Solve the system Ax=b, where A=\left[\begin{array}{ccc} 3&4\\ {5}&{6}\end{array}\right] and b=\left[\begin{array}{ccc} 5.2\\ {7}\end{array}\right]. What value is the first entry of x?

To solve the system Ax=b, we first need to find the inverse of matrix A. The formula for the inverse of a 2x2 matrix is:

A^(-1) = 1/(ad-bc) * \left[\begin{array}{ccc} d&-b\ {-c}&{a}\end{array}\right]

where A=\left[\begin{array}{ccc} a&b\ {c}&{d}\end{array}\right].

So, for A=\left[\begin{array}{ccc} 3&4\ {5}&{6}\end{array}\right], we have a=3, b=4, c=5, and d=6.

The determinant ad-bc = 36 - 45 = -2.

So, A^(-1) = -1/2 * \left[\begin{array}{ccc} 6&-4\ {-5}&{3}\end{array}\right] = \left[\begin{array}{ccc} -3&2\ {2.5}&{-1.5}\end{array}\right].

Now, we can find x by multiplying A^(-1) with b:

x = A^(-1)b = \left[\begin{array}{ccc} -3&2\ {2.5}&{-1.5}\end{array}\right] * \left[\begin{array}{ccc} 5.2\ {7}\end{array}\right].

To multiply these matrices, we do:

x_1 = -35.2 + 27 = -15.6 + 14 = -1.6 x_2 = 2.55.2 - 1.57 = 13 - 10.5 = 2.5

So, x = \left[\begin{array}{ccc} -1.6\ {2.5}\end{array}\right].

Therefore, the first entry of x is -1.6.