Building the adjoint, and why it multiplies back to a scalar times the identity

What to assume they know

• Minors and cofactors of every entry of a square array
• That a line's entries paired with another line's cofactors always sum to zero
• That a line's entries paired with their own cofactors give the determinant
• The transpose of a matrix, from Chapter 3, and what it does to subscripts
• Multiplying two matrices, row into column
• The identity matrix and the zero matrix at orders two and three
• Multiplying a matrix by a number, entry by entry

What they should be able to do

• Build the adjoint of a square array in two explicit moves and say what each move does
• Say which subscripts end up where after the transpose, and why the order of the two moves cannot be swapped
• Compute the adjoint of a two-by-two both longhand and by the chapter's own two-arrow shortcut
• Compute the adjoint of a three-by-three from all nine cofactors
• State the theorem relating an array, its adjoint and its determinant, in both orders of multiplication
• Explain, entry by entry, why the product is diagonal and why every diagonal entry is the same number
• Verify the theorem on a worked three-by-three, including a case where the determinant is zero
• State how the determinant of the adjoint relates to the determinant of the original, and name the exponent
• Say what the theorem is being used for on the next page

Where it usually goes wrong

• "The adjoint is the array of cofactors." It is that array transposed. Skipping the flip gives the right answer only when the cofactor array happens to be symmetric, which is rare and which will not be the exam question. The chapter's own side-by-side display on Part I p. 88 exists to prevent exactly this.
• "The two-arrow shortcut works at order three too." The chapter states it only for order two, and it is false at order three. At order three you must compute nine cofactors and then transpose. This is the most expensive single error a student can carry out of this section.
• "Adjoint means the same as inverse." They differ by a factor of the determinant, and the adjoint exists even when the inverse does not. Exercise 4.4 Q3 has an adjoint and no inverse at all.
• "If the determinant is zero there is no adjoint." There is. The construction never divides by anything. What collapses is the product, which becomes the zero matrix — and that is the theorem behaving correctly, not failing.
• "The product being diagonal is a fluke of the example." It is forced, entry by entry, by the matched and mismatched sums. Show the general grid filling in before showing any numbers.
• "The diagonal entries could be different from each other." Every one of them is a matched sum along a different line, and all six matched sums of an order-three array give the same number. That is exactly what §4.2.3 spent two pages establishing.
• "Order matters when you multiply by the adjoint." The theorem is stated in both orders and both give the same thing, which is unusual enough to be worth a beat — matrix multiplication does not normally commute, and Chapter 3 spent a section saying so.
• "The determinant of the adjoint is the determinant." It is the determinant raised to one less than the order. At order three that is a square, and the exercise set has a multiple choice devoted to the point.

Questions to check understanding

• Build the adjoint of a given two-by-two, longhand, and check it against the shortcut
• Build the adjoint of a given three-by-three from nine cofactors — the form of Exercise 4.4 Q1 and Q2
• Verify the theorem on a given array in both orders of multiplication — the form of Exercise 4.4 Q3 and Q4
• Verify the theorem on an array whose determinant is zero, and say what the product is
• Predict the determinant of the adjoint of an array of stated order — the form of Exercise 4.4 Q17
• Explain, without numbers, why the product of an array with its adjoint has zeros off the diagonal
• Given a cofactor array, produce the adjoint, and say what would go wrong if the flip were skipped

Examples worth working on the board

Values marked verified are worked out here from the chapter's own printed data; neither answers file was opened, and this chapter prints no answers to its exercises.

• The section's own statement of intent (§4.5, Part I p. 87). Two sentences: the inverse was met in the previous chapter, and what is wanted now is the condition under which one exists; to get there, the adjoint has to be defined first. That is an unusually clear signposting for this book.
• Definition 3 (§4.5.1, Part I p. 87). Take a square array, replace every entry with that entry's cofactor, and transpose what you get: the result is the adjoint. Read off the printed page: the arrays here sit inside square brackets, not bars — this whole section is about matrices, and the only bars that appear are around the determinant on the right of the theorem. That switch of delimiter happens exactly at the §4.4 to §4.5 boundary and is worth marking.
• The transpose display (§4.5.1, Part I p. 88). The chapter writes the cofactor array and then its transpose side by side, so the subscript swap is visible: the cofactor carrying subscripts one-two moves out of the top row into the second row. Freeze on that. The commonest error in the whole section is building the cofactor array and forgetting to flip it, and the chapter's own display is the antidote.
• Example 12 (§4.5.1, Part I p. 88). A two-by-two with entries two and three on top, one and four below. Verified: its four cofactors are four, minus one, minus three and two, taken row by row, and the adjoint is those four transposed — four and minus three on top, minus one and two below.
• The two-by-two shortcut, and it is a real figure (§4.5.1, Part I p. 88). The chapter draws the general two-by-two inside brackets with two long rounded loops crossing over it, one enclosing the two entries on the main diagonal and one enclosing the other two, and two downward arrows beneath, labelled with a change-of-sign instruction under the off-diagonal loop and a swap instruction under the main-diagonal loop; the result is printed to the right. Read closely. This is one of only two pieces of line art in the whole chapter, and it is the most reusable thing in the section — but see Misconceptions, because the shortcut is stated only for order two and students reach for it at order three.
• Theorem 1 (§4.5.1, Part I p. 88). Multiplying a square array by its adjoint gives the same result in either order, and that result is the determinant multiplying the identity of the same order. The chapter announces the theorem as being given without proof and then immediately supplies a Verification, which is a fair description — the verification is an argument for the order-three case rather than a general proof.
• The Verification, and the sentence that does all the work (§4.5.1, Part I pp. 88–89). One clause carries the whole thing: a line's entries against their own cofactors give the determinant, and against another line's cofactors give zero. Substituted into the matrix product, the first half fills the diagonal and the second half empties everything else. Run this slowly, entry by entry, on a three-by-three grid — it is the single best ninety seconds available in this module, and it is where the previous topic's zero finally gets spent.
• Example 13, first half (§4.5.1, Part I pp. 90–91). A three-by-three of ones and threes and fours whose determinant is one. Verified: the determinant is one times sixteen minus nine, minus three times four minus three, plus three times three minus four — that is seven minus three minus three, or one. The nine cofactors are seven, minus one, minus one; minus three, one, zero; minus three, zero, one — and the adjoint is that array transposed. Verified: the product with the original is the identity, which here is also the determinant times the identity because the determinant happens to be one. Say that out loud; an example whose determinant is one hides the very factor the theorem is about.
• Exercise 4.4 Q3, and it is the best item in the set (Part I p. 92). A two-by-two whose determinant is zero. Verified: its determinant is two times minus six minus three times minus four, which is minus twelve plus twelve, or zero; its adjoint has entries minus six and minus three on top, four and two below; and both products come out as the zero matrix. The theorem still holds, with the determinant on the right being zero. This is the case that makes the theorem feel like a theorem rather than a coincidence.
• Exercise 4.4 Q4 (Part I p. 92). A three-by-three with two zeros in its middle column. Verified: its determinant is eleven, its adjoint has rows zero, three, two; minus eleven, one, eight; zero, minus one, three; and both products come out as eleven times the identity. A clean, unrigged verification with a determinant that is neither zero nor one.
• Exercise 4.4 Q1 and Q2 (Part I p. 92). Two straight adjoint computations. Verified for Q1, whose entries are one and two on top and three and four below: the adjoint is four and minus two on top, minus three and one below. Verified for Q2, a three-by-three: its adjoint has rows three, one, minus eleven; minus twelve, five, minus one; six, two, five.
• The determinant of the adjoint (Part I pp. 89–90). Starting from the theorem, taking determinants of both sides and using that the determinant of a product is the product of the determinants, the chapter derives the order-three case and then states the general rule: the determinant of the adjoint is the determinant of the original raised to one less than the order. Verified on the chapter's own Example 13 array: its adjoint's determinant is seven, minus zero, minus three — that is one — and the original's determinant squared is also one. The chapter does not perform that check; it is cheap and it makes the exponent concrete.
• Exercise 4.4 Q17 (Part I p. 93). A multiple choice asking for the determinant of the adjoint of a non-singular three-by-three. Verified: the exponent is one less than three, so the answer is the determinant squared, and the intended option is the squared one. This item is the exponent rule and nothing else.
• Miscellaneous Exercise Q4 (Part I p. 99). A three-by-three with two verifications asked of it, the first about the inverse of the adjoint and the adjoint of the inverse. Verified: the array's determinant is minus five, so it is non-singular and every object in the item exists. The second half of that item belongs to the next topic.
• The Summary bullets (Part I p. 102). The adjoint definition and the theorem are both restated, the theorem with the order named. Read off the page image: the Summary carries the theorem and not the exponent rule for the determinant of the adjoint, which appears only in the body on Part I p. 90. A student revising from the Summary alone cannot answer Exercise 4.4 Q17.

Figures to have open

• The chapter's own two-by-two shortcut picture: two crossing loops over a bracketed array with two labelled downward arrows beneath. This is printed on Part I p. 88 and read; redraw it faithfully and add a visible order-two-only tag, which the book does not carry.
• A flip movement for sections 2 and 3: the cofactor array rotating about its main diagonal with one entry tracked through the move. The chapter shows the before and after side by side on Part I p. 88 but not the movement between them.
• Two three-by-three grids for sections 7 and 8, one filling on the diagonal and one filling off it. Build both with the repo's DataTable component. The sums are the chapter's own; the grid presentation is added here.
• No figure in this chapter is numbered, captioned or labelled Fig. Verified on the page image of every one of the twenty-eight pages. The two-arrow shortcut on Part I p. 88 and the crossing arrows inside the two-by-two determinant on Part I p. 77 are the only line art in the chapter, and both are uncaptioned.

Where this sits in the book

• NCERT Class 12 Mathematics, Chapter 4 "Determinants", Part I pp. 76–103
• §4.5 Adjoint and Inverse of a Matrix, the opening two sentences, Part I p. 87
• §4.5.1 Adjoint of a matrix, Definition 3, Part I p. 87
• The transpose display, Example 12 and the two-by-two Remark, Part I p. 88
• Theorem 1 and its Verification, Part I pp. 88–89
• The determinant of the adjoint, Part I pp. 89–90; Example 13, Part I pp. 90–91
• Exercise 4.4, questions 1 to 4 and 17, Part I pp. 92–93
• Miscellaneous Exercise, question 4, Part I p. 99
• Summary, the adjoint and theorem bullets, Part I p. 102