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Find the value of "k" in the equation:

$k\left(\begin{array}{cc}3 & -1 \\ 5 & -4\end{array}\right) = \left(\begin{array}{cc}-3/4 & 1/4 \\ -5/4 & 1\end{array}\right)$

do I multiply $3 * (-3/4)$ then $(-1) * 1/4$ then $5 *(-5/4)$ and $(-4) * 1$ ?

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Need to reformat the matrices. I would, but I'm brand new here and don't know how. –  Jack Radcliffe Jul 20 '12 at 16:49
    
The technique provided in my answer to your other question math.stackexchange.com/a/173309/31475 should point you in the right direction. –  Arkamis Jul 20 '12 at 16:59
    
It's non-singular, so you can also do this :) $$ k = \left(\begin{array}{cc}-3/4 & 1/4 \\ -5/4 & 1\end{array}\right) \left(\begin{array}{cc}3 & -1 \\ 5 & -4\end{array}\right)^{-1}.$$ –  user2468 Jul 20 '12 at 17:13
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2 Answers

No, you don't multiply the entries of the matrices. Remember how you multiply a matrix by a scalar:

$$\alpha\left(\begin{array}{cc} a &b\\ c& d \end{array}\right) = \left(\begin{array}{cc} \alpha a& \alpha b\\ \alpha c & \alpha d \end{array}\right)$$ that is, you multiply each entry by the scalar.

So here, you are looking to see if you can find a single number $k$ such that $$\left(\begin{array}{rr} k\cdot 3 & k\cdot (-1)\\ k\cdot 5 & k\cdot (-4) \end{array}\right) = \left(\begin{array}{rr} -\frac{3}{4} & \frac{1}{4}\\ -\frac{5}{4} & 1 \end{array}\right).$$ In other words, you are trying to solve four equations simultaneously: $$\begin{align*} 3k &= -\frac{3}{4}\\ -k &= \frac{1}{4}\\ 5k &=-\frac{5}{4}\\ -4k &= 1 \end{align*}$$

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Since it's just a case of scalar multiplication, you can do this:

$$3k = \frac{-3}{4} \tag{1}$$ $$-k = \frac{1}{4} \tag{2}$$ $$5k = \frac{-5}{4} \tag{3}$$ $$-4k = 1 \tag{4}$$

Solving any one of those four equations will get you the value of $k$, in this case, $-\frac{1}{4}.$

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