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Let $A$ be a $4\times 3$ matrix. Consider matrix $B$ which is a pre-multiplier of matrix $A$, that is, $BA$. Find matrix $B$ if it performs the following elementary row operation on $A$

Multiplies the second row by 4.

I let $C$ be the product after multiplying the second row by $4$ and put it in the equation $BA=C$.

I let $A=\begin{matrix}a&b&c\\d&e&f\\g&h&i\\j&k&l\\\end{matrix}$ and $C=\begin{matrix}a&b&c\\4d&4e&4f\\g&h&i\\j&k&l\\\end{matrix}$

My next step I thought would be to rearrange $BA=C$ into $B=CA^{-1}$. However this requires finding the inverse of a non-square matrix. How would I go about this? and how would I use this to find other operations such as:

Adds twice row $3$ to row $4$, or Interchanges rows $1$ and $3$.

Also my knowledge of matrices goes as far as Gaussian Elimination, Determinants and Cofactor/Adjoint Matrices. So I may not understand anything more advanced than this unless you can explain it well.

Rustyn
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VikeStep
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  • What does multiplying with a diagonal matrix do? – Algebraic Pavel Sep 09 '13 at 00:47
  • you mean the identity matrix? thats the only diagonal matrix I have heard of. All it does is the same as multiplying by 1 – VikeStep Sep 09 '13 at 00:50
  • Your next step is not quite right. $A$ does not have an inverse. There is no notion of inverse of a non-square matrix (only pseudoinverse, but you don't need that here). – Tunococ Sep 09 '13 at 00:51
  • What about an identity matrix, but with the second number replaced by $4$? – Tunococ Sep 09 '13 at 00:51
  • Tunococ I see what you mean, its like a 4x4 identity matrix, but the position in 2,2 is replaced by 4. How would I adapt this to the last two: adding twice row 3 to row 4, or Interchanging rows 1 and 3 – VikeStep Sep 09 '13 at 00:56

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You have good ideas, but in this case it's more effective to simply guess at what $B$ should be directly. Write $B = \left[ \begin{matrix} \_ & \_ & \_ & \_ \\ \_ & \_ & \_ & \_ \\ \_ & \_ & \_ & \_ \\ \_ & \_ & \_ & \_ \\ \end{matrix} \right]$ so that $BA = C$ becomes $$ \left[ \begin{matrix} \_ & \_ & \_ & \_ \\ \_ & \_ & \_ & \_ \\ \_ & \_ & \_ & \_ \\ \_ & \_ & \_ & \_ \\ \end{matrix} \right] \left[\begin{matrix}a&b&c\\d&e&f\\g&h&i\\j&k&l\\\end{matrix}\right] = \left[\begin{matrix}a&b&c\\4d&4e&4f\\g&h&i\\j&k&l\\\end{matrix}\right] $$

Now deduce the entries of $B$, keeping in mind that this has to be true for ALL $a,b,c,d,e,f,g,h,i,j,k,l$. For example, the top row of $B$, when multiplied with the first column of $A$, gives a linear combination of $a,d,g,$ and $j$, which must then equal entry $a$ in matrix $C$. So what does this say about the linear combination we had? Hence, what do the entries in the top row of $B$ have to be?

The exact same technique works for adding twice row 3 to row 4 or interchanging rows 1 and 3. Just write out matrix $A$ as $a,b,c,d,e,f,g,h,i,j,k,l$, and then write what matrix $C$ has to be. Then, fill in the values of $B$ to make it work.

Caleb Stanford
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    The entries in the top of row B have to be 1, 0, 0, 0 right? so I guess this is helping explain what Tunococ suggested in the message on the OP, the answer will be an identity matrix with the second entry being 4. I'm still not sure how I can add twice row 3 to row 4 or interchange rows 1 and 3 though – VikeStep Sep 09 '13 at 01:16
  • Yes, it has to be $1,0,0,0$. You can do the same for the rest of the rows. And (I've edited) you can do the same for the other problems too. – Caleb Stanford Sep 09 '13 at 06:28