Multivariable-Calculus · Unit 3 · Video 6 · Interactive Practice

Minors, Signs, Transpose, Divide: A Three-by-Three Inverse by Hand

IKey Formulas

Formula	Name	What you need
$A^{-1} = \dfrac{1}{\det(A)}\operatorname{adj}(A)$	Inverse by cofactors	$\det(A) \neq 0$
$M_{ij} = \det\big(A \text{ with row } i \text{ and column } j \text{ deleted}\big)$	Minor	A position $(i,j)$ in $A$
$C_{ij} = (-1)^{i+j} M_{ij}, \qquad \operatorname{adj}(A) = C^{\mathsf{T}}$	Cofactor, then adjoint	All nine minors
$A \operatorname{adj}(A) = \det(A)\, I$	Why the divisor is $\det(A)$	Holds for every square $A$

Key Insight: $(-1)^{i+j}$ is an instruction, not an answer. At a minus it flips whatever sign the minor already had: $M_{12} = -1$ sits under a minus, so $C_{12} = +1$ — while $M_{13} = -2$ sits under a plus and stays $-2$ .

IIVisualization 1 — Where Each Minor Comes From

Every entry of $A$ hides a $2 \times 2$ determinant: delete its row, delete its column.

IIIVisualization 2 — Reading the Checkerboard

Does the checkerboard give a cofactor's sign, or tell you what to do to the minor?

IVVisualization 3 — Transpose, Then Divide

Three more steps turn the nine minors into $A^{-1}$ .

Step 1

💡 Those nine numbers solve every system at once: if $AX = B$ , then $X = A^{-1}B$ , whatever the right-hand side $B$ happens to be.

VQuiz Questions

Problem 1 · One Minor of the Video's Matrix

Given: $A = \begin{bmatrix} 2 & 3 & 3 \\ 2 & 4 & 5 \\ 1 & 1 & 2 \end{bmatrix}$ — find the minor $M_{32}$ .

✅ Correct! Deleting row 3 and column 2 leaves

\begin{vmatrix} 2 & 3 \\ 2 & 5 \end{vmatrix} = 10 - 6 = 4

❌ Not quite.

2

M_{33}

— you deleted column 3. Column 2 is the one that goes.

❌ Close, but that is the cofactor.

(-1)^{3+2} = -1

turns

4

into

C_{32} = -4

; the minor itself carries no checkerboard sign.

❌ Not quite.

-1

M_{23}

. In

M_{ij}

the first index is the row you delete, the second the column.

Show solution

Delete row 3 and column 2 of $A$ . The surviving entries are the ones in rows 1 and 2, columns 1 and 3:

M_{32} = \begin{vmatrix} 2 & 3 \\ 2 & 5 \end{vmatrix} = 2 \cdot 5 - 3 \cdot 2 = 10 - 6 = 4

The minor is $4$ . Only in step 2 does the checkerboard sign $(-1)^{3+2} = -1$ arrive and turn it into the cofactor $C_{32} = -4$ .

Problem 2 · Instruction or Final Sign?

Given: the top row of the matrix of minors is $3,\; -1,\; -2$ , and the top row of the checkerboard is $+,\; -,\; +$ — find the top row of the cofactor matrix.

✅ Correct! The middle entry flips (

-1 \to +1

) and the two under a plus keep the signs they already had.

❌ That reads the diagram as the final signs. A minus means flip, so

-1

becomes

+1

; a plus means leave it, so

-2

stays

-2

❌ Those are still the minors. Position

(1,2)

sits under a minus, so its sign must flip.

❌ Check where the checkerboard starts. The top-left corner is a plus, so the first row reads

+,\; -,\; +

, not

-,\; +,\; -

Show solution

Apply each board entry as an instruction to the minor below it:

$(1,1)$ : sign $+$ , so leave $3$ alone $\Rightarrow C_{11} = 3$
$(1,2)$ : sign $-$ , so flip $-1 \Rightarrow C_{12} = +1$
$(1,3)$ : sign $+$ , so leave $-2$ alone $\Rightarrow C_{13} = -2$

The top row of cofactors is $3,\; 1,\; -2$ . Equivalently $C_{ij} = (-1)^{i+j} M_{ij}$ : the sign of a cofactor depends on two things — the instruction in the diagram and the sign the minor already had.

Problem 3 · One Entry of an Inverse

Given: $B = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{bmatrix}$ with $\det(B) = 1$ — find the entry in row 2, column 1 of $B^{-1}$ .

✅ Correct! The transpose sends

C_{12} = 20

to position

(2,1)

of the adjoint, and

\det(B) = 1

leaves it there.

❌ That is $C_{21}$ . You skipped the transpose: the

(2,1)

entry of

\operatorname{adj}(B)

C_{12}

, not

C_{21}

❌ That is the minor $M_{12}$ . Position

(1,2)

sits under a minus, so

C_{12} = -(-20) = +20

❌ That is the minor $M_{21}$ . Two steps are missing: the checkerboard flip, and the transpose that moves

C_{12}

into position

(2,1)

Show solution

Step 1 — which cofactor lands in position $(2,1)$ ? Since $\operatorname{adj}(B) = C^{\mathsf{T}}$ , the $(2,1)$ entry of the adjoint is $C_{12}$ .

Step 2 — the minor $M_{12}$ (delete row 1 and column 2):

M_{12} = \begin{vmatrix} 0 & 4 \\ 5 & 0 \end{vmatrix} = 0 \cdot 0 - 4 \cdot 5 = -20

Step 3 — the checkerboard. $(-1)^{1+2} = -1$ flips it: $C_{12} = +20$ .

Step 4 — divide by $\det(B) = 1$ :

\big(B^{-1}\big)_{21} = \frac{C_{12}}{\det(B)} = \frac{20}{1} = 20

The whole inverse is $B^{-1} = \begin{bmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{bmatrix}$ — and $18$ , the answer you get without the transpose, is sitting one flip away at position $(1,2)$ .

Problem 4 · The Same Four Steps on a $2 \times 2$

Given: $D = \begin{bmatrix} 4 & 7 \\ 2 & 6 \end{bmatrix}$ , whose minors are the single surviving entries — find $D^{-1}$ by minors, checkerboard, transpose, divide.

✅ Correct! This is the familiar

2 \times 2

rule — swap the diagonal, negate the off-diagonal, divide by

\det

— and it is just the four steps at size two.

❌ That is the cofactor matrix. One step left: transpose it, which swaps

-2

and

-7

❌ The diagonal did not move.

M_{11} = 6

and

M_{22} = 4

: deleting row 1 and column 1 leaves the other diagonal entry.

❌ The checkerboard was never applied. Positions

(1,2)

and

(2,1)

sit under a minus, so both off-diagonal entries flip sign.

Show solution

Step 1 — minors. Deleting a row and a column of a $2 \times 2$ leaves one entry, and that entry is the minor:

M = \begin{bmatrix} 6 & 2 \\ 7 & 4 \end{bmatrix}

Step 2 — checkerboard $\begin{bmatrix} + & - \\ - & + \end{bmatrix}$ , so the two off-diagonal entries flip:

C = \begin{bmatrix} 6 & -2 \\ -7 & 4 \end{bmatrix}

Step 3 — transpose:

\operatorname{adj}(D) = C^{\mathsf{T}} = \begin{bmatrix} 6 & -7 \\ -2 & 4 \end{bmatrix}

Step 4 — divide by $\det(D) = 4 \cdot 6 - 7 \cdot 2 = 10$ :

D^{-1} = \frac{1}{10}\begin{bmatrix} 6 & -7 \\ -2 & 4 \end{bmatrix}

Verify: $\begin{bmatrix} 4 & 7 \\ 2 & 6 \end{bmatrix}\begin{bmatrix} 6 & -7 \\ -2 & 4 \end{bmatrix} = \begin{bmatrix} 10 & 0 \\ 0 & 10 \end{bmatrix} = 10\,I$ , which is $\det(D)\, I$ exactly as in the $3 \times 3$ case.

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