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Let $A=\begin{bmatrix}6 & 5\\-7 & 9\end{bmatrix}$ and $C=\begin{bmatrix}1 & -2\\4 & -8\end{bmatrix}$. Find all matrices $B$ such that $AC=BC$.

$\begin{bmatrix}6&5\\-7&9\end{bmatrix} \times \begin{bmatrix}1&-2\\4&-8\end{bmatrix}$ $= \begin{bmatrix}6 \times 1 + 5 \times 4 & 6 \times (-2) + 5 \times (-8) \\ (-7) \times 1 + 9 \times 4 & (-7) \times (-2) + 9 \times (-8)\end{bmatrix}$ $= \begin{bmatrix}26 & -52 \\ 29 & -58\end{bmatrix}$

$\begin{bmatrix}6&5\\-7&9\end{bmatrix} \times \begin{bmatrix}b_1&b_2\\b_3&b_4\end{bmatrix} = \begin{bmatrix}26 & -52 \\ 29 & -58\end{bmatrix}$

$\begin{bmatrix}6b_1 + 5b_3 & 6b_2 + 5b_4 \\ -7b_1 + 9b_3 & -7b_2 + 9b_4 \end{bmatrix} = \begin{bmatrix}26 & -52 \\ 29 & -58\end{bmatrix}$

Equations:

$6b_1 + 5b_3 = 26$

$-7b_1 + 9b_3 = 29$

Eliminate $b_1$:

$42b_1 + 35b_3 = 182$

$-42b_1 + 54b_3 = 174\implies 89b_3 = 356 \implies b_3 = 4$

$6b_1 + 20 = 26 \implies b_1 = 1$

$6b_2 + 5b_4 = -52$

$-7b_2 + 9b_4 = -58$

Eliminate $b_2$:

$42b_2 + 35b_4 = -364$

$-42b_2 + 54b_4 = -348\implies89b_4 = -712 \implies b_4 = -8$

$6b_2 + 5(-8) = -52 \implies b_2 = -2$

Therefore, is it true that the only solution is when $A = B$? Or am I missing something?

EDIT:

Cameron Williams

You mentioned this, but it is not without problems.

$\begin{bmatrix}1&-2\\4&-8\end{bmatrix} \times \begin{bmatrix}b_1&b_2\\b_3&b_4\end{bmatrix} = \begin{bmatrix}26&-52\\29&-58\end{bmatrix}$

$\begin{bmatrix}b_1-2b_3 & b_2 - 2b_4 \\ 4b_1 - 8b_3 & 4b_2 - 8b_4\end{bmatrix} = \begin{bmatrix}26&-52\\29&-58\end{bmatrix}$

New equations:

$b_1 - 2b_3 = 26$

$4b_1 - 8b_3 = 29$

Eliminate $b_1$ and $b_3$ and make an invalid equation:

$(4b_1 - 8b_3) - 4(b_1 - 2b_3) = 29 - 104$

$0 = -75$

What does this mean?

I do not think I need to use the other 2 equations, but I do not understand what this invalid equality is supposed to mean.

Community
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user2213307
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    You made a mistake. What you did was computed $AB$, not $BC$. Your work here is correct ($A$ is invertible) but you were addressing the wrong problem. As for the problem you should be addressing: your matrix $C$ is not invertible, so you should expect that $B$ is not simply $A$ (and that there are, in fact, an infinite amount of matrices that satisfy the relationship $AC = BC$). – Cameron Williams Feb 01 '14 at 20:55

2 Answers2

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We have $$C=\begin{bmatrix}1 & -2 \\ 4 & -8 \end{bmatrix}$$ where we can see the second row of $C$ is a scalar multiple ($4$) of the first row of $C$.

So if we let $$D=\begin{bmatrix} 4a & -a \\ 4b & -b \end{bmatrix}$$ for any $a,b \in \mathbb{R}$, then we have $$DC=\mathbf{0}$$ where $\mathbf{0}$ is the $2 \times 2$ all-$0$ matrix.

Hence, if $AC=BC$, then $$AC=BC=BC+DC=(B+D)C,$$ and, since $a$ and $b$ were arbitrary, we have found an infinite number of solutions.

Rebecca J. Stones
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  • Do you imply that $B = A + D , \forall , D = \begin{bmatrix}4a & -a \ 4b & -b\end{bmatrix} , \forall a, , b , \in \mathbb{R}$? – user2213307 Feb 05 '14 at 20:47
  • I haven't eliminated the possibility of additional solutions, but, yes, setting $B=A+D$ for such $D$ gives a infinite family of solutions. – Rebecca J. Stones Feb 05 '14 at 21:19
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Two matrices are equal if, and only if, their action on any vector is the same (this is a special case of Yoneda's lemma which is really easy to prove using only things you know from linear algebra). Notice that $C$ has rank $1$ and its image is spanned by $(1,4)^T$. So the only thing we require from $B$ is to act exactly as $A$ on the vector $(1,4)^T$, i.e.: $$B\left(\begin{array}{c}1\\4\end{array}\right)=A\left(\begin{array}{c}1\\4\end{array}\right)=\left(\begin{array}{cc}6&5\\-7&9\end{array}\right)\left(\begin{array}{c}1\\4\end{array}\right)=\left(\begin{array}{c}26\\31\end{array}\right)$$ This gives you the $2$ equations in $4$ variables: $$\begin{array}{ll}b_{11}+4b_{12}=26\\b_{21}+4b_{22}=31\end{array}$$ which has a $2$-dimensional space of solutions.

Daniel Robert-Nicoud
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  • I don't think this qualifies as a proper answer. OP clearly knows what to do but simply made the mistake of computing $AB$ instead of $AC$. – Cameron Williams Feb 01 '14 at 21:03
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    @CameronWilliams Yes, but I wanted to point out a more elegant and easier (and much less computationally hard) solution to the problem, pointing out at the same time that there is some mistake in what the OP has done in his solution. – Daniel Robert-Nicoud Feb 01 '14 at 21:06
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    @CameronWilliams Also, I believe that understanding how the rank of $C$ affects the space of solutions of $AC=BC$ can be really useful in one's efforts in learning linear algebra, conceptually speaking. – Daniel Robert-Nicoud Feb 01 '14 at 21:08