Answer
$a=1$
$b = 1$
Work Step by Step
$(4+3i)a+(3-4i)b = 7-i$
Expand the above expression:
$4a+3ia+3b-4ib = 7-i$
Combine like terms:
$(4a+3b) + (3a-4b)i = 7-i$
In order for the complex number to equal $7-i$, the real component must equal $7$ and the imaginary component must equal $-1$.
Equation 1: This is an expression from the real component of the complex number.
$4a+3b=7$
Equation 2: This is an expression from the imaginary component of the complex number.
$3a-4b=-1$
Solve the above system of equations using a matrix:
Form a matrix from the two equations above:
$\begin{bmatrix}
4 & 3 & |7 \\
3 & -4 & |-1\\
\end{bmatrix}$
Simplify the matrix to a form where it would be easy to solve for a and b:
$\begin{bmatrix}
4 & 3 & |7 \\
3 & -4 & |-1\\
\end{bmatrix}$ ~ $\begin{bmatrix}
4 & 3 & |7 \\
0 & -25 & |-25\\
\end{bmatrix}$
From the simplified matrix, there are two equations:
Equation 3: This is from the first row of the matrix.
$4a+3b=7$
Equation 4: This is from the second row of the matrix.
$-25b = -25$
Solve Equation 4, to determine the value of b:
$b = 1$
Substitute the value of b into Equation 3 to solve for the value of a:
$4a+3(1)=7$
$4a+3=7$
$4a=4$
$a=1$
In summary:
$a=1$
$b = 1$
