# USMA linear equation for x y z

• MHB
• karush
In summary, the USMA solved the equation $R_1=R_3-3R_1\left[ \begin{array}{rrr|r} 1 & 0 & - \dfrac{28}{19} & \dfrac{3}{19} \\ 0 & 1 & \dfrac{34}{19} & \dfrac{14}{19} \\ 0 & 0 & 1 & \dfrac{13}{37} \end{array} \right]$ by subtracting the vector of solutions, $R_2=-\dfrac{2}{19}\cdot R_2\left[ \begin{array}{rrr|r} 1 & 0 & - \ karush Gold Member MHB https://www.physicsforums.com/attachments/9471​ ok lots of options to solve this but I would start by$R3-R1\to R3$if I remember correctly if get a diagonal of ones and the rest zeros in A we will have B from Ax=B$\tiny{USMA = United \,States\, Military\, Academy}$​ Last edited: karush said: ok lots of options to solve this but I would start by$R3-R1\to R3$You can start any way you like. if I remember correctly if get a diagonal of ones and the rest zeros in A we will have B from Ax=B ?? You already had Ax=B when you started. Maybe Ix = B', where B' is a vector of solutions? Ok W|A returned this but I don't know how you could get this in a few minutes on an intrance exam$A=\left[ \begin{array}{cccc}
1 & 0 &0 & \frac{25}{37} \\
0 & 1 & 0 & \frac{4}{37} \\
0 &0 &1 & \frac{13}{37}
\end{array} \right]$karush said: Ok W|A returned this but I don't know how you could get this in a few minutes on an intrance exam$A=\left[ \begin{array}{cccc}
1 & 0 &0 & \frac{25}{37} \\
0 & 1 & 0 & \frac{4}{37} \\
0 &0 &1 & \frac{13}{37}
\end{array} \right]$Practice? this is more practice than I'm in the mood for:($R_1=\dfrac{R_1}{2}\left[ \begin{array}{ccc|c} 1 & \frac{5}{2} & 3 & 2 \\ 7 & 8 & 4 & 7 \\ 3 & 2 & 5 & 4 \end{array} \right]R_2=R_2-7R_1\left[ \begin{array}{rrr|r} 1 & \frac{5}{2} & 3 & 2 \\ 0 & - \frac{19}{2} & -17 & -7 \\ 3 & 2 & 5 & 4 \end{array} \right]R_3=R_3-3R_1\left[ \begin{array}{rrr|r} 1 & \frac{5}{2} & 3 & 2 \\ 0 & - \frac{19}{2} & -17 & -7 \\ 0 & - \frac{11}{2} & -4 & -2 \end{array} \right] R_2=- \dfrac{2}{19}\cdot R_2\left[ \begin{array}{rrr|r} 1 & \frac{5}{2} & 3 & 2 \\0 & 1 & \frac{34}{19} & \frac{14}{19} \\ 0 & - \frac{11}{2} & -4 & -2 \end{array} \right]R_1=R_1-\dfrac{5}{2}\cdot R_2
\left[ \begin{array}{rrr|r} 1 & 0 & - \dfrac{28}{19} & \dfrac{3}{19} \\ 0 & 1 & \dfrac{34}{19} & \dfrac{14}{19} \\ 0 & - \dfrac{11}{2} & -4 & -2 \end{array} \right]R_3=R_3+\dfrac{11}{2}R_2
\left[ \begin{array}{rrr|r} 1 & 0 & - \dfrac{28}{19} & \dfrac{3}{19} \\ 0 & 1 & \dfrac{34}{19} & \dfrac{14}{19} \\ 0 & 0 & \dfrac{111}{19} & \dfrac{39}{19} \end{array} \right]R_3=\dfrac{19}{111}R_3
\left[ \begin{array}{rrr|r} 1 & 0 & - \dfrac{28}{19} & \dfrac{3}{19} \\ 0 & 1 & \dfrac{34}{19} & \dfrac{14}{19} \\ 0 & 0 & 1 & \dfrac{13}{37} \end{array} \right]
R_2=R_2-\left(\frac{34}{19}\right)R_3
\left[ \begin{array}{rrr|r} 1 & 0 & 0 & \dfrac{25}{37} \\ 0 & 1 & 0 & \dfrac{4}{37} \\ 0 & 0 & 1 & \dfrac{13}{37} \end{array} \right]\$

Last edited:

## 1. What is the USMA linear equation for x, y, and z?

The USMA linear equation for x, y, and z is a mathematical representation of a relationship between three variables. It is written in the form of ax + by + cz = d, where a, b, and c are coefficients and d is a constant term.

## 2. How do you solve a USMA linear equation for x, y, and z?

To solve a USMA linear equation for x, y, and z, you need to use the method of elimination or substitution. This involves manipulating the equation to isolate one variable, such as x, and then solving for its value. This process is repeated for the other variables until all three values are determined.

## 3. What are the applications of USMA linear equations for x, y, and z?

USMA linear equations for x, y, and z have a wide range of applications in science, engineering, and economics. They are used to model and analyze relationships between three variables, such as distance, time, and speed in physics, or supply, demand, and price in economics.

## 4. Can a USMA linear equation for x, y, and z have multiple solutions?

Yes, a USMA linear equation for x, y, and z can have multiple solutions. This means that there can be more than one set of values for x, y, and z that satisfy the equation. These solutions can be represented as points on a three-dimensional graph.

## 5. How does a USMA linear equation for x, y, and z differ from a two-variable linear equation?

A USMA linear equation for x, y, and z differs from a two-variable linear equation in that it involves three variables instead of two. This means that it represents a relationship in three-dimensional space, rather than on a two-dimensional graph. Additionally, solving a USMA linear equation for x, y, and z requires using more advanced methods, such as Gaussian elimination or Cramer's rule.

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