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Higher Mathematics 1st Paper · Classes 11-12 · Chapter 1 · 1.8.1

Adjoint, Finding the Inverse, and Properties of the Inverse.

Can a matrix be “turned upside down”? It can, as long as its determinant is not zero. And the job takes three moves: find the cofactors, transpose, divide by the determinant. The board asks about exactly these moves more than anything else, and the marks are usually lost at the transpose.

1. The adjoint and the formula

Every entry of a square matrix A = (aij) has a cofactor: Aij = (−1)i+j × the minor of aij. Arrange the cofactors as the matrix [Aij], then transpose it, so rows become columns and columns become rows. What you get is the adjoint adj A. For order 3 the first row of adj A is A₁₁, A₂₁, A₃₁; the second is A₁₂, A₂₂, A₃₂; the third is A₁₃, A₂₃, A₃₃. Notice that the first row holds A₂₁, not A₁₂. That is the transpose at work.

When A is non-singular, that is |A| ≠ 0, then A⁻¹ = (1/|A|) adj A. This is the adjoint method.

Four steps give an inverse: 1) find |A| and check it is not 0; 2) find the nine cofactors; 3) transpose the cofactor matrix to get adj A; 4) divide by |A|. The commonest slip is skipping step 3 and treating the cofactor matrix itself as adj A.

2. One full calculation

Walk the book's example in four steps. Each step produces one matrix, so you can see the cofactor matrix and adj A as two separate things.

step by step press next

The inverse in four steps

Step 1. The determinant. A =
Multiply the first row by its cofactors (5, −7, −1): |A| = 1×5 + 2(−7) + 4(−1) = −13. Not zero, so the inverse exists.

3. Diagonal matrices and more inverses

A diagonal matrix needs no cofactors at all. Replace each diagonal entry by its reciprocal: a, b, c become 1/a, 1/b, 1/c, and everything else stays 0. See the inverse of [[2, 0, 0], [0, 3, 0], [0, 0, 4]] under the first chip below.

The other chips are worked answers from exercise 1(B), questions 12 and 13. Each gives |A| and A⁻¹, so redo one or two in your own notebook and compare.

the book's answers tap a chip

Six inverse matrices

Diagonal. →
In general [[a, 0, 0], [0, b, 0], [0, 0, c]] has inverse [[1/a, 0, 0], [0, 1/b, 0], [0, 0, 1/c]].

4. Five properties of the inverse

The inverse of an inverse brings you back: (A⁻¹)⁻¹ = A. If A and B are non-singular of the same order, AB is non-singular too, and (AB)⁻¹ = B⁻¹A⁻¹. The order reverses, like a sock and a shoe: the shoe went on last, so it comes off first.

Transposing and inverting can be done in either order: (Aᵗ)⁻¹ = (A⁻¹)ᵗ. If an inverse exists it is unique: A⁻¹ = B and A⁻¹ = C give B = C. And if A is non-singular and symmetric, that is A' = A, then its inverse is symmetric too: (A⁻¹)' = A⁻¹.

There is a lookalike trap: (AB)' = B'A' has almost the same shape, but it is the rule for transposes. Drop each formula into the box with its name.

sort them formula, then box

Match the formula

Pick up a formula, then drop it in the box with its name. Six boxes, six formulas.

5. A·adj A and f(A) = 0

The key property of the adjoint: A·adj A = adj A·A = |A|·I. Divide both sides by |A| and A⁻¹ = adj A/|A| falls out. In the book's 14(ii), M = [[0, 1, 1], [1, 2, 0], [3, −1, 4]] has |M| = −11, and (adj M)M has −11 down the diagonal and zeros elsewhere, which is |M| I₃.

The quickest way to confirm an inverse is to multiply. In exercise 1(B) 14(i), A = [[1, 0, 2], [2, −1, 3], [4, 1, 8]] and B = [[−11, 2, 2], [−4, 0, 1], [6, −1, −1]] multiply to AB = BA = I₃, so B = A⁻¹ and A = B⁻¹.

Here is another route. If a matrix satisfies a polynomial equation f(A) = 0, you can get the inverse without cofactors. Factor out A to reach A·(…) = kI, and then A⁻¹ = (…)/k. In the book's 14(iii), A = [[1, 2], [4, −3]] and A² + 2A − 11I = 0, so A(A + 2I) = 11I and A⁻¹ = (1/11)(A + 2I). Change the example and the step below.

calculator change the step

Factor out A

The values come from the book; the steps are laid out for you.

Creative question 4(c) uses the same A = [[1, 2], [4, −3]] with f(x) = x² + 2x − 8. Then f(A) = A² + 2A − 8I = [[9, −4], [−8, 17]] + [[2, 4], [8, −6]] − [[8, 0], [0, 8]] = [[3, 0], [0, 3]]. So (1/3)f(A) = I₂, an identity matrix. The third example in the calculator above walks through it.

6. Three rules for determinants

The book's creative and multiple-choice questions ask for the determinants of adj A, kA and A⁻¹, yet the book never teaches the rules. So they get a box of their own.

Use the calculator below to change the order, |A| and k and watch the rules work.

calculator change the number

Determinant rules

The numbers are for practice. None of this goes anywhere; the notebook is yours.

7. Check yourself

Three questions in the exam's style. Start with the first.

check yourself pick one answer

Question 1

If I₃ is the 3 × 3 identity matrix, what is (I₃)⁻¹?

In the second, keep the order of the steps in mind.

check yourself pick one answer

Question 2

If A = [[−1, 3], [2, 5]], which of the following is adj A?

For the last, recall the formula and work it out.

check yourself pick one answer

Question 3

If A = [[−1, −3], [4, 2]], what is |adj(A)|?

The whole note on a few cards.

cut from the note tap to turn over

Seven cards

QuestionWhat is the adjoint matrix adj A?
AnswerThe transpose of the cofactor matrix [A_ij].
QuestionWhat are the four steps for finding an inverse?
AnswerDeterminant → cofactors → transpose (adj A) → divide by |A|.
QuestionWhat is A⁻¹ for the same A?
Answer(1/−13) adj A = [[−5/13, 2/13, 6/13], [7/13, 5/13, −11/13], [1/13, −3/13, 4/13]]
QuestionWhat is (AB)⁻¹ equal to?
AnswerB⁻¹A⁻¹ (the order reverses).
QuestionHow many inverses can a matrix have?
AnswerOnly one (unique): A⁻¹ = B and A⁻¹ = C give B = C.
QuestionWhat is A·adj A?
Answer|A|·I (and adj A·A = |A|·I).
QuestionIf A = [[1, 2], [4, −3]] and A² + 2A − 11I = 0, what is A⁻¹?
AnswerA(A + 2I) = 11I, so A⁻¹ = (1/11)[[3, 2], [4, −1]].

Source: Higher Mathematics 1st Paper, Classes 11-12 (NCTB), Chapter 1 “Matrices and Determinants”, section 1.8.1.