If I have a matrix in the form $Ax=B$, why must I have...
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If I have a matrix in the form $Ax=B$, why must I have $begin{bmatrix}i\jend{bmatrix}=βbegin{bmatrix}-2\1end{bmatrix}?$
The 2019 Stack Overflow Developer Survey Results Are InWhy are $A=begin{bmatrix} a & b \ c & d end{bmatrix}$ and $B=begin{bmatrix} 0 & 1 \ -det(A) & operatorname{tr}(A) end{bmatrix}$ similar?Solve $3X =begin{bmatrix}0&3\6&9end{bmatrix}$ for $X$The number of the solutions of $ x^{10}= begin{bmatrix}1&0\ 0&1 end{bmatrix}$Given an eigen values evaluate $S*tinybegin{bmatrix} 0\1\0 end{bmatrix}$Interpreting $begin{bmatrix} 1 & 2 & 3 end{bmatrix} cdot h $ as a scalar or matrix multiplication$begin{bmatrix}6&a\b&1end{bmatrix}begin{bmatrix}2\0end{bmatrix}=begin{bmatrix}4\-10end{bmatrix}$ Find $a$ and $b$Eigenvectors of $begin{bmatrix}a&-b\b&aend{bmatrix}$Is $begin{bmatrix}0&1\0&1end{bmatrix}$ linearly dependent?Find matrix $A$ of $Ax=b$ if $b=begin{bmatrix} 1 \ 2 \ 1 \ end{bmatrix} \$Finding all $3times3$ matrices such that $Abegin{bmatrix}x\y\zend{bmatrix}=begin{bmatrix}1\0\0end{bmatrix}$ has two distinct solutions
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I have inserted numbers in place of i and j, and have successfully deduced the value of β as a coefficient. However, i am unsure why we must have that equation if we know $Abegin{bmatrix}i\jend{bmatrix}=0$.
linear-algebra matrices
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add a comment |
$begingroup$
I have inserted numbers in place of i and j, and have successfully deduced the value of β as a coefficient. However, i am unsure why we must have that equation if we know $Abegin{bmatrix}i\jend{bmatrix}=0$.
linear-algebra matrices
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This question is very unclear - it in not clear what your vectors have to do with $Ax=B.$ What do you mean "you must have?"
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– Thomas Andrews
Mar 21 at 15:13
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If "$Abegin{bmatrix} i \ jend{bmatrix}= 0$" and "$Ax= B$" then $Aleft(x+ beta begin{bmatrix} i \ jend{bmatrix}right)= Ax+ beta Abegin{bmatrix} i \ j end{bmatrix}= B$ for any $beta$. Is that what you are asking?
$endgroup$
– user247327
Mar 21 at 15:35
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Yes thank you, sorry for the poor wording
$endgroup$
– MorpheusZion
Mar 21 at 23:37
add a comment |
$begingroup$
I have inserted numbers in place of i and j, and have successfully deduced the value of β as a coefficient. However, i am unsure why we must have that equation if we know $Abegin{bmatrix}i\jend{bmatrix}=0$.
linear-algebra matrices
$endgroup$
I have inserted numbers in place of i and j, and have successfully deduced the value of β as a coefficient. However, i am unsure why we must have that equation if we know $Abegin{bmatrix}i\jend{bmatrix}=0$.
linear-algebra matrices
linear-algebra matrices
edited Mar 21 at 15:16
Thomas Andrews
131k12147298
131k12147298
asked Mar 21 at 15:02
MorpheusZionMorpheusZion
84
84
$begingroup$
This question is very unclear - it in not clear what your vectors have to do with $Ax=B.$ What do you mean "you must have?"
$endgroup$
– Thomas Andrews
Mar 21 at 15:13
$begingroup$
If "$Abegin{bmatrix} i \ jend{bmatrix}= 0$" and "$Ax= B$" then $Aleft(x+ beta begin{bmatrix} i \ jend{bmatrix}right)= Ax+ beta Abegin{bmatrix} i \ j end{bmatrix}= B$ for any $beta$. Is that what you are asking?
$endgroup$
– user247327
Mar 21 at 15:35
$begingroup$
Yes thank you, sorry for the poor wording
$endgroup$
– MorpheusZion
Mar 21 at 23:37
add a comment |
$begingroup$
This question is very unclear - it in not clear what your vectors have to do with $Ax=B.$ What do you mean "you must have?"
$endgroup$
– Thomas Andrews
Mar 21 at 15:13
$begingroup$
If "$Abegin{bmatrix} i \ jend{bmatrix}= 0$" and "$Ax= B$" then $Aleft(x+ beta begin{bmatrix} i \ jend{bmatrix}right)= Ax+ beta Abegin{bmatrix} i \ j end{bmatrix}= B$ for any $beta$. Is that what you are asking?
$endgroup$
– user247327
Mar 21 at 15:35
$begingroup$
Yes thank you, sorry for the poor wording
$endgroup$
– MorpheusZion
Mar 21 at 23:37
$begingroup$
This question is very unclear - it in not clear what your vectors have to do with $Ax=B.$ What do you mean "you must have?"
$endgroup$
– Thomas Andrews
Mar 21 at 15:13
$begingroup$
This question is very unclear - it in not clear what your vectors have to do with $Ax=B.$ What do you mean "you must have?"
$endgroup$
– Thomas Andrews
Mar 21 at 15:13
$begingroup$
If "$Abegin{bmatrix} i \ jend{bmatrix}= 0$" and "$Ax= B$" then $Aleft(x+ beta begin{bmatrix} i \ jend{bmatrix}right)= Ax+ beta Abegin{bmatrix} i \ j end{bmatrix}= B$ for any $beta$. Is that what you are asking?
$endgroup$
– user247327
Mar 21 at 15:35
$begingroup$
If "$Abegin{bmatrix} i \ jend{bmatrix}= 0$" and "$Ax= B$" then $Aleft(x+ beta begin{bmatrix} i \ jend{bmatrix}right)= Ax+ beta Abegin{bmatrix} i \ j end{bmatrix}= B$ for any $beta$. Is that what you are asking?
$endgroup$
– user247327
Mar 21 at 15:35
$begingroup$
Yes thank you, sorry for the poor wording
$endgroup$
– MorpheusZion
Mar 21 at 23:37
$begingroup$
Yes thank you, sorry for the poor wording
$endgroup$
– MorpheusZion
Mar 21 at 23:37
add a comment |
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$begingroup$
This question is very unclear - it in not clear what your vectors have to do with $Ax=B.$ What do you mean "you must have?"
$endgroup$
– Thomas Andrews
Mar 21 at 15:13
$begingroup$
If "$Abegin{bmatrix} i \ jend{bmatrix}= 0$" and "$Ax= B$" then $Aleft(x+ beta begin{bmatrix} i \ jend{bmatrix}right)= Ax+ beta Abegin{bmatrix} i \ j end{bmatrix}= B$ for any $beta$. Is that what you are asking?
$endgroup$
– user247327
Mar 21 at 15:35
$begingroup$
Yes thank you, sorry for the poor wording
$endgroup$
– MorpheusZion
Mar 21 at 23:37