Draw the unit square with vertices O=(0,0), P=(1,0), Q=(1,1), R=(0,1). We use column vectors: a point’s x coordinate sits above its y coordinate. The matrix A=[[1,2],[0,1]] sends (x,y) to (x+2y,y). Compute each vertex before drawing the new edges. You obtain (0,0),(1,0),(3,1),(2,1), a parallelogram.
This operation is a shear: horizontal position changes by twice the height. The bottom edge stays fixed. The top moves two units right, but the vertical distance between the horizontal edges remains one. The parallelogram therefore has area one. A slanted edge does not automatically mean a larger area: base times perpendicular height, not sloping side length, gives this area.
Predict
The first column of A is (1,0), the image of e₁=(1,0). The second column is (2,1), the image of e₂=(0,1). Predict A(−1,2) by taking −1 times the first image plus twice the second. The result is (3,2). Row arithmetic gives the same answer: (−1+2·2,2).
Now let B=[[0,−1],[1,0]], a quarter-turn counterclockwise. Predict whether rotating after the shear gives the same result as shearing after the rotation. Use the point (1,1), and write the intermediate point. A picture of only the final endpoints can conceal which operation was applied first.
Experiment
Keep the shear as A, the quarter-turn as B and input (1,1). Reveal both traces. A then B gives (1,1)→(3,1)→(−1,3). B then A gives (1,1)→(−1,1)→(1,1). The disagreement is evidence that matrix multiplication is not commutative. BA means B acting on the result of A, because the rightmost action reaches the input first.
Build a matrix witness
Column-vector convention: A acts first in BA. Predict the square image and both composition endpoints before revealing. Presets have integer entries from −2 to 2; this is a bounded explorer, not a general matrix editor.
Manual, temporary calculations only. Edits clear results. Nothing is saved or counted as mastery.
The displayed square, image and both traces always share identical x/y scales. When a changed input needs a larger window, the whole comparison expands together. Connecting trace segments record algebraic intermediate states; they do not claim a physical object travels along those lines. The target reconstruction is a separate question, not another point in that drawing.
Choose the Collapse preset A=[[1,1],[2,2]]. Try target (3,6), then (3,5). Predict which has predecessors and whether one predecessor would establish uniqueness. Finally select Zero and target (0,0). Compare a collapsed line with a collapsed point rather than treating every determinant-zero case as the same geometry.
Understand
Every point decomposes as (x,y)=x e₁+y e₂. A linear action preserves addition and scalar multiplication, so A(x,y)=x Ae₁+y Ae₂. This proves why the two basis images determine the entire map and why they form its columns. It also explains why all linear matrices fix the origin. A nonzero translation requires an additional vector; a plain 2×2 matrix cannot perform it.
For the worked shear and rotation, direct multiplication gives BA=[[0,−1],[1,2]] and AB=[[2,−1],[1,0]]. Substituting (1,1) verifies the two endpoints above. A single point can sometimes agree even when matrices differ, so equality at one point would not prove equality of actions. Agreement on both standard basis vectors would: their columns would then agree.
For a general matrix [[a,b],[c,d]], the determinant is ad−bc. The two column vectors span the image parallelogram; its area is |ad−bc|, as the coordinate base-height/dissection formula shows. The sign records whether orientation is preserved or reversed. A reflection has negative determinant but positive area. Zero determinant makes the columns dependent, so the square loses area and distinct inputs can have the same image.
When ad−bc is nonzero, multiplying by [[d,−b],[−c,a]]/(ad−bc) gives the identity: the diagonal numerators are ad−bc and the off-diagonal numerators cancel. This proves the inverse formula and uniqueness of reconstruction. For our shear the inverse is [[1,−2],[0,1]]; it subtracts twice the unchanged height from the horizontal coordinate, recovering every original vertex.
For the collapse preset every output is (x+y,2x+2y). Target (3,6) has the whole predecessor line x+y=3, while target (3,5) has none. Neither conclusion is an inverse matrix. For the zero matrix, target zero has every point as a predecessor; every nonzero target has none. The numerical engine uses bounded integer cross-products for these consistency decisions, not a visual guess from nearly overlapping lines.
Apply
On paper, change the shear to C=[[1,−1],[0,1]]. Construct the same square’s image, infer the inverse and verify a reconstructed target (2,3). Then test whether a map sending e₁ to (2,1) and e₂ to (−1,2) is this shear or a different linear action. These changed matrices are written practice, not additional widget presets.
Check the construction and reconstruction
The square images are (0,0),(1,0),(0,1),(−1,1). The inverse is [[1,1],[0,1]], so target (2,3) comes from (5,3); forward calculation gives (5−3,3)=(2,3). The supplied basis images instead define [[2,−1],[1,2]], a different matrix. Matching one image or one visual feature would not justify identifying them.
Retain an equal-scale drawing, a row-arithmetic trace, both operation orders and a full reconstruction classification. “No inverse” is incomplete if the question asks about a particular target: explain whether it has no predecessors or more than one, and exhibit the whole line or plane when appropriate.
Continue
For a multiplication recap, revisit matrix products and composition. For larger systems, try matrix encoding and verified inverses. These are optional depth routes; two-dimensional reasoning here needs only school algebra and coordinates. This Unit 4 extension is course enrichment, not AP examination content or evidence of independently assessed proficiency.
Apply · explain · revise
Practice and reasoning
Drafts save locally as you type. Reveal worked solutions when useful; the site remembers that exposure. Automatic matches check the requested answer, while written arguments require review against the rubric.
Construct the image of triangle O=(0,0), P=(2,0), Q=(0,1) under A=[[1,−1],[1,1]]. Use equal coordinate scales and show the row arithmetic for P+Q=(2,1). Compare triangle areas.
Structure your argument and preserve revisions
These optional planning fields save locally. Recording your response saves an immutable copy of the argument and these fields. Older records may contain only first/latest responses; missing intermediate revisions are not reconstructed.
Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
A linear map sends e₁ to (2,−1) and e₂ to (1,3). Recover its matrix using column vectors, and find the image of (−2,1). Explain why these two images determine every point.
Structure your argument and preserve revisions
These optional planning fields save locally. Recording your response saves an immutable copy of the argument and these fields. Older records may contain only first/latest responses; missing intermediate revisions are not reconstructed.
Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
For A=[[1,1],[0,1]], B=[[−1,0],[0,1]] and p=(2,−1), compute A then B and B then A, retaining each intermediate point. Find BA and AB.
Structure your argument and preserve revisions
These optional planning fields save locally. Recording your response saves an immutable copy of the argument and these fields. Older records may contain only first/latest responses; missing intermediate revisions are not reconstructed.
Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
For A=[[2,1],[1,1]], reconstruct the input that produces target (5,3). Derive an inverse and verify the target by forward multiplication.
Structure your argument and preserve revisions
These optional planning fields save locally. Recording your response saves an immutable copy of the argument and these fields. Older records may contain only first/latest responses; missing intermediate revisions are not reconstructed.
Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Let S=[[2,−1],[−2,1]]. Classify predecessors of (3,−3) and (3,−2); describe all predecessors in the possible case. Also classify target (0,0) under the zero matrix.
Structure your argument and preserve revisions
These optional planning fields save locally. Recording your response saves an immutable copy of the argument and these fields. Older records may contain only first/latest responses; missing intermediate revisions are not reconstructed.
Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
For D=[[−2,0],[0,1]], describe the image of the unit square, its orientation and its area. A learner proposes a 2×2 matrix for translation (x,y)↦(x+1,y−2). Refute the proposal for all possible matrices.
Structure your argument and preserve revisions
These optional planning fields save locally. Recording your response saves an immutable copy of the argument and these fields. Older records may contain only first/latest responses; missing intermediate revisions are not reconstructed.
Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Reflect and continue
Before marking this lesson complete, explain one key idea without the worked solution and identify a condition that matters. Marking complete records participation, not independent mastery.
