Adjoint and Inverse of a Matrix

4.5 Adjoint and Inverse of a Matrix

The inverse of a matrix was introduced in the previous chapter. Here we establish exactly when an inverse exists and how to find it, using a special matrix called the adjoint of the original matrix.

The Adjoint of a Matrix

For any square matrix $A$, we first find the cofactor of each element. If $A = [a_{ij}]$ is an $n \times n$ matrix, the cofactor of the element $a_{ij}$ is denoted by $C_{ij}$ and is defined as:

where $M_{ij}$ is the minor of $a_{ij}$ (the determinant of the submatrix obtained by deleting the $i$-th row and $j$-th column).

Now, form a new matrix by replacing every element $a_{ij}$ of $A$ with its cofactor $C_{ij}$. This new matrix is called the cofactor matrix of $A$.

The adjoint of $A$, written as $\text{adj } A$, is defined as the transpose of this cofactor matrix.

In other words, the $(i,j)$-th element of $\text{adj } A$ is the cofactor $C_{ji}$ (note the swapped indices).

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A Fundamental Property of the Adjoint

The adjoint is not just a formal construction. It has a direct and powerful relationship with the original matrix $A$ and its determinant $|A|$.

Proof (for a $3 \times 3$ matrix):

Let $A = [a_{ij}]_{3 \times 3}$ and let $C_{ij}$ be the cofactor of $a_{ij}$. The $(i,j)$-th element of the product $A (\text{adj } A)$ is given by:

Since $(\text{adj } A)_{kj} = C_{jk}$, we have:

Now, consider two cases:

• Case 1: $i = j$. The sum becomes $\sum_{k=1}^{3} a_{ik} C_{ik}$. This is exactly the expansion of the determinant $|A|$ along the $i$-th row. So, $(A (\text{adj } A))_{ii} = |A|$.

• Case 2: $i \neq j$. The sum is $\sum_{k=1}^{3} a_{ik} C_{jk}$. This is the sum of the products of elements of the $i$-th row with the cofactors of a different row (the $j$-th row). A fundamental property of determinants states that such a sum is always zero. So, $(A (\text{adj } A))_{ij} = 0$ for $i \neq j$.

Therefore, the product $A (\text{adj } A)$ is a diagonal matrix with all diagonal entries equal to $|A|$. This is precisely $|A| I_3$.

A similar argument using columns shows that $(\text{adj } A) A$ also equals $|A| I_3$. The proof generalises to any order $n$.

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Inverse of a Matrix

We can now use the property $A (\text{adj } A) = |A| I$ to define the inverse.

If $A$ is a square matrix and $|A| \neq 0$, then we can divide both sides of the equation $A (\text{adj } A) = |A| I$ by $|A|$:

This shows that the matrix $\frac{1}{|A|} \text{adj } A$ is the multiplicative inverse of $A$.

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Properties of the Inverse (Letter-coded Properties)

The textbook lists several important properties that follow from the definition.

(I) If $A$ is a non-singular square matrix, then $|A^{-1}| = \frac{1}{|A|}$.

Proof: We know $A A^{-1} = I$. Taking determinants on both sides:

Since $|AB| = |A||B|$ and $|I| = 1$, we get:

Since $A$ is non-singular, $|A| \neq 0$, so we can divide:

(II) If $A$ and $B$ are non-singular matrices of the same order, then $(AB)^{-1} = B^{-1} A^{-1}$.

Proof: We need to show that $B^{-1} A^{-1}$ is the inverse of $AB$. Multiply:

Similarly,

Since the product with $AB$ gives the identity matrix in both orders, $B^{-1} A^{-1}$ is indeed the inverse of $AB$.

(III) If $A$ is a non-singular square matrix, then $(A^T)^{-1} = (A^{-1})^T$.

Proof: We know $A A^{-1} = I$. Taking the transpose of both sides:

Using the property $(AB)^T = B^T A^T$ and $I^T = I$, we get:

This equation shows that $(A^{-1})^T$ is the inverse of $A^T$. Therefore:

(IV) If $A$ is a non-singular square matrix, then $\text{adj } (A^T) = (\text{adj } A)^T$. …