Question

Examine the consistency of the system of equation

1. x + 2y = 2
2x + 3y = 3

Note :- Solve by Matrix Method ( Gauss - Jordan Method)​

Answer 1

Explanation:

The given system of equation is,

$\tt x + 2y = 2 \\ \tt 2x + 3y = 3$

The given system of equations can be written in the form of AX = B, Where,

$\tt \green{ A = \tt{\begin{bmatrix}1 & 2 \\2 & 3 \end{bmatrix}} \:, \: \: X = {\begin{bmatrix} x \\ y \end{bmatrix}} \: \: and \: \: B = {\begin{bmatrix} 2 \\ 3 \end{bmatrix}}}$

Now,

$\rightarrow \tt \: |A| = 1 \times 3 - 2 \times 2 \\ \\ \rightarrow \tt \: |A| =3 - 4 \\ \\ \rightarrow \tt \red{|A| = - 1}$

$\tt \therefore \: \: A \: \: is \: \: non - singular.$

$\tt \: \therefore \: \: A {}^{ - 1} \: exists.$

$\underline{ \bold{ \: Hence, \: the \: given \: system \: of \: equations \: is \: consistent.}}$

Answer 2

$\tt{\:x + 2y =2}$

$\tt{\:2x + 3y =3}$

Write equation AX = B

$\left[ \begin{array}{c c}\tt 1 & \tt 2 \\\tt 2 & \tt 3\end{array}\right]$ $\left[ \begin{array}{c}\tt x \\ \tt y\end{array}\right]$ $=$ $\left[ \begin{array}{c} \tt 2 \\ \tt 3\end{array} \right]$

Hence,

A = $\left[ \begin{array}{c c}\tt 1 & \tt 2 \\ \tt 2 & \tt 3\end{array} \right]$, B = $\left[ \begin{array}{c}\tt x \\ \tt y\end{array} \right]$, C = $\left[ \begin{array}{c} \tt 2 \\ \tt 3\end{array}\right]$

$\:$

|A| = $\left| \begin{array}{c c}1 & 2 \\ 2 &3\end{array}\right|$

= (1 × 3) - (2 × 2) = -1 ≠ 0

Since, |A| ≠ 0

The system of equations is consistent.