Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Determinants

Question:

The following system of equations has :

$x+2 y+3 z=1$

$x-y+4 z=0$

$2 x+y+7 z=1$

Options:

a unique solution

no solution

infinitely many solutions

only two solutions

Correct Answer:

no solution

Explanation:

$x+2 y+3 z=1$

$x-y+4 z=0$

$2 x+y+7 z=1$

$A=\left[\begin{array}{ccc}1 & 2 & 3 \\ 1 & -1 & 4 \\ 2 & 1 & 7\end{array}\right]$        $B=\left[\begin{array}{l}1 \\ 0 \\ 1\end{array}\right]$

finding $|A|$

$|A|=\left|\begin{array}{ccc}
1 & 2 & 3 \\
1 & -1 & 4 \\
2 & 1 & 7
\end{array}\right|$

$=1\left[\begin{array}{rr}-1 & 4 \\ 1 & 7\end{array}\right]-2\left[\begin{array}{ll}1 & 4 \\ 2 & 7\end{array}\right]+3\left[\begin{array}{cc}1 & -1 \\ 2 & 1\end{array}\right]$

$=1(-7-4)-2(7-8)+3(1+2)$

$=1(-11)-2(-1)+3(3)$

$-11+2+9$

$=-11+11=0$

as |A| = 0 and every element of B is not zero

option → 2, no solution