Find the inverse of the following, if it exists, using elementary row (column) transformations :
`[(2,0,-1),(5,1,0),(0,1,3)]`

Answer

To find the inverse of the matrix \( A = \begin{pmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{pmatrix} \) using elementary row transformations, we will follow these steps:

### Step 1: Form the augmented matrix
We start by forming the augmented matrix \([A | I]\), where \(I\) is the identity matrix of the same order (3x3):
\[ \begin{pmatrix} 2 & 0 & -1 & | & 1 & 0 & 0 \\ 5 & 1 & 0 & | & 0 & 1 & 0 \\ 0 & 1 & 3 & | & 0 & 0 & 1 \end{pmatrix} \]

### Step 2: Make the leading coefficient of the first row equal to 1
To make the leading coefficient of the first row (which is currently 2) equal to 1, we can multiply the first row by \(\frac{1}{2}\):
\[ R_1 \rightarrow \frac{1}{2} R_1 \]
This gives us:
\[ \begin{pmatrix} 1 & 0 & -\frac{1}{2} & | & \frac{1}{2} & 0 & 0 \\ 5 & 1 & 0 & | & 0 & 1 & 0 \\ 0 & 1 & 3 & | & 0 & 0 & 1 \end{pmatrix} \]

### Step 3: Eliminate the first column below the leading 1
Next, we will eliminate the entries below the leading 1 in the first column. We can do this by replacing \(R_2\) and \(R_3\):
\[ R_2 \rightarrow R_2 - 5R_1 \quad \text{and} \quad R_3 \rightarrow R_3 - 0R_1 \]
After performing these operations, we have:
\[ \begin{pmatrix} 1 & 0 & -\frac{1}{2} & | & \frac{1}{2} & 0 & 0 \\ 0 & 1 & \frac{5}{2} & | & -\frac{5}{2} & 1 & 0 \\ 0 & 1 & 3 & | & 0 & 0 & 1 \end{pmatrix} \]

### Step 4: Make the leading coefficient of the second row equal to 1
The leading coefficient of the second row is already 1. Now we will eliminate the entry in the second column of the third row:
\[ R_3 \rightarrow R_3 - R_2 \]
This gives us:
\[ \begin{pmatrix} 1 & 0 & -\frac{1}{2} & | & \frac{1}{2} & 0 & 0 \\ 0 & 1 & \frac{5}{2} & | & -\frac{5}{2} & 1 & 0 \\ 0 & 0 & \frac{1}{2} & | & \frac{5}{2} & -1 & 1 \end{pmatrix} \]

### Step 5: Make the leading coefficient of the third row equal to 1
To make the leading coefficient of the third row equal to 1, we multiply the third row by 2:
\[ R_3 \rightarrow 2R_3 \]
This results in:
\[ \begin{pmatrix} 1 & 0 & -\frac{1}{2} & | & \frac{1}{2} & 0 & 0 \\ 0 & 1 & \frac{5}{2} & | & -\frac{5}{2} & 1 & 0 \\ 0 & 0 & 1 & | & 5 & -2 & 2 \end{pmatrix} \]

### Step 6: Eliminate entries above the leading 1 in the third row
Now we will eliminate the entries above the leading 1 in the third row:
\[ R_1 \rightarrow R_1 + \frac{1}{2}R_3 \quad \text{and} \quad R_2 \rightarrow R_2 - \frac{5}{2}R_3 \]
This gives us:
\[ \begin{pmatrix} 1 & 0 & 0 & | & 3 & -1 & 1 \\ 0 & 1 & 0 & | & -15 & 6 & -5 \\ 0 & 0 & 1 & | & 5 & -2 & 2 \end{pmatrix} \]

### Step 7: Read the inverse from the augmented matrix
The right side of the augmented matrix now represents the inverse of \(A\):
\[ A^{-1} = \begin{pmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{pmatrix} \]

### Final Answer
Thus, the inverse of the matrix \( A \) is:
\[ A^{-1} = \begin{pmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{pmatrix} \]