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

Find the matrix of cofactors for the matrix \[ {\bf A} = [value] \]
Feedback: The matrix of cofactors of matrix \({\bf A}\) is found by replacing each element of \({\bf A}\) with its cofactor. The cofactor of \({\bf A}_{ij}\) is the determinant of the matrix that remains when row \(i\) and column \(j\) are removed, multiplied by \((-1)^{i+j}\)

Use Cramer's rule to find the y-coordinate of the solution of the system:

\[[value]\ =\ [value] , \]
\[ [value]\ =\ [value] , \]
\[ [value]\ =\ [value] . \]

y=
Feedback: Cramer's rule states that y = \(\frac{D_{y}}{D}\) where \(D\) is the determinant of the matrix and \(D_{y}\) is the determinant of the matrix with the y column replaced by the numbers on the right hand side of the equations. Therefore for the system \[[value]\ =\ [value] , \] \[ [value]\ =\ [value] , \] \[ [value]\ =\ [value] , \] \[ D = det[value]=[value]\] \[D_{y} = det[value]=[value]\] and hence, \[y=\frac{D_{y}}{D}=\frac[value][value] = [value]\]

Use Cramer's rule to find the solution of the system:

\[ [value]\ =\ [value] , \]
\[ [value]\ =\ [value] , \]
\[ [value]\ =\ [value] , \]
x=
y=
z=
Feedback: Cramer's rule states that x = \(\frac{D_{x}}{D}\) where \(D\) is the determinant of the matrix and \(D_{x}\) is the determinant of the matrix with the x column replaced by the numbers on the right hand side of the equations and similarly y = \(\frac{D_{y}}{D}\) and z = \(\frac{D_{z}}{D}\). Therefore for the system, \[[value]\ =\ [value] , \] \[ [value]\ =\ [value] , \] \[ [value]\ =\ [value] , \] \[ D = det[value]=[value],\] \[D_{x} = det[value]=[value],\] \[D_{y} = det[value]=[value],\] \[D_{z} = det[value]=[value].\] Hence, \[x=\frac[value][value] = [value],\] \[y=\frac[value][value] = [value],\] \[z=\frac[value][value] = [value].\]

A 2 by 2 matrix \[{\bf A} = [a_{ij}]\] has \[a_{ij}= [value] i - [value] j.\] What is \(\det({\bf A})\)?
Feedback: The matrix \({\bf A}\) is \([value]=[value]\) The determinant is defined as \(ad-bc\). Therefore \(\det({\bf A})=[value]\)

Evaluate \(\det({\bf A})\) for the matrix \({\bf A}\) given below:
\[ {\bf A} = [value]\]
\(\det({\bf A})\)=
Feedback: The matrix \(\bf A\) is \([value]\) The determinant is easiest calculated by expanding along the third column, giving \([value]\cdot (-1)^{2+3} \cdot ([value]\cdot[value] - [value]\cdot [value])\). Therefore \(\det({\bf A})=[value]\)

Write down the 3 by 3 elementary matrix \({\bf E}_{[value] [value]}([value])\) that (when premultiplying a given matrix) adds \([value]\) times row \([value]\) to row \([value]\):
Feedback: The required matrix is the 3 by 3 identity matrix with element [[value],[value]] set to [value].

If a [value] by [value] matrix \({\bf A}\) has \(\det({\bf A}) = [value]\), find \[\det([value]{\bf A}),\det([value]{\bf A}),\det({\bf A}^[value]),\det(({\bf A}^{-1})^[value])\]
Enter your answers in the right order in the boxes below.

\(\det([value]{\bf A}) = \)
\(\det([value]{\bf A}) = \)
\(\det({\bf A}^[value]) = \)
\(\det({\bf A}^{-1})^[value]) = \)
Feedback: If a \([value]\) by \([value]\) matrix \({\bf A}\) has \(\det({\bf A}) = [value]\), then \[\det([value]{\bf A}) = [value]^[value]\cdot[value] = [value]\] \[\det(\[value]{\bf A}) = ([value])^[value]\cdot[value] = [value]\] \[\det({\bf A}^[value])= ([value])^[value] = [value]\] \[\det(({\bf A}^{-1})^[value]) =[value]^[value] = [value]\]

A 2 by 2 matrix \[{\bf A} = [value].\] What is \(\det({\bf A})\)?
Feedback: The matrix \({\bf A}\) is \([value]=[value]\) The determinant is defined as \(ad-bc\). Therefore \(\det({\bf A})=[value]\)

Evaluate \(\det({\bf A})\) for the matrix \({\bf A}\) given below:
\[ {\bf A} = [value]\]
\(\det({\bf A})\)=
Feedback: The matrix \(\bf A\) is \([value]\) The determinant is easiest calculated by expanding along the first row, giving \([value]\cdot (-1)^{1+1} \cdot ([value]\cdot[value] - [value]\cdot [value])\). Therefore \(\det({\bf A})=[value]\)

Find the inverse of the matrix \({\bf A}\) : \[ {\bf A} = [value] . \]
Feedback: The inverse of the diagonal matrix \({\bf A} = [value] \), is a diagonal matrix whose elements are the inverses of the original elements. Therefore \[{\bf A}^{-1} = [value]\]

Give the elementary matrix which, when multiplied to the left of a 3 by 3 matrix, will replace row 2 by row 2 + [value] times row 3.

\(\textbf{E} = \)

Give the elementary matrix which, when multiplied to the left of a 3 by 3 matrix, will replace row 1 by row 1 + [value] times row 3.

\(\textbf{E} = \)

Give the elementary matrix which, when multiplied to the left of a 3 by 3 matrix, will replace row 2 by row 2 + [value] times row 1.

\(\textbf{E} = \)

Give the elementary matrix which, when multiplied to the left of a 3 by 3 matrix, will replace row 1 by row 1 - [value] times row 2.

\(\textbf{E} = \)

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

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

Find the inverse of \({\bf A}\), if it is invertible.
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 \({\bf A^{-1}} = [value]\)