Chapter 3.5, Problem 30PS

To determine

(a)

The matrix that leads X to $AX=$zero matrix.

$AX=0$when X is in the form of $\left[\begin{array}{l}abc\\ abc\\ abc\end{array}\right]$or when each column of X is a multiple of $\left(1,1,1\right)$

Given:

M is the space of 3 by 3 matrices and multiply every matrix X in M by

$A=\left[\begin{array}{l}10-1\\ -110\\ 0-11\end{array}\right]$. Notice $A\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right]=\left[\begin{array}{l}0\\ 0\\ 0\end{array}\right]$

Calculation:

Consider row reduction method to check that the rows of A are dependent and A has rank 2.

The nullspace of A has basis: $\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right]$

Thus, each column of X is in the nullspace of A, that means when they are multiples of the basis of $N\left(A\right)$, then $AX=0$ will have the form: $\left[\begin{array}{l}abc\\ abc\\ abc\end{array}\right]$having $\text{dim}\left(\text{nullspace}\right)=3$.

(b)

The matrix that have the form AX for some matrix X.

$AX=B$when all the columns of matrix add to zero : $\left[\begin{array}{l}abc\\ def\\ -a-d-b-e-c-f\end{array}\right]$

Given:

M is the space of 3 by 3 matrices and multiply every matrix X in M by

$A=\left[\begin{array}{l}10-1\\ -110\\ 0-11\end{array}\right]$. Notice $A\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right]=\left[\begin{array}{l}0\\ 0\\ 0\end{array}\right]$

Calculation:

The columns of any matrix form AX are linear combinations of the columns of matrix A. Thus, these are the vectors whose components add to zero, that means,

$AX=B$only when $\left[\begin{array}{l}abc\\ def\\ -a-d-b-e-c-f\end{array}\right]$because adding these columns will result in zero and thus the $\text{dim}\left(B\right)=6$.

(c)

The dimensions of nullspace and column space of M. Also, reason behind $\left(n-r\right)+r=9$.

$\text{dim}\left(B\right)=6$and $\text{dim}\left(\text{nullspace}\right)=3$

$\left(n-r\right)+r=9$because there are 9 entries in the matrix

Given:

M is the space of 3 by 3 matrices and multiply every matrix X in M by

$A=\left[\begin{array}{l}10-1\\ -110\\ 0-11\end{array}\right]$. Notice $A\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right]=\left[\begin{array}{l}0\\ 0\\ 0\end{array}\right]$

Calculation:

It is noted that dimension of the nullpace of M is 3 and the dimension of column space B is 6.

Thus, dimension of a 3 by 3 matrix M will be $3+6=9$

Therefore, $\dim \left(M^{\text{3x3}}\right)=9$which means there are 9 entries in the 3 by 3 matrix.

To determine

(b)

The matrix that have the form AX for some matrix X.

$AX=B$when all the columns of matrix add to zero : $\left[\begin{array}{l}abc\\ def\\ -a-d-b-e-c-f\end{array}\right]$

Given:

M is the space of 3 by 3 matrices and multiply every matrix X in M by

$A=\left[\begin{array}{l}10-1\\ -110\\ 0-11\end{array}\right]$. Notice $A\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right]=\left[\begin{array}{l}0\\ 0\\ 0\end{array}\right]$

Calculation:

The columns of any matrix form AX are linear combinations of the columns of matrix A. Thus, these are the vectors whose components add to zero, that means,

$AX=B$only when $\left[\begin{array}{l}abc\\ def\\ -a-d-b-e-c-f\end{array}\right]$because adding these columns will result in zero and thus the $\text{dim}\left(B\right)=6$.

(c)

The dimensions of nullspace and column space of M. Also, reason behind $\left(n-r\right)+r=9$.

$\text{dim}\left(B\right)=6$and $\text{dim}\left(\text{nullspace}\right)=3$

$\left(n-r\right)+r=9$because there are 9 entries in the matrix

Given:

M is the space of 3 by 3 matrices and multiply every matrix X in M by

$A=\left[\begin{array}{l}10-1\\ -110\\ 0-11\end{array}\right]$. Notice $A\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right]=\left[\begin{array}{l}0\\ 0\\ 0\end{array}\right]$

Calculation:

It is noted that dimension of the nullpace of M is 3 and the dimension of column space B is 6.

Thus, dimension of a 3 by 3 matrix M will be $3+6=9$

Therefore, $\dim \left(M^{\text{3x3}}\right)=9$which means there are 9 entries in the 3 by 3 matrix.

To determine

(c)

The dimensions of nullspace and column space of M. Also, reason behind $\left(n-r\right)+r=9$.

$\text{dim}\left(B\right)=6$and $\text{dim}\left(\text{nullspace}\right)=3$

$\left(n-r\right)+r=9$because there are 9 entries in the matrix

Given:

M is the space of 3 by 3 matrices and multiply every matrix X in M by

$A=\left[\begin{array}{l}10-1\\ -110\\ 0-11\end{array}\right]$. Notice $A\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right]=\left[\begin{array}{l}0\\ 0\\ 0\end{array}\right]$

Calculation:

It is noted that dimension of the nullpace of M is 3 and the dimension of column space B is 6.

Thus, dimension of a 3 by 3 matrix M will be $3+6=9$

Therefore, $\dim \left(M^{\text{3x3}}\right)=9$which means there are 9 entries in the 3 by 3 matrix.