How do you find the values of x, y and z given 3[(x, y-1), (4, 3z)]=[(15, 6), (6z, 3x+y)]?

Dec 1, 2016

See explanation.

Explanation:

First step is to multiply the matrix on the left by $3$ (to multiply a matrix by a number you have to multiply a;; elements of the matrix by the number):

$3 \times \left[\begin{matrix}x & y - 1 \\ 4 & 3 z\end{matrix}\right] = \left[\begin{matrix}15 & 6 \\ 6 z & 3 x + y\end{matrix}\right]$

$\left[\begin{matrix}3 x & 3 y - 3 \\ 12 & 9 z\end{matrix}\right] = \left[\begin{matrix}15 & 6 \\ 6 z & 3 x + y\end{matrix}\right]$

To check when 2 square matrices are equal we have to check if all their elements are equal.

So in this example we get such set of equations.

$\left\{\begin{matrix}3 x = 15 \\ 3 y - 3 = 6 \\ 6 z = 12 \\ 9 z = 3 x + y\end{matrix}\right.$

From the first three equations we can calculate the values of each variable:

1. $3 x = 15 \implies x = 5$

2. $3 y - 3 = 6 \implies 3 y = 9 \implies y = 3$

3. $6 z = 12 \implies z = 2$

Now we have to check if the calculated values fulfill the last equation:

$9 \cdot 2 = 3 \cdot 5 + 3$

$18 = 18$

Left side equals right side, so the calculated values are solutions to all 4 equations.

Answer: The equation is true for $x = 5 , y = 3 , z = 2$