# Chapter 2 - Matrices - 2.1 Operations with Matrices - 2.1 Exercises - Page 48: 15

$$x=3, \quad y=2, \quad z=1.$$

#### Work Step by Step

We have \begin{align*} &4\left[\begin{array}{rrr}{x} & {y} \\ {z} & {-1} \end{array}\right]=2\left[\begin{array}{rrr}{y} & {z} \\ {-x} & {1} \end{array}\right]+2\left[\begin{array}{rrr}{4} & {x} \\ {5} & {-x} \end{array}\right]\\ \Longrightarrow & \left[\begin{array}{rrr}{4x} & {4y} \\ {4z} & {-4} \end{array}\right]=\left[\begin{array}{rrr}{2y} & {2z} \\ {-2x} & {2} \end{array}\right]+\left[\begin{array}{rrr}{8} & {2x} \\ {10} & {-2x} \end{array}\right]\\ \Longrightarrow & \left[\begin{array}{rrr}{4x} & {4y} \\ {4z} & {-4} \end{array}\right]=\left[\begin{array}{rrr}{2y+8} & {2z+2x} \\ {-2x+10} & {2-2x} \end{array}\right]. \end{align*} By comparing the corresponding entries of the matrices in both sides of the above equation, we get the following system of equations \begin{align*} 4x&=2y+8\\ 4y&=2z+2x\\ 4z&=-2x+10\\ -4&=2-2x. \end{align*} From which, we have the solution $$x=3, \quad y=2, \quad z=1.$$

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