Adjoint and Inverse of a Matrix

Adjoint of a Matrix

The adjoint of a square matrix $A = [a_{ij}]_{n \times n}$ is defined as the transpose of the cofactor matrix of $A$. If $A_{ij}$ denotes the cofactor of the element $a_{ij}$, then the adjoint of $A$ is written as $\operatorname{adj} A$.

For a $3 \times 3$ matrix, $$A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix},$$ its cofactor matrix is $$\begin{bmatrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{bmatrix},$$ and therefore $$\operatorname{adj} A = \left(\begin{bmatrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{bmatrix}\right)^T = \begin{bmatrix} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{bmatrix}.$$

So, to find the adjoint of a matrix, we follow these steps:

1. Find the minor of each element.
2. Convert each minor into a cofactor using the sign rule $(-1)^{i+j}$.
3. Form the cofactor matrix.
4. Take its transpose.

Shortcut for $2 \times 2$ matrices: If $$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix},$$ then $$\operatorname{adj} A = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}.$$ That is, interchange the diagonal elements and change the signs of the off-diagonal elements.

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Singular and Non-Singular Matrices

A square matrix is classified using its determinant.

1. Singular Matrix

A square matrix $A$ is called singular if $$|A| = 0.$$ A singular matrix does not have an inverse.

2. Non-Singular Matrix

A square matrix $A$ is called non-singular if $$|A| \neq 0.$$ A non-singular matrix always has an inverse.

Important Results

1. If $A$ and $B$ are square matrices of the same order, then $$|AB| = |A||B|.$$
2. If $A$ and $B$ are both non-singular, then $AB$ and $BA$ are also non-singular because $$|AB| = |A||B| \neq 0.$$
3. If either $A$ or $B$ is singular, then the product may become singular because its determinant may become zero.

These ideas are important while studying inverse, matrix equations, and properties of matrix products.

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Properties of Adjoint

The adjoint of a matrix satisfies some very important properties.

Theorem 1: If $A$ is any square matrix of order $n$, then $$A(\operatorname{adj} A) = (\operatorname{adj} A)A = |A|I,$$ where $I$ is the identity matrix of the same order.

This result is extremely important because it directly leads to the inverse formula for a non-singular matrix. If $|A| \neq 0$, then dividing both sides by $|A|$ gives $$A\left(\frac{1}{|A|}\operatorname{adj} A\right)=I,$$ which shows that $$A^{-1}=\frac{1}{|A|}\operatorname{adj} A.$$

Theorem 2: If $A$ is a non-singular square matrix of order $n$, then $$|\operatorname{adj} A| = |A|^{n-1}.$$

For example, if $A$ is a $3 \times 3$ matrix and $|A|=k$, then $$|\operatorname{adj} A| = k^2.$$ This is a very frequently used result in determinant-based problems.

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Inverse of a Matrix

A square matrix $A$ is invertible if and only if it is non-singular, that is, $$|A| \neq 0.$$ The inverse of $A$, denoted by $A^{-1}$, is defined as the matrix satisfying $$AA^{-1} = A^{-1}A = I.$$ For a non-singular matrix, $$A^{-1} = \frac{1}{|A|} \operatorname{adj} A.$$

Properties of Inverse

1. $$\left(A^{-1}\right)^{-1} = A$$
2. $$(AB)^{-1} = B^{-1}A^{-1}$$ This is called the reversal law.
3. $$(A^T)^{-1} = (A^{-1})^T$$
4. $$|A^{-1}| = \frac{1}{|A|}$$

These properties are valid whenever the inverse exists, that is, whenever $A$ is non-singular.

Example 1: Adjoint of a $2 \times 2$ Matrix

Find the adjoint of the matrix $A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}$.

Solution: For a $2 \times 2$ matrix $$\begin{bmatrix} a & b \\ c & d \end{bmatrix},$$ we use the shortcut $$\operatorname{adj} A = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}.$$ Here, $a=2$, $b=3$, $c=1$, and $d=4$. So, $$\operatorname{adj} A = \begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix}.$$

Verification using cofactors: $$A_{11}=4, \quad A_{12}=-1, \quad A_{21}=-3, \quad A_{22}=2.$$ So the cofactor matrix is $$\begin{bmatrix} 4 & -1 \\ -3 & 2 \end{bmatrix},$$ and its transpose is $$\begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix}.$$ Hence the result is correct.

Answer: $$\operatorname{adj} A = \begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix}.$$

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Example 2: Checking Singularity

Show that the matrix $A = \begin{bmatrix} 1 & 2 \\ 3 & 6 \end{bmatrix}$ is a singular matrix.

Solution: A matrix is singular if its determinant is zero. So, $$|A| = \begin{vmatrix} 1 & 2 \\ 3 & 6 \end{vmatrix} = (1)(6) - (3)(2) = 6 - 6 = 0.$$ Since the determinant is zero, the matrix is singular.

Answer: Proved that $A$ is singular.

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Example 3: Inverse of a $2 \times 2$ Matrix

Find the inverse of the matrix $A = \begin{bmatrix} 2 & -2 \\ 4 & 3 \end{bmatrix}$, if it exists.

Solution: First compute the determinant: $$|A| = \begin{vmatrix} 2 & -2 \\ 4 & 3 \end{vmatrix} = (2)(3) - (4)(-2) = 6 + 8 = 14.$$ Since $|A| \neq 0$, the inverse exists.

Now find the adjoint. For a $2 \times 2$ matrix, $$\operatorname{adj} A = \begin{bmatrix} 3 & 2 \\ -4 & 2 \end{bmatrix}.$$ Therefore, $$A^{-1} = \frac{1}{|A|} \operatorname{adj} A = \frac{1}{14}\begin{bmatrix} 3 & 2 \\ -4 & 2 \end{bmatrix}.$$ We may also write it as $$A^{-1} = \begin{bmatrix} 3/14 & 1/7 \\ -2/7 & 1/7 \end{bmatrix}.$$

Answer: $$A^{-1} = \frac{1}{14}\begin{bmatrix} 3 & 2 \\ -4 & 2 \end{bmatrix}.$$

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Example 4: Determinant of Adjoint

If $A$ is a square matrix of order $3$ such that $|A| = 5$, find the value of $|\operatorname{adj} A|$.

Solution: For a non-singular matrix of order $n$, $$|\operatorname{adj} A| = |A|^{n-1}.$$ Here $n=3$ and $|A|=5$. Thus, $$|\operatorname{adj} A| = 5^{3-1} = 5^2 = 25.$$

Answer: $$|\operatorname{adj} A| = 25.$$

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Example 5: Finding Adjoint of a $3 \times 3$ Matrix

Find the adjoint of $A = \begin{bmatrix} 1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1 \end{bmatrix}$.

Solution: First find all cofactors.

$$A_{11} = (+1)\begin{vmatrix} 3 & 5 \\ 0 & 1 \end{vmatrix} = 3,$$ $$A_{12} = (-1)\begin{vmatrix} 2 & 5 \\ -2 & 1 \end{vmatrix} = -\big(2 - (-10)\big) = -12,$$ $$A_{13} = (+1)\begin{vmatrix} 2 & 3 \\ -2 & 0 \end{vmatrix} = 0 - (-6) = 6,$$ $$A_{21} = (-1)\begin{vmatrix} -1 & 2 \\ 0 & 1 \end{vmatrix} = -(-1)=1,$$ $$A_{22} = (+1)\begin{vmatrix} 1 & 2 \\ -2 & 1 \end{vmatrix} = 1 - (-4)=5,$$ $$A_{23} = (-1)\begin{vmatrix} 1 & -1 \\ -2 & 0 \end{vmatrix} = -\big(0 - 2\big)=2,$$ $$A_{31} = (+1)\begin{vmatrix} -1 & 2 \\ 3 & 5 \end{vmatrix} = -5 - 6 = -11,$$ $$A_{32} = (-1)\begin{vmatrix} 1 & 2 \\ 2 & 5 \end{vmatrix} = -(5-4)=-1,$$ $$A_{33} = (+1)\begin{vmatrix} 1 & -1 \\ 2 & 3 \end{vmatrix} = 3 - (-2)=5.$$

Hence the cofactor matrix is $$\begin{bmatrix} 3 & -12 & 6 \\ 1 & 5 & 2 \\ -11 & -1 & 5 \end{bmatrix}.$$ Therefore, the adjoint is its transpose: $$\operatorname{adj} A = \begin{bmatrix} 3 & 1 & -11 \\ -12 & 5 & -1 \\ 6 & 2 & 5 \end{bmatrix}.$$

Answer: $$\operatorname{adj} A = \begin{bmatrix} 3 & 1 & -11 \\ -12 & 5 & -1 \\ 6 & 2 & 5 \end{bmatrix}.$$

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Example 6: Inverse of a $3 \times 3$ Matrix

Find the inverse of the matrix $A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{bmatrix}$.

Solution: First find the determinant. Expanding along the first row, $$|A| = 1\begin{vmatrix} 1 & 4 \\ 6 & 0 \end{vmatrix} - 2\begin{vmatrix} 0 & 4 \\ 5 & 0 \end{vmatrix} + 3\begin{vmatrix} 0 & 1 \\ 5 & 6 \end{vmatrix}.$$ So, $$|A| = 1(0-24) - 2(0-20) + 3(0-5) = -24 + 40 - 15 = 1.$$ Since $|A|=1 \neq 0$, the inverse exists.

Now compute cofactors: $$A_{11}=-24, \quad A_{12}=20, \quad A_{13}=-5,$$ $$A_{21}=18, \quad A_{22}=-15, \quad A_{23}=4,$$ $$A_{31}=5, \quad A_{32}=-4, \quad A_{33}=1.$$ Thus the adjoint is $$\operatorname{adj} A = \begin{bmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{bmatrix}.$$ Finally, $$A^{-1} = \frac{1}{|A|}\operatorname{adj} A = \operatorname{adj} A,$$ because $|A|=1$. So, $$A^{-1} = \begin{bmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{bmatrix}.$$

Answer: $$A^{-1} = \begin{bmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{bmatrix}.$$

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Example 7: Finding Unknown in a Singular Matrix

If the matrix $A = \begin{bmatrix} 2 & x \\ -3 & 6 \end{bmatrix}$ is singular, find the value of $x$.

Solution: Since the matrix is singular, its determinant must be zero: $$\begin{vmatrix} 2 & x \\ -3 & 6 \end{vmatrix} = 0.$$ Now evaluate: $$(2)(6) - (-3)(x) = 0,$$ $$12 + 3x = 0,$$ $$3x = -12,$$ $$x = -4.$$

Answer: $$x=-4.$$

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Example 8: Verifying Inverse Reversal Law

If $A = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}$ and $B = \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix}$, verify that $(AB)^{-1} = B^{-1}A^{-1}$.

Solution: First compute determinants: $$|A| = 3\cdot 5 - 7\cdot 2 = 15 - 14 = 1,$$ $$|B| = 6\cdot 9 - 8\cdot 7 = 54 - 56 = -2.$$ Both are non-zero, so both matrices are invertible.

Now, $$A^{-1} = \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}$$ because $|A|=1$. Also, $$\operatorname{adj} B = \begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix},$$ so $$B^{-1} = -\frac{1}{2}\begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix} = \begin{bmatrix} -9/2 & 4 \\ 7/2 & -3 \end{bmatrix}.$$

Now calculate the right-hand side: $$B^{-1}A^{-1} = \begin{bmatrix} -9/2 & 4 \\ 7/2 & -3 \end{bmatrix}\begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix} = \begin{bmatrix} -61/2 & 87/2 \\ 47/2 & -67/2 \end{bmatrix}.$$

Next compute $AB$: $$AB = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}\begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix} = \begin{bmatrix} 67 & 87 \\ 47 & 61 \end{bmatrix}.$$ Then, $$|AB| = 67\cdot 61 - 87\cdot 47 = 4087 - 4089 = -2.$$ Hence, $$\operatorname{adj}(AB) = \begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix},$$ and $$(AB)^{-1} = -\frac{1}{2}\begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix} = \begin{bmatrix} -61/2 & 87/2 \\ 47/2 & -67/2 \end{bmatrix}.$$ Thus, $$(AB)^{-1} = B^{-1}A^{-1}.$$

Answer: Verified.

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Example 9: Matrix Polynomial to Find Inverse

Show that $A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}$ satisfies the equation $A^2 - 4A + I = O$, where $I$ is the identity matrix and $O$ is the zero matrix. Using this equation, find $A^{-1}$.

Solution: First compute $$A^2 = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}\begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix}.$$ Now, $$-4A = \begin{bmatrix} -8 & -12 \\ -4 & -8 \end{bmatrix}.$$ So, $$A^2 - 4A + I = \begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix} + \begin{bmatrix} -8 & -12 \\ -4 & -8 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}.$$ Thus the equation is verified.

Now multiply $$A^2 - 4A + I = O$$ by $A^{-1}$: $$(A^2 - 4A + I)A^{-1} = O.$$ This gives $$A - 4I + A^{-1} = O,$$ so $$A^{-1} = 4I - A.$$ Hence, $$A^{-1} = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} - \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}.$$

Answer: $$A^{-1} = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}.$$

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Example 10: Inverse of a Diagonal Matrix

Find the inverse of the diagonal matrix $D = \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix}$, where $x, y, z$ are non-zero.

Solution: For a diagonal matrix, $$|D| = xyz.$$ Since $x,y,z \neq 0$, the determinant is non-zero, so the inverse exists.

The cofactors are: $$A_{11}=yz, \quad A_{22}=xz, \quad A_{33}=xy,$$ and all off-diagonal cofactors are zero. Thus, $$\operatorname{adj} D = \begin{bmatrix} yz & 0 & 0 \\ 0 & xz & 0 \\ 0 & 0 & xy \end{bmatrix}.$$ Now, $$D^{-1} = \frac{1}{xyz}\begin{bmatrix} yz & 0 & 0 \\ 0 & xz & 0 \\ 0 & 0 & xy \end{bmatrix} = \begin{bmatrix} 1/x & 0 & 0 \\ 0 & 1/y & 0 \\ 0 & 0 & 1/z \end{bmatrix}.$$

Answer: $$D^{-1} = \begin{bmatrix} 1/x & 0 & 0 \\ 0 & 1/y & 0 \\ 0 & 0 & 1/z \end{bmatrix}.$$