-1

$ \left( \begin{array}{cc} 1 & 2 \\ 2 & 1 \end{array} \right) % \left( \begin{array}{cc} a \\ b \end{array} \right) % = % \left( \begin{array}{cc} 3 - X \\ 6 - X \end{array} \right) $

Can anyone please verify if my answer is correct:

$a = 1-0.66X$

$b = 1+0.32X$

flyhigh1
  • 11

1 Answers1

0

The matrix equation $\begin{bmatrix}1&2\\2&1\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}=\begin{bmatrix}3-X\\6-X\end{bmatrix}$

is equivalent to the system of equations:

$$\begin{cases}a+2b=3-X\\ 2a+b=6-X\end{cases}$$

Taking twice the second line and subtracting the first implies the equation: $2(2a+b)-(a+2b)=2(6-X)-(3-X)$, or simplified $3a = 9-X$. In other words, $a=3-\frac{1}{3}X$

Do so similarly for finding $b$.

JMoravitz
  • 79,518
  • JMoravitz: b = -0.33X Could you please confirm? – flyhigh1 Jul 12 '16 at 02:36
  • @flyhigh1 and how did you come upon that? Did you do like I did and take the equations in the implied system and combined them? You should have done twice the first equation minus the second equation to get $2(a+2b)-(2a+b)=2(3-x)-(6-x)$ to get $3b = -x$. Note that $\frac{1}{3}$ is not the same thing as $0.33$. If you are doing work in mathematical contexts, be exact with numbers, don't round unless explicitly told to or unless completely unavoidable. $0.33=\frac{33}{100}\neq \frac{33}{99}=\frac{1}{3}$ – JMoravitz Jul 12 '16 at 02:39