Adjoint And Inverse Of Matrices
In this chapter we will extend our knowledge of determinants
I. Introduction To Adjoint
In the previous chapter we learnt about minor and cofactor of an element $a_{i,j}$ of a square matrix $A$.
For any element $a_{i,j}$ of $A$, the cofactor is given by eliminating the $i \xasuper{th}$ row and $j \xasuper{th}$ column of $A$, thus obtaining a submatrix $S_{i, j}$ of order $(n - 1)$, and then calculating the determinant of $S_{i, j}$
Let us understand the adjoint matrix better using the following example of a $3 \times 3$ matrix.
Let $A = \begin{bmatrix} 2 & 3 & 4 \\ 4 & 9 & 16 \\ 5 & 6 & 7 \end{bmatrix}$
$a_{1,1} = 2$
Eliminating the $1\xasuper{st}$ row and the $1\xasuper{st}$ column we get:
$S_{1, 1} = \begin{bmatrix} 9 & 16 \\ 6 & 7\end{bmatrix}$
$\therefore$ The minor $m_{1,1} = \left\lvert S_{1, 1} \right\rvert = (9 \times 7 - 16 \times 6) = -33$
$\therefore$ The cofactor of $a_{1, 1}$ is $(-1)^{1 + 1} \cdot \left\lvert S_{1, 1} \right\rvert = -33$
Let us denote the cofactor of any element $a_{i, j}$ as $c_{i, j}$, which gives us $c_{1,1} = -33$
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For $a_{1,2} = 3$ , eliminating the $1\xasuper{st}$ row and the $2\xasuper{nd}$ column we get:
$S_{1, 2} = \begin{bmatrix} 4 & 16 \\ 5 & 7\end{bmatrix}$
$\therefore c_{1,2} = (-1)^{1 + 2} \cdot \left\lvert S_{1, 2} \right\rvert = (-1)^{1 + 2}(4 \times 7 - 16 \times 5) = 52$
Repeating the same steps for all the remaining elements, we get:
$c_{1,3} = (-1)^{1 + 3} \cdot \left\lvert S_{1, 3} \right\rvert = (-1)^{1 + 3}(4 \times 6 - 9 \times 5) = -21$
$c_{2,1} = (-1)^{2 + 1} \cdot \left\lvert S_{2, 1} \right\rvert = (-1)^{2 + 1}(3 \times 7 - 4 \times 6) = 3$
$c_{2,2} = (-1)^{2 + 2} \cdot \left\lvert S_{2, 2} \right\rvert = (-1)^{2 + 2}(2 \times 7 - 4 \times 5) = -6$
$c_{2,3} = (-1)^{2 + 3} \cdot \left\lvert S_{2, 3} \right\rvert = (-1)^{2 + 3}(2 \times 6 - 3 \times 5) = 3$
$c_{3,1} = (-1)^{3 + 1} \cdot \left\lvert S_{3, 1} \right\rvert = (-1)^{3 + 1}(3 \times 16 - 4 \times 9) = -12$
$c_{3,2} = (-1)^{3 + 2} \cdot \left\lvert S_{3, 2} \right\rvert = (-1)^{3 + 2}(2 \times 16 - 4 \times 4) = -16$
$c_{3,3} = (-1)^{3 + 3} \cdot \left\lvert S_{3, 3} \right\rvert = (-1)^{3 + 3}(2 \times 16 - 4 \times 4) = 16$
If we replace each element $a_{i,j}$ of the matrix by the corresponding cofactor $c_{i,j}$ of that element, we get the following $3 \times 3$ matrix.
$C = \begin{bmatrix} -33 & 52 & -21 \\ 3 & -6 & 3 \\ -12 & -16 & 16 \end{bmatrix}$
This is called the cofactor matrix of $A$.
The transpose of a cofactor matrix is called the adjoint matrix, and is denoted by $adj(A)$ where $A$ is the original square matrix. This is also known as adjugate or adjunct matrix.
Therefore, in our above example:
$adj(A) = C^T = \begin{bmatrix} -33 & 3 & -12 \\ 52 & -6 & -16 \\ -21 & 3 & 16 \end{bmatrix}$
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II. Properties Of Adjoint
$i)\ adj (I) = I$ where $I$ is the identity matrix.
$ii)\ adj (kA) = k^{n-1} adj (A)$ where $n$ is the order of $A$ and $k$ is a scalar quantity.
Let us consider a matrix $A$ of order $n$, and let $C$ be the cofactor matrix of $A$
$C = \begin{bmatrix}c_{1,1} & c_{1,2} & \cdots & c_{1,n} \\ c_{2,1} & c_{2,2} & \cdots & c_{2,n} \\ \vdots & \vdots & \ddots & \vdots \\ c_{n,1} & c_{n,2} & \cdots & c_{n,n} \end{bmatrix}$
$= \begin{bmatrix}(-1)^{1 + 1}|S_{1,1}| & (-1)^{1 + 2}|S_{1,2}| & \cdots & (-1)^{1 + n}|S_{1,n}| \\ (-1)^{2 + 1}|S_{2,1}| & (-1)^{2 + 2}|S_{2,2}| & \cdots & (-1)^{2 + n}|S_{2,n}| \\ \vdots & \vdots & \ddots & \vdots \\ (-1)^{n + 1}|S_{n,1}| & (-1)^{n + 2}|S_{n,2}| & \cdots & (-1)^{n + n}|S_{n,n}| \end{bmatrix}$
where, $S_{ij}$ is the submatrix of $A$ obtained by eliminating the $i \xasuper{th}$ row and $j \xasuper{th}$ column from $A$ and has an order of $n-1$
When we multiply each element of $A$ by $k$, each element of $S_{ij}$ also gets multiplied by $k$.
Using the properties of determinants, since $S_{ij}$ has an order of $n - 1$
$|kS_{ij}| = k^{n-1} |S_{ij}|$
Therefore,
$S'_{ij = } = |kS_{ij}| = k^{n-1} |S_{ij}|$ where $S'_{ij}$ is a submatrix of $kA$
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Let $C'$ be the cofactor matrix of $kA$, then each of its elements,
$c'_{ij} = (-1)^{i + j} |S'_{ij}| = (-1)^{i + j} |kS_{ij}| = k^{n - 1} \cdot (-1)^{i + j} |S_{ij}| = k^{n-1} c_{ij}$
$C' = \begin{bmatrix}k^{n-1} c_{1,1} & k^{n-1}c_{1,2} & \cdots & k^{n-1}c_{1,n} \\ k^{n-1} c_{2,1} & k^{n-1} c_{2,2} & \cdots & k^{n-1} c_{2,n} \\ \vdots & \vdots & \ddots & \vdots \\ k^{n-1} c_{n,1} & k^{n-1} c_{n,2} & \cdots & k^{n-1} c_{n,n} \end{bmatrix}$
$= k^{n-1} \begin{bmatrix}c_{1,1} & c_{1,2} & \cdots & c_{1,n} \\ c_{2,1} & c_{2,2} & \cdots & c_{2,n} \\ \vdots & \vdots & \ddots & \vdots \\ c_{n,1} & c_{n,2} & \cdots & c_{n,n} \end{bmatrix} = k^{n-1}C$
Therefore,
$adj(kA) = (C')^T = (k^{n - 1}C)^T = k^{n-1} C^T = k^{n-1} adj(A)$
$iii)\ adj (A^T) = [adj (A)]^T$
For this proof we will use $C_A$ to denote the cofactor matrix $C$ of $A$.
If we consider the transpose $T$ of a matrix $A$, then $t_{ji} = a_{ij}$.
Therefore, each submatrix $S'_{ji}$ of $T$ is the transpose of the corresponding submatrix $S_{ij}$ of $A$.
Therefore, $|S'_{ji}| = |S_{ij}|$ and $C_{A^T} = (C_A)^T$
$\therefore adj(A^T) = (C_{A^T})^T = [(C_A)^T]^T = [adj(A)]^T$
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$iv)\ det (adj(A)) = (det (A))^{n - 1}$
$v)$ If $A$ and $B$ are two square matrices of the same order, then
$adj(AB) = adj(B)adj(A)$
III. Inverse Of Matrices
Another very important property of adjoints of matrices is:
$adj(A) \times A = A \times adj(A) = det(A) \cdot I_n$ where $n$ is the order of $A$ and $det(A) \ne 0$
For $\det(A) \ne 0$, dividing the above set of equations by $\det(A)$, we get:
$\dfrac{adj(A)}{det(A)} \times A = A \times \dfrac{adj(A)}{det(A)} = I_n$
Now we have a matrix which, when multiplied by the original matrix we get the unit matrix of the same order.
Therefore, the matrix $\dfrac{1}{det(A)} \cdot adj(A)$ is called the inverse of $A$ and is denoted by $A^{-1}$.
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IV. Properties Of Inverses
(Note that all the following properties of inverses apply only if the matrix has a non-zero determinant.)
i) Inverting a matrix twice gives the original matrix, that is $(A^{-1})^{-1} = A$
ii) For any scalar $k \ne 0$, $(kA)^{-1} = k^{-1}A^{-1}$
iii) Inverse of a transpose is equal to the transpose of inverse, that is, $(A^T)^{-1} = (A^{-1})^T$
iv) $det (A^{-1}) = \dfrac{1}{det(A)}$
v) $(AB)^{-1} = B^{-1}A^{-1}$. In general, $\left(\prod\limits_{i = 1}^{n} A_i \right)^{-1} = \prod\limits_{i = n}^{1} A_i^{-1}$