# Chapter 2 - Matrices - 2.1 Operations with Matrices - 2.1 Exercises - Page 49: 52

$$4\left[\begin{array}{l}{-3} \\ {3}\\{4} \end{array}\right]-2\left[\begin{array}{l}{5} \\ {4} \\{-8} \end{array}\right]=\left[\begin{array}{l}{-22} \\ {4}\\{32} \end{array}\right],$$

#### Work Step by Step

Assume that $$x_1\left[\begin{array}{l}{-3} \\ {3}\\{4} \end{array}\right]+x_2\left[\begin{array}{l}{5} \\ {4} \\{-8} \end{array}\right]=\left[\begin{array}{l}{-22} \\ {4}\\{32} \end{array}\right],$$ then we have the system \begin{aligned} -3x_{1}+5x_{2}&=-22 \\ 3x_{1} +4x_2 &=4\\ 4x_1-8x_{2} &=32 \end{aligned}. The augmented matrix is given by $$\left[ \begin {array}{ccc} -3&5&-22\\ 3&4&4 \\ 4&-8&32\end {array} \right] .$$ Using Gauss-Jordan elimination, we get the row-reduced echelon form as follows $$\left[ \begin {array}{ccc} 1&0&4\\ 0&1&-2 \\ 0&0&0\end {array} \right] .$$ From which the solution is $$x_1=4, \quad x_2=-2 .$$ Hence, we have $$4\left[\begin{array}{l}{-3} \\ {3}\\{4} \end{array}\right]-2\left[\begin{array}{l}{5} \\ {4} \\{-8} \end{array}\right]=\left[\begin{array}{l}{-22} \\ {4}\\{32} \end{array}\right],$$

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