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If there exists a matrix $\mathbf{A}$ such that \[ \mathbf{A} \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} -3 \\ 4 \\ 0 \end{pmatrix},\mathbf{A} \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}, \mathbf{A} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}\]calculate \[\mathbf{A} \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}.\]If there's no such matrix, answer with $\begin{pmatrix}? \\ ? \\ ? \end{pmatrix}.$

 

 

also 

 

 

If there exists a matrix $\mathbf{A}$ such that \[\mathbf{A} \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}, \mathbf{A} \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} -3 \\ 4 \\ 0 \end{pmatrix}, \mathbf{A} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} -2 \\ 6 \\ 3 \end{pmatrix}\]calculate \[\mathbf{A} \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}.\]If there's no such matrix, answer with $\begin{pmatrix}? \\ ? \\ ? \end{pmatrix}.$

 Jul 30, 2019
 #1
avatar+1710 
-1

First, they are the same question.

Second, I present- the un-confusingified problem-

\(If\ there\ exists\ a\ matrix\ \mathbf{A}\ such\ that\ \mathbf{A} \\\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} -3 \\ 4 \\ 0 \end{pmatrix},\mathbf{A} \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}, \mathbf{A} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix} \\calculate\ \mathbf{A} \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}. \\If\ there's\ no\ such\ matrix,\ answer\ with\\ \begin{pmatrix}? \\ ? \\ ? \end{pmatrix}.\)

 

Hope This Helps!

😁😁😁

 

Sorry the mathbf stuff were on my nerves.

 Jul 30, 2019
edited by tommarvoloriddle  Jul 30, 2019
 #2
avatar+6193 
0

There is no matrix A that satisfies the 3 equations.

 

(1,1,1) = (1,0,0)+(0,1,1)

A(1,1,1) = A((1,0,0)+(0,1,1)) = A(1,0,0) + A(0,1,1) = (-3,4,0) + (1,2,3) = (-2, 6, 3) != (3,2,1)

.
 Jul 30, 2019
edited by Rom  Jul 30, 2019
edited by Rom  Jul 30, 2019
 #3
avatar+25228 
+1

1.

If there exists a matrix \(\mathbf{A}\) such that
\(\mathbf{A} \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} -3 \\ 4 \\ 0 \end{pmatrix},\ \mathbf{A} \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix},\ \mathbf{A} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}\)
calculate
\(\mathbf{A} \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}\).

 

\(\begin{array}{|c|c|c|c|} \hline & I. & II. & III. \\ \hline \mathbf{A} & \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} & \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} & \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \\ \hline \begin{pmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33} \end{pmatrix} & \begin{pmatrix} -3 \\ 4 \\ 0 \end{pmatrix} & \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} & \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix} \\ \hline \end{array} \)

\(\begin{array}{|l|l|l|} \hline I. & II.& III. \\ \hline a_{11} = -3 & \mathbf{a_{12}+a_{12} = \color{red}1} & \underbrace{a_{11}}_{-3}+a_{12}+a_{13} = 3 \qquad \mathbf{a_{12}+a_{13} = \color{red}6}\quad \text{contradiction!} \\ a_{21} = 4 & a_{22}+a_{23} = 2 & a_{21}+a_{22}+a_{23} = 2 \\ a_{31} = 0 & a_{32}+a_{33} = 3 & a_{31}+a_{32}+a_{33} = 1 \\ \hline \end{array}\)

 

There's no such matrix \(\mathbf{A}\)

 

laugh

 Jul 30, 2019
 #4
avatar+25228 
+1

2.

\(\mathbf{A} \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix},\ \mathbf{A} \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} -3 \\ 4 \\ 0 \end{pmatrix},\ \mathbf{A} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} -2 \\ 6 \\ 3 \end{pmatrix}\)

calculate
\(\mathbf{A} \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}\).

 

\(\begin{array}{|c|c|c|c|} \hline & I. & II. & III. & IV. \\ \hline \mathbf{A} & \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} & \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} & \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} & \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix} \\ \hline \begin{pmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33} \end{pmatrix} & \begin{pmatrix} -3 \\ 4 \\ 0 \end{pmatrix} & \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} & \begin{pmatrix} -2 \\ 6 \\ 3 \end{pmatrix} \\ \hline \end{array}\)

\(\begin{array}{|l|l|lll|} \hline I. & II.& III. \\ \hline a_{11} = -3 & a_{12}+a_{12} = 1 & \underbrace{a_{11}}_{-3}+a_{12}+a_{13} = -2 & -3+ a_{12}+a_{13} = -2 & a_{12}+a_{13}=1\ \checkmark \\ a_{21} = 4 & a_{22}+a_{23} = 2 & \underbrace{a_{21}}_{4}+a_{22}+a_{23} = 6 & 4+ a_{22}+a_{23} = 6 & a_{22}+a_{23}=2\ \checkmark \\ a_{31} = 0 & a_{32}+a_{33} = 3 & \underbrace{a_{31}}_{0}+a_{32}+a_{33} = 3 & 0+ a_{32}+a_{33} = 3 & a_{32}+a_{33}=3\ \checkmark \\ \hline \end{array} \)

\(\begin{array}{|l|l|lll|} \hline IV. \\ \hline a_{11}-a_{21}-a_{13} = a_{11}-(a_{21}+a_{13}) = -3-1 = -4 \\ a_{21}-a_{22}-a_{23} = a_{21}-(a_{22}+a_{23}) = 4-2 = 2 \\ a_{31}-a_{32}-a_{33} = a_{31}-(a_{32}+a_{33}) = 0-3=-3 \\ \hline \end{array}\)

 

\(\mathbf{A} \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix} = \begin{pmatrix} -4 \\ 2 \\ -3 \end{pmatrix}\)

 

laugh

 Jul 30, 2019
 #5
avatar+30278 
+1

Question 2.

 

question 2

 Jul 30, 2019

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