Title: Matrix Inverse
Course: Demonstrating STACK (v4.3.7)

Let \({\bf A}\) be the matrix

 \[{\bf A}=[value]\]

Find the inverse of \({\bf A}\), if it is invertible. If \({\bf A}\) is not invertible, return the two by two zero matrix.
Feedback: The inverse of a 2 by 2 matrix \[[value]\] is \[\frac{1}{ad-bc}[value]\] Therefore the inverse of matrix \({\bf A}\) = \([value]\) is \[\frac{1}{([value]).([value]) -([value]).([value])}[value] = \frac{1}{([value]) -([value])}[value] =[value]\]

Let \({\bf A}\) be the matrix

 \[{\bf A}=[value]\]

Find the inverse of \({\bf A}\), if it is invertible. If \({\bf A}\) is not invertible, return the two by two zero matrix.
Feedback: The inverse of a 2 by 2 matrix \[[value]\] is \[\frac{1}{ad-bc}[value]\] If ad-bc = 0 the matrix cannot be inverted For matrix \({\bf A}\) = \([value]\) \[ad-bc =([value]).([value]) -([value]).([value]) = ([value]) -([value]) = 0.\] Therefore matrix \({\bf A}\) is not invertible.

Let \({\bf A}\) be the matrix

 \[{\bf A}=[value]\]

Find the inverse of \({\bf A}\), if it is invertible. If \({\bf A}\) is not invertible, return the zero matrix \({\bf O}\).
Feedback: Augment matrix \({\bf A}\) with a 3 by 3 identity matrix: \[[value]\] There is +1 in the upper left-hand corner so create 0's below it by adding \([value]\) times row 1 to row 2 and \([value]\) times row 1 to row 3 \[[value]\] Now there is +1 in entry [2,2] of the matrix so create 0's above and below it by adding \([value]\) times row 2 to row 1 and \([value]\) times row 2 to row 3 \[[value]\] Now there is +1 in entry [3,3] of the matrix so create 0's above it by adding \([value]\) times row 3 to row 1 and \([value]\) times row 3 to row 2 \[[value]\] The left half of this matrix is the identity matrix and the right half is the inverse of \({\bf A}\) Therefore \({\bf A^{-1}} = [value]\)

Let \({\bf A}\) be the matrix

 \[{\bf A}=[value]\]

Find the inverse of \({\bf A}\), if it is invertible. If \({\bf A}\) is not invertible, return the zero matrix \({\bf O}\).
Feedback: A matrix is not invertible if its determinant is zero. For matrix \({\bf A}\) = \([value]\) it can be seen that row 2 is \([value]\) times row 1. Therefore the determinant of \({\bf A}\) is zero and the matrix is not invertible.