Examples of Methods to Find Inverse of a Matrix

Example 1: Find the inverse of the matrix P = [Tex]\begin {bmatrix} 2 & 3 \\ 5 & 1 \end{bmatrix}[/Tex] by Direct Method.

Solution:

We can find the inverse of matrix P by following formula

P^-1 = [Tex]\bold{\frac{1}{ad – bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}}[/Tex]

P^-1 = [Tex]\frac{1}{(2\times1 )- (3\times 5)}\begin{bmatrix} 1 & -3 \\ -5 & 2 \end{bmatrix}[/Tex]

P^-1 = [Tex]\frac{1}{2- 15}\begin{bmatrix} 1 & -3 \\ -5 & 2 \end{bmatrix}[/Tex]

P^-1 = [Tex]\frac{1}{- 13}\begin{bmatrix} 1 & -3 \\ -5 & 2 \end{bmatrix}[/Tex]

Example 2: Find the inverse of matrix Q = [Tex]\begin {bmatrix} 1 & 0 & 4 \\ 6& 1 & 0\\ 5&2&3 \end{bmatrix}[/Tex] using inverse matrix formula.

Solution:

We can find the inverse of matrix Q by following formula.

Q^-1 = adj(Q) / |Q|

First, we find |Q|.

|Q| =[Tex] \begin {vmatrix} 1 & 0 & 4 \\ 6& 1 & 0\\ 5&2&3 \end{vmatrix} [/Tex]

|Q| = 1[Tex] \begin {vmatrix} 1 & 0\\ 2&3 \end{vmatrix} [/Tex] – 0 [Tex] \begin {vmatrix} 6 & 0\\ 5&3 \end{vmatrix} [/Tex]+ 4 [Tex] \begin {vmatrix} 6& 1 \\ 5&2 \end{vmatrix} [/Tex]

|Q| = 1[3 – 0] – 0 + 4[12 – 5]

|Q| = 1(3) + 4(7)

|Q| = 3 + 28

|Q| = 31

Now we will find adj(Q)

To find adj(Q) we find the cofactor matrix of Q as

adj Q = Transpose of cofactor matrix of Q

C_ij = (-1)^{i + j} Mij

where, M_ij is minor.

Cofactor(Q) = [Tex] \begin {bmatrix}(-1)^{1+1}\begin {vmatrix} 1 & 0\\ 2&3\\ \end{vmatrix} (-1)^{1+2}\begin {vmatrix} 6 & 0\\ 5&3\\ \end{vmatrix} (-1)^{1+3}\begin {vmatrix} 1 & 0\\ 2&3\\ \end{vmatrix} \\ \\ (-1)^{2+1}\begin {vmatrix} 0 & 4\\ 2&3\\ \end{vmatrix} (-1)^{2+2}\begin {vmatrix} 1 & 4\\ 5&3\\ \end{vmatrix} (-1)^{2+3}\begin {vmatrix} 1 & 0\\ 5&2\\ \end{vmatrix} \\ \\ (-1)^{3+1}\begin {vmatrix} 0 & 4\\ 1&0\\ \end{vmatrix} (-1)^{3+2}\begin {vmatrix} 1 & 4\\ 6&0\\ \end{vmatrix} (-1)^{3+3}\begin {vmatrix} 1 & 0\\ 6&1\\ \end{vmatrix} \\ \\ \end {bmatrix}[/Tex]

On solving above matrix we get

Cofactor of Q = [Tex]\begin {bmatrix} 3 & -18 & 7 \\ 8& -17 & -2\\ -4&24&1 \end{bmatrix} [/Tex]

adj(Q) = [Cofactor(Q)] ^T

adj(Q) = [Tex]\begin {bmatrix} 3 & -18 & 7 \\ 8& -17 & -2\\ -4&24&1 \end{bmatrix}^T[/Tex]

adj(Q) = [Tex]\begin {bmatrix} 3 & 8 & -4 \\ -1 8& -17 & 24\\ 7&-2&1 \end{bmatrix} [/Tex]

So, the inverse of matrix Q is given by

Q^-1 = (1 / 31) [Tex]\begin {bmatrix} 3 & 8 & -4 \\ -1 8& -17 & 24\\ 7&-2&1 \end{bmatrix} [/Tex]

Example 3: Find the inverse of matrix V = [Tex]\begin {bmatrix} 6 & 2 & 3 \\ 0 & 1 & 4\\ 5 &0 & 2 \end{bmatrix}[/Tex] by elementary transformations.

Solution:

To find the inverse of the matrix V = [Tex]\begin {bmatrix} 6 & 2 & 3 \\ 0 & 1 & 4\\ 5&0 & 2 \end{bmatrix}[/Tex] we will use row operation.

V = IV

[Tex]\begin {bmatrix} 6 & 2 & 3 \\ 0 & 1 & 4\\ 5&0 & 2 \end{bmatrix}[/Tex] = [Tex]\begin {bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0\\ 0&0 & 1 \end{bmatrix}[/Tex] [Tex]\begin {bmatrix} 6 & 2 & 3 \\ 0 & 1 & 4\\ 5&0 & 2 \end{bmatrix}[/Tex]

R1 ← R1 – R3

[Tex]\begin {bmatrix} 1 & 2 & 1 \\ 0 & 1 & 4\\ 5&0 & 2 \end{bmatrix}[/Tex] = [Tex]\begin {bmatrix} 1 & 0 & -1 \\ 0 & 1 & 0\\ 0&0 & 1 \end{bmatrix}[/Tex] V

R3 ← R3 – 5R1

[Tex]\begin {bmatrix} 1 & 2 & 1 \\ 0 & 1 & 4\\ 4&-10 &-3 \end{bmatrix}[/Tex] = [Tex]\begin {bmatrix} 1 & 0 & -1 \\ 0 & 1 & 0\\ -5&0 & 6 \end{bmatrix}[/Tex] V

R1← R1 – 2R2

[Tex]\begin {bmatrix} 1 & 0 & -7 \\ 0 & 1 & 4\\ 0&-10 & -3 \end{bmatrix}[/Tex] = [Tex]\begin {bmatrix} 1 & -2 & -1 \\ 0 & 1 & 0\\ -5&0 & 6 \end{bmatrix}[/Tex] V

R3 ← R3 + 10R1

[Tex]\begin {bmatrix} 1 & 0 & -7 \\ 0 & 1 & 4\\ 0&0 & 37 \end{bmatrix}[/Tex] = [Tex]\begin {bmatrix} 1 & -2 & -1 \\ 0 & 1 & 0\\ -5&10 & 6 \end{bmatrix}[/Tex] V

R3 ← R3/37

[Tex]\begin {bmatrix} 1 & 0 & -7 \\ 0 & 1 & 4\\ 0&0 & 1 \end{bmatrix}[/Tex] = [Tex]\begin {bmatrix} 1 & -2 & -1 \\ 0 & 1 & 0\\ -5/37&10/37 & 6/37 \end{bmatrix}[/Tex] V

R1 ← R1 + 7R3

[Tex]\begin {bmatrix} 1 & 0 & 0 \\ 0 & 1 & 4\\ 0&0 & 1 \end{bmatrix}[/Tex] = [Tex]\begin {bmatrix} 2/37 & -4/37 & 5/37 \\ 0 & 1 & 0\\ -5/37&10/37 & 6/37 \end{bmatrix}[/Tex] V

R2 ← R2 – 4R3

[Tex]\begin {bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0\\ 0&0 & 1 \end{bmatrix}[/Tex] = [Tex]\begin {bmatrix} 2/37 & -4/37 & 5/37 \\ 9/37 & -3/37 & -24/37\\ -5/37&10/37 & 6/37 \end{bmatrix}[/Tex] V

Since, the above expression is of form I = BV

Inverse of matrix V = [Tex]\begin {bmatrix} 2/37 & -4/37 & 5/37 \\ 9/37 & -3/37 & -24/37\\ -5/37&10/37 & 6/37 \end{bmatrix}[/Tex]

Methods to find the inverse of a matrix involve the inverse of a matrix formula and by elementary operations. The inverse of matrix A is represented as A^-1 which when multiplied by matrix A gives an identity matrix.

In this article, we will explore different methods to find the inverse of a matrix in detail along with the inverse of matrix definition and inverse of matrix properties.