456

1. 1x + (1)y + (0)z = 3
2. 0x + (1)y + (1)z = 5
3. 1x + (0)y + (1)z = 4

Calculate and check

Results are hidden until you calculate. Editing any coefficient or right-side entry clears the previous result.

Use these outputs as calculation evidence, then explain your encoding and verification in the lesson task. This bounded tool handles 3×3 integer matrices only; it is not a general numerical solver. Inputs and results are temporary and never count as mastery.

Change one row to be dependent on another, first with a compatible right side and then with an incompatible one. Compare the reported classifications. A zero determinant means an inverse is unavailable; it does not decide by itself whether the system has no solutions or infinitely many.

Understand

For the worked matrix, an inverse is one half of the array with rows (1,−1,1), (1,1,−1), (−1,1,1). Verify it by multiplication. The first row of A selects the first two rows of B, whose sum is (1,0,0). The second selects rows two and three, producing (0,1,0). The third selects rows one and three, producing (0,0,1). Thus AB=I. Checking BA in the same way also gives I.

Computation	First coordinate	Second coordinate	Third coordinate
Bb	(4−6+6)/2=2	(4+6−6)/2=2	(−4+6+6)/2=4
A(Bb)	2+2=4	2+4=6	2+4=6
A(Bb)−b	0	0	0

Why does this establish a unique solution? Any solution must satisfy B(Ax)=Bb, hence x=Bb because BA=I. Conversely the candidate Bb satisfies A(Bb)=b because AB=I. These are necessity and sufficiency, so they establish both existence and uniqueness.

A singular system behaves differently. If one equation is twice another and its right side is also doubled, that row adds no new restriction. If its right side differs from the doubled value, the equations contradict. A tool may describe this using rank, the number of independent rows after reversible elimination; the concrete row comparison already explains the distinction without a prior linear-algebra course.

Apply

The fresh matrix task stays within the tool’s input limits and asks for an independent product check, not only a copied screen. Record the variable order, entered A and b, inverse and candidate solution. Multiply at least one complete row of AB and substitute the candidate into every original equation. A residual of zero means those equations are satisfied; it does not establish that your original contextual model was appropriate.

Practice also supplies a plausible but wrong inverse. A single failed identity entry disproves it. Conversely, one successful right-side test is not enough to prove an inverse: a wrong matrix can agree on one special vector. The identity check examines the operation itself.

Elimination remains useful when no inverse exists. It can expose a contradiction or identify free variables, while an “inverse unavailable” message alone cannot describe all solutions.

Continue

Matrix encoding connects school systems to later linear maps and least-squares models. It is optional machinery for many small systems, not a reason to stop checking equations. Preserve the distinction between a numerical answer, an equation solution and a justified contextual decision.

For further study, ask why adding a multiple of one equation to another preserves the solution set. The operation can be undone by subtracting the same multiple. That reversibility is the common idea behind ordinary algebra, elimination and inverse verification; the compact notation does not replace it.