# USMA linear systems of equation solve for x y z

#### karush

##### Well-known member

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:

#### tkhunny

##### Well-known member
MHB Math Helper
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?

#### karush

##### Well-known member
Ok W|A returned this but I don't know how you could get this in a few minutes on an intrance exam

$A=\left[ \begin{array}{cccc} 1 & 0 &0 & \frac{25}{37} \\ 0 & 1 & 0 & \frac{4}{37} \\ 0 &0 &1 & \frac{13}{37} \end{array} \right]$

#### tkhunny

##### Well-known member
MHB Math Helper
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?

#### karush

##### Well-known member
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: