Question

Consider the following system of equations:

$\begin{array}{rcl}x+2y–3z& =& a\\ 2x+6y–11z& =& b\\ x–2y+7z& =& c,\end{array}$

Where $a,b$ and $c$ are real constants. Then the system of equations

• A. has a unique solution when $5a=2b+c$
• B. has an infinite number of solutions when $5a=2b+c$
• C. has no solution for all $a,b$ and $c$
• D. has a unique solution for all $a,b$ and $c$

Answer 1

The correct option is B

has an infinite number of solutions when $5a=2b+c$

Explanation for the correct option:

Step 1. Given the System of equations:

$\begin{array}{rcl}x+2y–3z& =& a\\ 2x+6y–11z& =& b\\ x–2y+7z& =& c,\end{array}$

Step 2. Compute $\begin{array}{rcl}\mathbf{∆}& \mathbf{,}& {\mathbf{∆}}_{\mathbf{1}}\mathbf{,}{\mathbf{∆}}_{\mathbf{2}}\mathbf{}\mathbf{a}\mathbf{n}\mathbf{d}{\mathbf{∆}}_{\mathbf{3}}\mathbf{}\end{array}$:

$\begin{array}{rcl}∆& =& \left|\begin{array}{ccc}1& 2& -3\\ 2& 6& -11\\ 1& -2& 7\end{array}\right|\\ & =& 20-2\left(25\right)-3(-10)\\ & =& 20-50+30\\ & =& 0\end{array}$

$\begin{array}{rcl}{∆}_1& =& \left|\begin{array}{ccc}a& 2& -3\\ b& 6& -11\\ c& -2& 7\end{array}\right|\\ & =& 20a–2(7b+11c)–3(-2b–6c)\\ & =& 20a–14b–22c+6b+18c\\ & =& 20a–8b–4c\\ & =& 4(5a–2b–c)\end{array}$

$\begin{array}{rcl}{∆}_2& =& \left|\begin{array}{ccc}1& a& -3\\ 2& b& -11\\ 1& c& 7\end{array}\right|\\ & =& 7b+11c–a\left(25\right)–3(2c–b)\\ & =& 7b+11c–25a–6c+3b\\ & =& –25a+10b+5c\\ & =& –5(5a–2b-c)\end{array}$

$\begin{array}{rcl}{∆}_3& =& \left|\begin{array}{ccc}1& 2& a\\ 2& 6& b\\ 1& -2& c\end{array}\right|\\ & =& 6c+2b-2(2c–b)–10a\\ & =& –10a+4b+2c\\ & =& –2(5a–2b-c)\end{array}$

Step 3. Find the relation for infinite solution :

For infinite solutions,

$\begin{array}{rcl}∆& =& {∆}_1={∆}_2={∆}_3=0\\ & \Rightarrow & 5a=2b+c\\ & & \end{array}$

Hence, the correct option is B.