Inverse product order: repair the undo sequence

Inverse product order determines whether an undo calculation returns the starting input. In this original exact-number exercise, two individually reversible stages turn the column (1, 0) into (2, 1). Applying their inverses in the same written order returns (3, -1), which is wrong. Reversing the written inverse factors restores (1, 0). We can identify the mistake without changing either forward stage.

Fix the two forward stages

Let A be the two-by-two matrix with rows 1, 1 and 0, 1. Let B have rows 1, 0 and 1, 1. All entries are stipulated exact real numbers. A column has two ordered entries, x and y. Matrix multiplication pairs each left row with a right column, multiplies corresponding entries and adds the results. Row breaks and coordinate positions remain part of the input.

A sends (x, y) to (x + y, y). B sends it to (x, x + y). Each output remains a two-entry column in the same real coordinate space. These rules are mathematical definitions, without measured quantities, recorded device output or a performance claim. In particular, an entry equal to zero is a supplied value rather than an empty worksheet position.

Use AB to mean A after B: the rightmost factor acts first. Thus B performs the first forward stage and A performs the last. Parentheses around a coordinate pair identify a column throughout this article. Write this convention beside the two matrices before evaluating a product; reading the written factors from left to right as actions would silently change the problem.

Calculate the unchanged forward product

Starting with (1, 0), B gives (1, 1). Applying A then gives (2, 1). Keep the intermediate column as well as the start and finish. The forward path is (1, 0), then (1, 1), then (2, 1). A proposed return rule must restore that original start from the stated finish, rather than choose a more convenient input.

Multiplying the arrays gives C = AB with rows 2, 1 and 1, 1. Its upper-left entry is 1 times 1 plus 1 times 1, giving 2. Its upper-right entry is 1 times 0 plus 1 times 1, giving 1. The second row of A selects the second row of B, supplying the remaining row 1, 1.

For an arbitrary column (x, y), B first gives (x, x + y), and A then gives (2x + y, x + y). That agrees with multiplication by C. This formula checks that the product represents the specified forward stages on every input. It also supplies a way to inspect the result independently of the particular starting column (1, 0).

Verify each individual inverse in both directions

Define Ai with rows 1, -1 and 0, 1. It sends (u, v) to (u - v, v). Applying it after A gives (x + y - y, y), or (x, y). Applying A after Ai gives (u - v + v, v), or (u, v). Both compositions return every original column, so Ai is a two-sided inverse of A.

Define Bi with rows 1, 0 and -1, 1. It sends (u, v) to (u, v - u). After B, this becomes (x, x + y - x), restoring (x, y). In the other direction, B applied after Bi gives (u, u + v - u), restoring (u, v). Thus Bi is a two-sided inverse of B on the same full real coordinate space.

The letters Ai and Bi name whole matrices, not reciprocals of their individual entries. The notation A to the power minus one and B to the power minus one may also name these inverses. Keep the actual return formulas beside that notation. Their two-sided identities have been proved here, rather than assumed because a proposed matrix happens to have the right dimensions.

Test the same-written-order proposal

The mistaken candidate is W = Ai Bi. Its first row is 2, -1 and its second row is -1, 1. For example, the upper-left entry is 1 times 1 plus negative 1 times negative 1, giving 2. This is a legitimate two-by-two product. Its failure will come from the order of its actions, rather than an undefined multiplication.

Apply W to the saved finish (2, 1). It returns (3, -1): the first entry is 2 times 2 minus 1, and the second is negative 2 plus 1. That differs from the original (1, 0) in both positions. One failed original input is sufficient to reject W as a rule that restores every column after C.

Its staged calculation explains the result. In Ai Bi, Bi acts first. Applied to (2, 1), Bi gives (2, -1). Ai then subtracts negative 1 from 2 and retains negative 1, producing (3, -1). The first undo action addressed B even though A had acted last in the forward path. Keep this failed path visible when correcting the formula.

Reverse the inverse factors

The corrected candidate is R = Bi Ai, with rows 1, -1 and -1, 2. Ai now acts first on the finish (2, 1), giving (1, 1). Bi follows and gives (1, 0). These intermediate columns retrace the original forward path in reverse. We have restored the actual saved input while keeping A, B and C unchanged.

Check more than that one successful column. Multiplying C by R gives rows 1, 0 and 0, 1: its entries are 2 minus 1, negative 2 plus 2, 1 minus 1, and negative 1 plus 2. This is the two-by-two identity I. It returns any real column unchanged, so C followed after R restores every column in C's output space.

The reverse product RC also equals I. Its first row gives 2 minus 1 and 1 minus 1, or 1, 0. Its second row gives negative 2 plus 2 and negative 1 plus 2, or 0, 1. Both identity checks belong in the record. R is a two-sided inverse of the complete forward product, not merely a candidate that happened to decode (2, 1).

Explain why reversal works beyond these entries

Suppose A and B are any two same-size square matrices with two-sided inverses Ai and Bi. Associativity allows parentheses to move while preserving the written order of factors. Consequently (AB)(Bi Ai) can be grouped as A(B Bi)Ai. Since B Bi equals I, the expression becomes A Ai, which also equals I. No interchange of neighboring factors occurred in that calculation.

In the other direction, (Bi Ai)(AB) can be grouped as Bi(Ai A)B. The middle product Ai A equals I, leaving Bi B = I. Together these equations establish that Bi Ai is an inverse of AB in both directions. The argument depends on the stated two-sided inverse identities and associativity, without requiring A and B to commute.

The action order is consistent with the formula. The forward product AB applies B and then A. Its inverse Bi Ai applies Ai and then Bi. Thus the last forward action is undone first. Distinguish the written order of factors from the order in which a column encounters them; under our convention, each rightmost factor acts before the factor to its left.

Keep successful shortcuts separate

There are inputs for which the same-written-order candidate works. Take diagonal matrices with diagonal entries 2, 3 and 5, 7, respectively. Their inverses have reciprocal diagonal entries, and multiplying those diagonal inverses in either order gives the same matrix. This changed control does not validate W for the original shear matrices, whose entries and failed decoding remain fixed.

Another exact control uses A(a) with rows 1, a and 0, 1, and B(b) with rows 1, 0 and b, 1. Their individual inverses replace a or b with its negative. If a or b is zero, one forward factor is I, so either written inverse order succeeds. That restricted result explains why checking only identity factors could miss the mistake.

For these shear controls, the two forward products agree exactly when ab = 0. Their diagonal entries are 1 + ab, 1 in one order and 1, 1 + ab in the other, with matching off-diagonal entries. Exact enumeration checked all 121 ordered choices with a and b from negative five through five. Reverse-order inversion always passed both identities; the same-order candidate passed precisely when ab was zero.

A separate control evaluated the original C and R on 81 rational columns. The first coordinate ranged over x divided by three and the second over y divided by five, with integers x and y from negative four through four. Every column returned unchanged after the forward product and corrected inverse. These finite checks detect arithmetic errors within their stated range; the two associative identity arguments establish the broader result.

Distinguish the neighboring inverse and expansion problems

The one-sided inverse example uses rectangular maps that discard or append a third coordinate. One composition restores every pair while the reverse fails on triples with a nonzero third coordinate. Here both factors already have two-sided inverses on the same space; the problem is ordering their return stages.

The matrix square expansion repair retains both ordered cross products when squaring a sum. Its correction concerns omitted or improperly combined summands. This exercise has no sum to expand. It keeps the forward product fixed and checks which composition of individually valid inverses restores every original column.

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Undo the final forward stage before undoing the first.

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Save the complete correction

Preserve the original A and B, their product C, and both individual two-sided inverse checks. Save W and its failed path from (2, 1) to (3, -1), followed by R and both identity products. That record identifies the changed operation precisely: reverse the inverse factors while retaining the forward problem and its original input.