Adjoint of a Matrix

DEFINITION 2.6.9 (Minor, Cofactor of a Matrix)   The number $\det \left(A(i\vert j)\right)$ is called the $(i,j)^{\mbox{th}}$ minor of $A$ . We write $A_{ij} = \det \left(A(i\vert j)\right).$ The $(i,j)^{\mbox{th}}$ cofactor of $A,$ denoted $C_{ij},$ is the number $(-1)^{i+j} A_{ij}.$

DEFINITION 2.6.10 (Adjoint of a Matrix)   Let $A$ be an $n \times n$ matrix. The matrix $B= [b_{ij}]$ with $b_{ij} = C_{ji},$ for $1 \leq i, j \leq n$ is called the Adjoint of $A,$ denoted $Adj (A).$

EXAMPLE 2.6.11   Let $A = \begin{bmatrix}1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & 2 & 2 \end{bmatrix}.$ Then $Adj(A) = \begin{bmatrix}4 & 2 & -7 \\ -3 & -1& 5 \\ 1 & 0 & -1 \end{bmatrix};$
as $C_{11} = (-1)^{1+1}A_{11} = 4, C_{12} = (-1)^{1+2} A_{12} = -3, C_{13} = (-1)^{1+3} A_{13} = 1,$ and so on.

THEOREM 2.6.12   Let $A$ be an $n \times n$ matrix. Then

1. for $1 \leq i \leq n,$ $\; \sum\limits_{j=1}^n a_{ij} \; C_{ij} = \sum\limits_{j=1}^n a_{ij} (-1)^{i+j} \; A_{i j} = \det(A),$
2. for $i \neq \ell, \; \sum\limits_{j=1}^n a_{ij} \; C_{\ell j} = \sum\limits_{j=1}^n a_{ij} (-1)^{\ell+j} \; A_{\ell j} = 0,$ and
3. $\; A (Adj (A) ) = \det(A) I_n.$ Thus,

   $$\det (A) \neq 0 \Rightarrow A^{-1} = \frac{1}{\det(A)} Adj (A).$$ (2.6.2)

   
   

Proof. Let $B= [b_{ij}]$ be a square matrix with

• the $\ell^{\mbox{th}}$ row of $B$ as the $i^{\mbox{th}}$ row of $A,$
• the other rows of $B$ are the same as that of $A.$

By the construction of $B,$ two rows ( $i^{\mbox{th}}$ and $\ell^{\mbox{th}}$ ) are equal. By Part 5 of Lemma 2.6.6, $\det (B) = 0.$ By construction again, $\det\bigl(A(\ell\vert j)\bigr) = \det\bigl(B(\ell\vert j)\bigr)$ for $1 \leq j \leq n.$ Thus, by Remark 2.6.7, we have

$$0 = \det (B) = \sum_{j=1}^n (-1)^{\ell + j} b_{\ell j} \det\bigl(B(\ell\vert j)\bigr) = \sum_{j=1}^n (-1)^{\ell + j} a_{ij} \det\bigl(B(\ell\vert j)\bigr)$$

$$= \sum_{j=1}^n (-1)^{\ell + j} a_{ij} \det\bigl(A(\ell\vert j)\bigr) = \sum_{j=1}^n a_{ij} C_{\ell j}.$$

Now,

$$\biggl(A\bigl( {\mbox{Adj}}(A) \bigr)\biggr)_{ij} = \sum_{k=1}^n a_{ik} \bigl( {\mbox{Adj}}(A)\bigr)_{kj} = \sum_{k=1}^n a_{ik} C_{jk}$$

$$= \left\{\begin{array}{cc} 0 & {\mbox{ if }} i \neq j \\ \det(A) & {\mbox{ if }} i = j \end{array}\right.$$

Thus, $\; A (Adj (A) ) = \det(A) I_n.$ Since, $\det(A) \neq 0,$ $\; A \displaystyle \frac{1}{\det(A)} Adj(A) = I_n.$ Therefore, $A$ has a right inverse. Hence, by Theorem 2.5.9 $A$ has an inverse and

$$A^{-1} = \frac{1}{\det(A)} Adj (A).$$

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EXAMPLE 2.6.13   Let $A= \begin{bmatrix}1 & -1 & 0 \\ 0 & 1 & 1\\ 1 & 2 & 1 \end{bmatrix}.$ Then

$$Adj (A) = \begin{bmatrix} -1 & 1 & -1 \\ 1 & 1 & -1\\ -1 &-3 & 1 \end{bmatrix}$$

and $\det (A) = -2.$ By Theorem 2.6.12.3, $A^{-1} = \begin{bmatrix}1/2 & -1/2 & 1/2 \\ -1/2 & -1/2 & 1/2\\ 1/2 & 3/2 & -1/2 \end{bmatrix}.$

The next corollary is an easy consequence of Theorem 2.6.12 (recall Theorem 2.5.9).

COROLLARY 2.6.14   If $A$ is a non-singular matrix, then
$\bigl(Adj(A) \bigr) A = \det(A) I_n \;\;$ and $\;\; \sum\limits_{i=1}^n a_{ij} \; C_{ik} = \left\{\begin{array}{cc} \det (A) & {\mbox{ if }} j = k \\ 0 & {\mbox{if }} j \neq k \end{array}\right..$

THEOREM 2.6.15   Let $A$ and $B$ be square matrices of order $n.$ Then $\;\det (A B) = \det (A) \det (B).$

Proof. Step 1. Let $\det (A) \neq 0.$
This means, $A$ is invertible. Therefore, either $A$ is an elementary matrix or is a product of elementary matrices (see Theorem 2.5.8). So, let $E_1, E_2, \ldots, E_k$ be elementary matrices such that $A = E_1 E_2 \cdots E_k.$ Then, by using Parts 1, 2 and 4 of Lemma 2.6.6 repeatedly, we get

$$\det(AB) = \det (E_1 E_2 \cdots E_k B) =\det (E_1) \det( E_2 \cdots E_k B)$$

$$= \det(E_1) \det( E_2) \det(E_3 \cdots E_k B)$$

$$= \det(E_1 E_2) \det(E_3 \cdots E_k B)$$

$$= \vdots$$

$$= \det(E_1 E_2 \cdots E_k) \det( B)$$

$$= \det(A) \det(B).$$

Thus, we get the required result in case $A$ is non-singular.

Step 2. Suppose $\det (A) = 0.$
Then $A$ is not invertible. Hence, there exists an invertible matrix $P$ such that $P A = C,$ where $C = \begin{bmatrix}C_1 \\ {\mathbf 0} \end{bmatrix}.$ So, $A = P^{-1} C,$ and therefore

$$\det(AB) = \det ((P^{-1} C) B) = \det (P^{-1} (C B)) = \det \left( P^{-1} \begin{bmatrix}C_1 B \\ {\mathbf 0}\end{bmatrix}\right)$$

$$= \det( P^{-1} ) \cdot \det \left( \begin{bmatrix}C_1 B \\ {\mathbf 0} \end{bmatrix}\right) \;\; {\mbox{ as }} P^{-1} {\mbox{ is non-singular}}$$

$$= \det (P) \cdot 0 = 0 = 0 \cdot \det (B) = \det(A) \det(B).$$

Thus, the proof of the theorem is complete. height6pt width 6pt depth 0pt

COROLLARY 2.6.16   Let $A$ be a square matrix. Then $A$ is non-singular if and only if $A$ has an inverse.

Proof. Suppose $A$ is non-singular. Then $\det(A) \neq 0$ and therefore, $A^{-1} = \displaystyle\frac{1}{\det(A)} Adj(A).$ Thus, $A$ has an inverse.

Suppose $A$ has an inverse. Then there exists a matrix $B$ such that $A B = I = BA.$ Taking determinant of both sides, we get

$$\det(A) \det (B) = \det(AB) = \det(I) = 1.$$

This implies that $\det (A) \neq 0.$ Thus, $A$ is non-singular. height6pt width 6pt depth 0pt

THEOREM 2.6.17   Let $A$ be a square matrix. Then $\det (A) = \det(A^t).$

Proof. If $A$ is a non-singular Corollary 2.6.14 gives $\det (A) = \det(A^t).$

If $A$ is singular, then $\det (A) = 0.$ Hence, by Corollary 2.6.16, $A$ doesn't have an inverse. Therefore, $A^t$ also doesn't have an inverse (for if $A^t$ has an inverse then $A^{-1} = \bigl((A^t)^{-1}\bigr)^t).$ Thus again by Corollary 2.6.16, $\det(A^t) = 0.$ Therefore, we again have $\det(A) = 0 = \det(A^t).$

Hence, we have $\det (A) = \det(A^t).$ height6pt width 6pt depth 0pt