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A right inverse is a matrix that, when multiplied by the given matrix on the right, results in the identity matrix. Essentially, if you have a matrix \( A \), a right inverse matrix \( B \) must satisfy the equation: \( A \cdot B = I \), where \( I \) is the identity matrix. This identity matrix is specific to the resulting dimensions of the operation.
In the original exercise, it is shown that \( \begin{bmatrix} 1 & 1 \end{bmatrix} \cdot \begin{bmatrix} 2 \ -1 \end{bmatrix} = \begin{bmatrix} 1 \end{bmatrix} \), which is indeed an identity matrix of size 1x1. Thus, \( \begin{bmatrix} 2 \ -1 \end{bmatrix} \) acts as a right inverse for \( \begin{bmatrix} 1 & 1 \end{bmatrix} \).
Right inverses are particularly useful in solving matrix equations where multiplication order matters. However, a matrix might have multiple right inverses or none at all, especially if it has more rows than columns.

A left inverse is a concept similar to a right inverse, but it involves pre-multiplying the given matrix. If you have a matrix \( A \), a left inverse matrix \( B \) must satisfy the equation: \( B \cdot A = I \). Here, \( I \) is again the identity matrix that matches the dimensions of the operation.
In the problem, they demonstrated that \( \begin{bmatrix} 2 \ -1 \end{bmatrix} \cdot \begin{bmatrix} 1 & 1 \end{bmatrix} = \begin{bmatrix} 2 & -1 \end{bmatrix} \), which is not an identity matrix of any size. Therefore, \( \begin{bmatrix} 2 \ -1 \end{bmatrix} \) is not a left inverse for \( \begin{bmatrix} 1 & 1 \end{bmatrix} \).
Finding a left inverse is generally applicable to cases where a matrix has more columns than rows. A critical takeaway is that a matrix can only have a left inverse if its columns are linearly independent.

The identity matrix is a foundational concept in understanding matrix inverses, whether left or right. It is a square matrix with ones on the diagonal and zeros elsewhere. Multiplying any matrix by the identity matrix of compatible size leaves the original matrix unchanged. This is similar to multiplying a number by 1 in arithmetic.
Consider the identity matrix \( I \), where for any matrix \( A \), the equation \( A \cdot I = A \) and \( I \cdot A = A \) hold true. The exercise shows how achieving the identity matrix is necessary when finding matrix inverses.
For an identity matrix of size 1x1, like \( \begin{bmatrix} 1 \end{bmatrix} \), its role is crucial in assessing whether a matrix is an inverse. In matrices where the sizes aren't square, an identity matrix can't be achieved, which means an inverse (either left or right) is nonexistent. This is evident in the part of the exercise where \( \begin{bmatrix} 1 & 1 \end{bmatrix} \) cannot have a left inverse because it doesn't result in an appropriate identity matrix when multiplied by any other matrix.