LINEAR-ALGEBRA · Unit 19 · Video 1 · Interactive Practice

How Cofactors Secretly Encode the Inverse of Any Matrix

IKey Formulas

Formula	Name	Description
$C_{ij} = (-1)^{i+j} M_{ij}$	Cofactor	Signed minor determinant
$A^{-1} = \frac{1}{\det A}\, C^T$	Cofactor inverse formula	Inverse via adjugate (transpose of cofactor matrix)
$A \cdot C^T = \det(A) \cdot I$	Key identity	Diagonal entries $= \det(A)$ , off-diagonal $= 0$

IIVisualization 1 — Cofactors assemble the inverse

Step by step, the four cofactors of $A = \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}$ become its inverse.

Step 1

💡 When $\det A = 0$ there is nothing to divide by — the adjugate exists but $A^{-1}$ does not.

IIIVisualization 2 — The adjugate identity $A\,C^{T} = \det(A)\,I$

For every matrix, $A$ times its adjugate puts $\det A$ on the diagonal and $0$ everywhere else.

Matrix

💡 The zero off-diagonal entries are no accident — the next panel shows exactly why they must vanish.

IVVisualization 3 — Why the off-diagonal entries are zero

Dotting one row's entries with another row's cofactors secretly evaluates a matrix with two identical rows.

Entries from Cofactors from

VQuiz Questions

Question 1

Consider $A = \begin{bmatrix} 3 & 1 \\ 2 & 4 \end{bmatrix}$ . What is the cofactor $C_{12}$ ?

Recall: $C_{ij} = (-1)^{i+j} M_{ij}$ , where $M_{ij}$ is the determinant of the submatrix after deleting row $i$ and column $j$ .

✅ Correct! The sign factor $(-1)^{1+2} = -1$ flips the minor from $2$ to $-2$ .

❌ Not quite. Delete row 1 and column 2 to get minor $M_{12} = 2$ , then apply the sign $(-1)^{1+2} = -1$ .

Show solution

Solution:

To find $C_{12}$ : delete row 1 and column 2 from $A = \begin{bmatrix} 3 & 1 \\ 2 & 4 \end{bmatrix}$ .

The remaining entry is $a_{21} = 2$ , so $M_{12} = 2$ .

Apply the sign factor: $(-1)^{1+2} = (-1)^3 = -1$ .

$C_{12} = (-1)^{1+2} \cdot M_{12} = (-1)(2) = \boxed{-2}$

Question 2

In the cofactor inverse formula $A^{-1} = \frac{1}{\det A}\, C^T$ , which cofactor appears in the $(1, 2)$ entry of $A^{-1}$ ?

Hint: Pay close attention to the transpose.

✅ Correct! The transpose swaps indices: $(C^T)_{12} = C_{21}$ .

❌ That's the classic mistake! Don't forget the transpose — entry $(i,j)$ of $C^T$ is $C_{ji}$ , not $C_{ij}$ .

❌ Not quite. Think about what transposing does to row and column indices.

Show solution

Solution:

The formula is $A^{-1} = \frac{1}{\det A}\, C^T$ .

The $(i, j)$ entry of $C^T$ equals $C_{ji}$ (the transpose swaps indices).

So the $(1, 2)$ entry of $C^T$ is $C_{21}$ .

Therefore: $(A^{-1})_{12} = \frac{C_{21}}{\det A}$ .

Common mistake: Choosing $C_{12}$ by forgetting the transpose!

Question 3

When computing the off-diagonal entry $(A \cdot C^T)_{ij}$ with $i \neq j$ , you dot row $i$ of $A$ with the cofactors of row $j$ . Why does this always equal zero?

✅ Correct! The "fake matrix" trick: replacing one row with another creates two identical rows, forcing det = 0.

❌ Not quite. Think: if you use row $i$ 's entries in a cofactor expansion meant for row $j$ , what matrix's determinant are you really computing?

Show solution

Solution:

The sum $\displaystyle\sum_k a_{ik}\, C_{jk}$ uses entries from row $i$ but cofactors from row $j$ .

This is exactly the cofactor expansion along row $j$ of a "fake" matrix where row $j$ has been replaced by a copy of row $i$ .

This fake matrix has two identical rows (row $i$ appears in both position $i$ and position $j$ ).

A matrix with two identical rows is always singular, so its determinant is $0$ .

Therefore $(A \cdot C^T)_{ij} = 0$ whenever $i \neq j$ .

Question 4

Suppose $A$ is a $4 \times 4$ matrix with all integer entries and $\det(A) = 1$ . What can we conclude about $A^{-1}$ ?

✅ Correct! Cofactors of an integer matrix are integers, and dividing by det(A) = 1 preserves this. A deep structural insight from the cofactor formula!

❌ Not quite. Use $A^{-1} = \frac{1}{\det A}\, C^T$ . Cofactors of an integer matrix are always integers — what happens when you divide by $\det(A) = 1$ ?

Show solution

Solution:

From the cofactor formula: $A^{-1} = \frac{1}{\det A}\, C^T$ .

Each cofactor $C_{ij}$ is a signed determinant of a $3 \times 3$ submatrix of $A$ . Since $A$ has integer entries, every cofactor is an integer (determinants of integer matrices are always integers).

Since $\det(A) = 1$ : $A^{-1} = \frac{1}{1}\, C^T = C^T$

Every entry of $A^{-1}$ is an integer cofactor divided by $1$ , so all entries of $A^{-1}$ are integers.

This beautiful structural insight — invisible to row reduction — comes directly from the cofactor formula! (Such matrices are called unimodular.)

Solved: 0 / 4

How Cofactors Secretly Encode the Inverse of Any Matrix

IKey Formulas

IIVisualization 1 — Cofactors assemble the inverse

IIIVisualization 2 — The adjugate identity A CT=det⁡(A) IA\,C^{T} = \det(A)\,IACT=det(A)I

IVVisualization 3 — Why the off-diagonal entries are zero

VQuiz Questions

IIIVisualization 2 — The adjugate identity $A\,C^{T} = \det(A)\,I$