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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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    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}.2012-07-20

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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, $-\frac{1}{4}.$