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Let $$B = \begin{bmatrix} 1 & 2 & 3 \\ 1 & 1 & 1 \\ 1 & 3 & 5 \end{bmatrix} $$ and $$C = \begin{bmatrix} 1 & 3 & 5 \\ 1 & 2 & 3 \\ 1 & 1 & 1 \end{bmatrix}. $$

Find a matrix $A$ that satisfies the equation AB = C. I tried to do A = C*B^(-1) but found the determinant is 0 for B. Is there another way to solve this?

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  • $\begingroup$ sure. Multiplying on the left by an "elementary"matrix $A_1$ causes a row operation on $B.$ Multiplying on the left by $A_2$ causes a second row operation. Instead of trying to get reduced form, you try to get $C.$ The result is $A = A_k A_{k-1} \cdots A_2 A_1$ works $\endgroup$ – Will Jagy Feb 7 '18 at 2:20
  • $\begingroup$ Have you noticed that $C$ is just $B$ with the rows permuted? Do you know how to find a matrix such that multiplying by it permutes the rows? $\endgroup$ – Gerry Myerson Feb 8 '18 at 6:06
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Use row-operations on $B^\top A^\top = C^\top$.

Since $B$ is singular, there is no guarantee that the equation has a solution (it will depend on the columns of $C^\top$).

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