Title: Matrices
Course: Demonstrating STACK (v4.3.7)

Let \({\bf A}\) and \({\bf B}\) be the following matrices: \[{\bf A} = [value]\]
\[{\bf B} = [value]\]

Find the sum \({\bf A + B:}\)
Feedback: \[{\bf A} = [value]\] and \[{\bf B} = [value]\] Therefore \[\begin{array}{rcc} {\bf A} + {\bf B} & =& \begin{bmatrix} [value] +[value] & [value] +[value] \\ [value] +[value] & [value] +[value] \end{bmatrix} \end{array}\] \[=[value]\]

Let \({\bf A}\) and \({\bf B}\) be the following matrices: \[{\bf A} = [value]\]
\[{\bf B} = [value]\] Find the matrix product \({\bf A} \bullet {\bf B}\):
Feedback: \[{\bf A} = [value]\] \[{\bf B} = [value]\] \[\begin{array}{rccc} {\bf A} \bullet {\bf B} & =& [value] \\ & = &[value].\end{array}\]

Let \({\bf A}\) be the following matrix: \[{\bf A} = [value]\]
Find the scalar multiple \([value] {\bf A}\) :
Feedback: Each element of the matrix \({\bf A}\) is multiplied by the scale factor \([value]\) Therefore \([value] {\bf A} = [value]\)

If \[{\bf A}=[value],  {\bf B} = [value],\]

find a 2 by 2 matrix \({\bf X}\) such that

 \[{\bf A}\cdot{\bf X} = {\bf B}.\]
Feedback: For matrix \[{\bf A} = [value]\] find the inverse \[{\bf A^{-1}} = [value]\] Then \[{\bf X} = {\bf A^{-1}}\cdot{\bf B} =[value] \cdot [value]=[value]\]

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

find a non-zero 2 by 2 matrix \({\bf X}\) such that

 \[{\bf A}\cdot{\bf X} = {\bf O}_{2x2}\]

\({\bf O}_{2x2}\) is the 2 by 2 zero matrix.
Feedback: Let \({\bf X}\) = \([value]\) \[{\bf A}\cdot{\bf X} = [value].[value]=[value]\] Therefore \[[value] = 0 \mbox{ , } [value] = 0\] \[[value] = 0 \mbox{ , } [value] = 0\] The 2 equations in \(w\) and \(y\) are linearly dependent as are the 2 equations in \(x\) and \(z\). Therefore there is no unique solution but \[w=[value] \mbox{ and } x=[value] \] So \({\bf X}\) could be \([value]\)