Let A be an `mxxn` matrix. If there exists a matrix L of type `nxxm` such that `LA=I_(n)`, then L is called left inverse of A. Similarly, if there exists a matrix R of type `nxxm` such that `AR=I_(m)`, then R is called right inverse of A.
For example, to find right inverse of matrix
`A=[(1,-1),(1,1),(2,3)]`, we take `R=[(x,y,x),(u,v,w)]`
and solve`AR=I_(3)`, i.e.,
`[(1,-1),(1,1),(2,3)][(x,y,z),(u,v,w)]=[(1,0,0),(0,1,0),(0,0,1)]`
`{:(implies,x-u=1,y-v=0,z-w=0),(,x+u=0,y+v=1,z+w=0),(,2x+3u=0,2y+3v=0,2z+3w=1):}`
As this system of equations is inconsistent, we say there is no right inverse for matrix A.
The number of right inverses for the matrix `[(1,-1,2),(2,-1,1)]` is

Options

A. 0
B. 1
C. 2
D. infinite

Explanation

To find the number of right inverses for the matrix \( A = \begin{pmatrix} 1 & -1 & 2 \\ 2 & -1 & 1 \end{pmatrix} \), we will follow these steps:

### Step 1: Define the Right Inverse
Let \( R \) be the right inverse of \( A \). Since \( A \) is a \( 2 \times 3 \) matrix, \( R \) must be a \( 3 \times 2 \) matrix. We can denote \( R \) as:
\[ R = \begin{pmatrix} x & y \\ u & v \\ w & z \end{pmatrix} \]

### Step 2: Set Up the Equation
We need to solve the equation \( AR = I_2 \), where \( I_2 \) is the \( 2 \times 2 \) identity matrix:
\[ AR = \begin{pmatrix} 1 & -1 & 2 \\ 2 & -1 & 1 \end{pmatrix} \begin{pmatrix} x & y \\ u & v \\ w & z \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

### Step 3: Perform the Matrix Multiplication
Calculating \( AR \):
\[ AR = \begin{pmatrix} 1x - 1u + 2w & 1y - 1v + 2z \\ 2x - 1u + 1w & 2y - 1v + 1z \end{pmatrix} \]
This gives us the following equations:
1. \( x - u + 2w = 1 \) (Equation 1)
2. \( y - v + 2z = 0 \) (Equation 2)
3. \( 2x - u + w = 0 \) (Equation 3)
4. \( 2y - v + z = 1 \) (Equation 4)

### Step 4: Solve the System of Equations
We have 4 equations and 6 variables (\( x, y, u, v, w, z \)). This means we can express some variables in terms of others.

From Equation 1:
\[ u = x + 2w - 1 \]

From Equation 3:
\[ u = 2x + w \]

Equating the two expressions for \( u \):
\[ x + 2w - 1 = 2x + w \implies w = 1 - x \]

Substituting \( w \) back into the expression for \( u \):
\[ u = 2x + (1 - x) = x + 1 \]

Now substituting \( u \) into Equation 2:
\[ y - v + 2z = 0 \implies v = y + 2z \]

Substituting \( v \) into Equation 4:
\[ 2y - (y + 2z) + z = 1 \implies y - z = 1 \implies y = z + 1 \]

### Step 5: Count the Free Variables
Now we have:
- \( x \) and \( z \) are free variables.
- \( w = 1 - x \)
- \( u = x + 1 \)
- \( y = z + 1 \)
- \( v = (z + 1) + 2z = 3z + 1 \)

Thus, we can express all variables in terms of \( x \) and \( z \). Since both \( x \) and \( z \) can take any real value, there are infinitely many solutions.

### Conclusion
The number of right inverses for the matrix \( A \) is infinite.