Mathematics · Ch 4 — Determinants

Adjoint of a Matrix

4.5.1

Adjoint of a Matrix

The adjoint of a matrix is the stepping stone to finding its inverse. It is built directly from the cofactors you have already learned to compute.

Definition of the Adjoint

For a square matrix $A = [a_{ij}]_{n \times n}$ , the adjoint of $A$ , written as $\text{adj } A$ , is defined as the transpose of the matrix of cofactors.

Let $A_{ij}$ be the cofactor of the element $a_{ij}$ . First, form the cofactor matrix $[A_{ij}]_{n \times n}$ — this is a matrix where you replace each element $a_{ij}$ with its cofactor $A_{ij}$ . Then, take the transpose of this cofactor matrix. The result is the adjoint.

\text{adj } A = \text{Transpose of } [A_{ij}]_{n \times n} = [A_{ji}]_{n \times n}

Note

The notation $[A_{ji}]$ means the element in the $i$ -th row and $j$ -th column of the adjoint is the cofactor $A_{ji}$ from the original matrix. This swapping of indices is exactly what the transpose does.

Adjoint for a $3 \times 3$ Matrix

If $A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$ , then the cofactor matrix is $\begin{bmatrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{bmatrix}$ .

Taking its transpose gives the adjoint:

\text{adj } A = \begin{bmatrix} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{bmatrix}

Adjoint for a $2 \times 2$ Matrix — A Useful Shortcut

For a $2 \times 2$ matrix $A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}$ , the cofactors are:

$A_{11} = a_{22}$
$A_{12} = -a_{21}$
$A_{21} = -a_{12}$
$A_{22} = a_{11}$

The cofactor matrix is $\begin{bmatrix} a_{22} & -a_{21} \\ -a_{12} & a_{11} \end{bmatrix}$ . Transposing it gives the adjoint:

\text{adj } A = \begin{bmatrix} a_{22} & -a_{12} \\ -a_{21} & a_{11} \end{bmatrix}

Tip

For a $2 \times 2$ matrix, you can find the adjoint directly without computing cofactors: swap the diagonal elements $a_{11}$ and $a_{22}$ , and change the signs of the off-diagonal elements $a_{12}$ and $a_{21}$ .

Theorem 1: The Fundamental Adjoint-Product Relation

Important

For any square matrix $A$ of order $n$ ,

$A(\text{adj } A) = (\text{adj } A)A = |A| I$

where $I$ is the identity matrix of order $n$ .

›Proof

Verification for a $3 \times 3$ matrix

Let $A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$ and $\text{adj } A = \begin{bmatrix} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{bmatrix}$ .

Consider the product $A(\text{adj } A)$ . The element in the $i$ -th row and $j$ -th column of this product is:

$a_{i1}A_{j1} + a_{i2}A_{j2} + a_{i3}A_{j3}$

This is the sum of the products of elements of the $i$ -th row of $A$ with the cofactors of the $j$ -th row of $A$ .

Case 1: $i = j$

The sum becomes $a_{i1}A_{i1} + a_{i2}A_{i2} + a_{i3}A_{i3}$ , which is exactly the expansion of $|A|$ along the $i$ -th row. So each diagonal entry equals $|A|$ .

Case 2: $i \neq j$

The sum becomes $a_{i1}A_{j1} + a_{i2}A_{j2} + a_{i3}A_{j3}$ . This is the sum of products of elements of the $i$ -th row with the cofactors of a different row $j$ . By a property of determinants, this sum is zero.

Therefore,

$A(\text{adj } A) = \begin{bmatrix} |A| & 0 & 0 \\ 0 & |A| & 0 \\ 0 & 0 & |A| \end{bmatrix} = |A| \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = |A| I$

The same reasoning applied to $(\text{adj } A)A$ (using columns instead of rows) gives the same result.

Singular and Non-Singular Matrices

These definitions classify matrices based on whether their determinant is zero.

Definition 4 — Singular Matrix: A square matrix $A$ is called singular if $|A| = 0$ .

Example: $A = \begin{bmatrix} 1 & 2 \\ 4 & 8 \end{bmatrix}$ . Here $|A| = 1(8) - 2(4) = 8 - 8 = 0$ , so $A$ is singular.

Definition 5 — Non-Singular Matrix: A square matrix $A$ is called non-singular if $|A| \neq 0$ .

Example: $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ . Here $|A| = 1(4) - 2(3) = 4 - 6 = -2 \neq 0$ , so $A$ is non-singular.

Theorem 2 and Theorem 3 (Stated Without Proof)

Theorem 2: If $A$ and $B$ are non-singular matrices of the same order, then $AB$ and $BA$ are also non-singular matrices of the same order.

Theorem 3: The determinant of the product of matrices equals the product of their determinants:

|AB| = |A| \, |B|

where $A$ and $B$ are square matrices of the same order.

Determinant of the Adjoint

From Theorem 1, we have $A(\text{adj } A) = |A| I$ . Taking determinants on both sides:

|A(\text{adj } A)| = \big| |A| I \big|

Using Theorem 3 on the left: $|A| \, |\text{adj } A| = \big| |A| I \big|$

For a $3 \times 3$ matrix, $|A| I = \begin{bmatrix} |A| & 0 & 0 \\ 0 & |A| & 0 \\ 0 & 0 & |A| \end{bmatrix}$ . The determinant of this diagonal matrix is $|A|^3$ .

Therefore: $|A| \, |\text{adj } A| = |A|^3$

If $|A| \neq 0$ , we can divide both sides by $|A|$ to get:

|\text{adj } A| = |A|^2

A

Theorem 4: Condition for Invertibility

Important

A square matrix $A$ is invertible if and only if $A$ is non-singular (i.e., $|A| \neq 0$ ).

›Proof

Part 1: If $A$ is invertible, then $A$ is non-singular.

Since $A$ is invertible, there exists a matrix $B$ such that $AB = BA = I$ .

Taking determinants: $|AB| = |I|$ …

Definition 3Adjoint of a Matrix

Definition: Adjoint of a Matrix

For a square matrix $A = [a_{ij}]_{n \times n}$ , the adjoint (denoted $\text{adj } A$ ) is defined as the transpose of the cofactor matrix of $A$ .

That is:

First, find the cofactor $A_{ij}$ for each element $a_{ij}$ of $A$ .
Arrange all these cofactors into a matrix $[A_{ij}]_{n \times n}$ (the cofactor matrix).
Then, take the transpose of this cofactor matrix. The result is $\text{adj } A$ .

In symbols:

\text{adj } A = \big( [A_{ij}]_{n \times n} \big)^T

For a $3 \times 3$ matrix:

A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}

the adjoint is:

\text{adj } A = \begin{bmatrix} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{bmatrix}

Intuition

The adjoint is a way to "package" all the cofactors of a matrix so that when multiplied by the original matrix, the result is a scalar (the determinant) times the identity matrix. This makes the adjoint a key step in finding the inverse of a matrix.

Concrete Example (for a $2 \times 2$ matrix)

Let:

A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}

Cofactors:

$A_{11} = 4$
$A_{12} = -1$
$A_{21} = -3$
$A_{22} = 2$

Cofactor matrix:

\begin{bmatrix} 4 & -1 \\ -3 & 2 \end{bmatrix}

Transpose to get adjoint: …

Definition 4Singular

Definition

A square matrix $A$ is called singular if its determinant is zero:

|A| = 0

Intuition

A singular matrix has no inverse — it "collapses" space, losing information. You cannot reverse its effect.

Example

For

A = \begin{bmatrix} 1 & 2 \\ 4 & 8 \end{bmatrix}

the determinant is …

Definition 5Non-singular

Definition

A square matrix $A$ is called non-singular if its determinant is not zero.

In symbols:

$A$ is non-singular $\iff |A| \neq 0$ .

This is Definition 5 from the textbook. It is the exact opposite of a singular matrix, which has $|A| = 0$ .

Intuition

Think of the determinant as a "scale factor" or a "test of invertibility."

If $|A| \neq 0$ , the matrix has a unique inverse — it can "undo" its own transformation.

If $|A| = 0$ , the matrix collapses space (loses information) and cannot be reversed.

Concrete Example

Let …

Theorem 1

Theorem 4: Invertibility and Non-Singularity

A square matrix $A$ is invertible if and only if $A$ is non-singular.

A \text{ is invertible} \iff |A| \neq 0

What This Means

The theorem gives us a clean, practical test for whether a matrix has an inverse. Instead of searching for some matrix $B$ such that $AB = I$ , we simply compute the determinant. If it is zero, the matrix is singular and has no inverse. If it is non-zero, the matrix is non-singular and an inverse exists — and we even have a formula for it.

The Two Key Definitions

Before we prove the theorem, recall two definitions from the textbook:

Singular matrix: A square matrix $A$ is singular if $|A| = 0$ .
Non-singular matrix: A square matrix $A$ is non-singular if $|A| \neq 0$ .

Watch out

A common mistake is to confuse "singular" with "invertible". They are opposites: singular means no inverse exists; non-singular means an inverse does exist.

The Complete Proof

›Proof

We must prove two directions: (1) if $A$ is invertible, then $A$ is non-singular; (2) if $A$ is non-singular, then $A$ is invertible.

Forward direction (invertible $\Rightarrow$ non-singular)

Suppose $A$ is an invertible matrix of order $n$ . Then there exists a square matrix $B$ of order $n$ such that

$AB = BA = I$

where $I$ is the identity matrix of order $n$ .

Take determinants on both sides of $AB = I$ :

$|AB| = |I|$

By Theorem 3 (the determinant of a product equals the product of determinants), we have

$|A|\,|B| = 1$

Since $|I| = 1$ for any identity matrix. Now $|A|\,|B| = 1$ implies $|A| \neq 0$ (if $|A|$ were zero, the product would be zero, not 1). Therefore $A$ is non-singular.

Backward direction (non-singular $\Rightarrow$ invertible)

Suppose $A$ is non-singular, so $|A| \neq 0$ . From Theorem 1, we know that for any square matrix $A$ ,

$A(\text{adj }A) = (\text{adj }A)A = |A|\,I$

Since $|A| \neq 0$ , we can divide both sides by $|A|$ :

$A\left(\frac{1}{|A|}\text{adj }A\right) = \left(\frac{1}{|A|}\text{adj }A\right)A = I$

Define $B = \frac{1}{|A|}\text{adj }A$ . Then $AB = BA = I$ , which means $B$ is the inverse of $A$ . Hence $A$ is invertible, and moreover

$A^{-1} = \frac{1}{|A|}\text{adj }A$

This completes the proof.

Note

The proof gives us more than just the theorem — it provides the explicit formula for the inverse: $A^{-1} = \frac{1}{|A|}\text{adj }A$ . This is the method you will use to compute inverses for $2 \times 2$ and $3 \times 3$ matrices.

When to Use This Theorem

You use this theorem whenever you need to decide whether a matrix has an inverse. The procedure is simple: …

Theorem 2

Theorem 4: Invertibility and Non-Singularity

A square matrix $A$ is invertible if and only if $A$ is non-singular.

A \text{ is invertible} \iff |A| \neq 0

What This Means

The Two Key Definitions

Before we prove the theorem, recall two definitions from the textbook:

Singular matrix: A square matrix $A$ is singular if $|A| = 0$ .
Non-singular matrix: A square matrix $A$ is non-singular if $|A| \neq 0$ .

Watch out

A common mistake is to confuse "singular" with "invertible". They are opposites: singular means no inverse exists; non-singular means an inverse does exist.

The Complete Proof

›Proof

We must prove two directions: (1) if $A$ is invertible, then $A$ is non-singular; (2) if $A$ is non-singular, then $A$ is invertible.

Forward direction (invertible $\Rightarrow$ non-singular)

Suppose $A$ is an invertible matrix of order $n$ . Then there exists a square matrix $B$ of order $n$ such that

$AB = BA = I$

where $I$ is the identity matrix of order $n$ .

Take determinants on both sides of $AB = I$ :

$|AB| = |I|$

By Theorem 3 (the determinant of a product equals the product of determinants), we have

$|A|\,|B| = 1$

Since $|I| = 1$ for any identity matrix. Now $|A|\,|B| = 1$ implies $|A| \neq 0$ (if $|A|$ were zero, the product would be zero, not 1). Therefore $A$ is non-singular.

Backward direction (non-singular $\Rightarrow$ invertible)

Suppose $A$ is non-singular, so $|A| \neq 0$ . From Theorem 1, we know that for any square matrix $A$ ,

$A(\text{adj }A) = (\text{adj }A)A = |A|\,I$

Since $|A| \neq 0$ , we can divide both sides by $|A|$ :

$A\left(\frac{1}{|A|}\text{adj }A\right) = \left(\frac{1}{|A|}\text{adj }A\right)A = I$

Define $B = \frac{1}{|A|}\text{adj }A$ . Then $AB = BA = I$ , which means $B$ is the inverse of $A$ . Hence $A$ is invertible, and moreover

$A^{-1} = \frac{1}{|A|}\text{adj }A$

This completes the proof.

Note

When to Use This Theorem

You use this theorem whenever you need to decide whether a matrix has an inverse. The procedure is simple: …

Theorem 3

Theorem 4: Invertibility and Non-Singularity

A square matrix $A$ is invertible if and only if $A$ is non-singular.

A \text{ is invertible} \iff |A| \neq 0

What This Means

The Two Key Definitions

Before we prove the theorem, recall two definitions from the textbook:

Singular matrix: A square matrix $A$ is singular if $|A| = 0$ .
Non-singular matrix: A square matrix $A$ is non-singular if $|A| \neq 0$ .

Watch out

A common mistake is to confuse "singular" with "invertible". They are opposites: singular means no inverse exists; non-singular means an inverse does exist.

The Complete Proof

›Proof

We must prove two directions: (1) if $A$ is invertible, then $A$ is non-singular; (2) if $A$ is non-singular, then $A$ is invertible.

Forward direction (invertible $\Rightarrow$ non-singular)

Suppose $A$ is an invertible matrix of order $n$ . Then there exists a square matrix $B$ of order $n$ such that

$AB = BA = I$

where $I$ is the identity matrix of order $n$ .

Take determinants on both sides of $AB = I$ :

$|AB| = |I|$

By Theorem 3 (the determinant of a product equals the product of determinants), we have

$|A|\,|B| = 1$

Since $|I| = 1$ for any identity matrix. Now $|A|\,|B| = 1$ implies $|A| \neq 0$ (if $|A|$ were zero, the product would be zero, not 1). Therefore $A$ is non-singular.

Backward direction (non-singular $\Rightarrow$ invertible)

Suppose $A$ is non-singular, so $|A| \neq 0$ . From Theorem 1, we know that for any square matrix $A$ ,

$A(\text{adj }A) = (\text{adj }A)A = |A|\,I$

Since $|A| \neq 0$ , we can divide both sides by $|A|$ :

$A\left(\frac{1}{|A|}\text{adj }A\right) = \left(\frac{1}{|A|}\text{adj }A\right)A = I$

Define $B = \frac{1}{|A|}\text{adj }A$ . Then $AB = BA = I$ , which means $B$ is the inverse of $A$ . Hence $A$ is invertible, and moreover

$A^{-1} = \frac{1}{|A|}\text{adj }A$

This completes the proof.

Note

When to Use This Theorem

You use this theorem whenever you need to decide whether a matrix has an inverse. The procedure is simple: …

Theorem 4

Theorem 4: Invertibility and Non-Singularity

A square matrix $A$ is invertible if and only if $A$ is non-singular.

A \text{ is invertible} \iff |A| \neq 0

What This Means

The Two Key Definitions

Before we prove the theorem, recall two definitions from the textbook:

Singular matrix: A square matrix $A$ is singular if $|A| = 0$ .
Non-singular matrix: A square matrix $A$ is non-singular if $|A| \neq 0$ .

Watch out

A common mistake is to confuse "singular" with "invertible". They are opposites: singular means no inverse exists; non-singular means an inverse does exist.

The Complete Proof

›Proof

We must prove two directions: (1) if $A$ is invertible, then $A$ is non-singular; (2) if $A$ is non-singular, then $A$ is invertible.

Forward direction (invertible $\Rightarrow$ non-singular)

Suppose $A$ is an invertible matrix of order $n$ . Then there exists a square matrix $B$ of order $n$ such that

$AB = BA = I$

where $I$ is the identity matrix of order $n$ .

Take determinants on both sides of $AB = I$ :

$|AB| = |I|$

By Theorem 3 (the determinant of a product equals the product of determinants), we have

$|A|\,|B| = 1$

Since $|I| = 1$ for any identity matrix. Now $|A|\,|B| = 1$ implies $|A| \neq 0$ (if $|A|$ were zero, the product would be zero, not 1). Therefore $A$ is non-singular.

Backward direction (non-singular $\Rightarrow$ invertible)

Suppose $A$ is non-singular, so $|A| \neq 0$ . From Theorem 1, we know that for any square matrix $A$ ,

$A(\text{adj }A) = (\text{adj }A)A = |A|\,I$

Since $|A| \neq 0$ , we can divide both sides by $|A|$ :

$A\left(\frac{1}{|A|}\text{adj }A\right) = \left(\frac{1}{|A|}\text{adj }A\right)A = I$

Define $B = \frac{1}{|A|}\text{adj }A$ . Then $AB = BA = I$ , which means $B$ is the inverse of $A$ . Hence $A$ is invertible, and moreover

$A^{-1} = \frac{1}{|A|}\text{adj }A$

This completes the proof.

Note

When to Use This Theorem

You use this theorem whenever you need to decide whether a matrix has an inverse. The procedure is simple: …

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Adjoint of a Matrix