Exams Joint Entrance Examination Mathematics Matrices and Determinants Question 105 Question 105 of 472 Medium MCQ
If the system of equations
x + y + z = 5
x + 2y + 3z = 9
x + 3y + az = \[\beta \]
has infinitely many solutions, then \[\beta \] - \[\alpha \] equals -
Solutions · 1 Post your solution Official solution
\[D = \left| { ( 1 1 1 1 2 3 1 3 α ) \begin{pmatrix} 1 & 1 & 1 \\
1 & 2 & 3 \\
1 & 3 & \alpha \\
\end{pmatrix} 1 1 1 1 2 3 1 3 α } \right| = \left| { ( 1 1 1 0 1 2 0 2 α − 1 ) \begin{pmatrix} 1 & 1 & 1 \\
0 & 1 & 2 \\
0 & 2 & {\alpha - 1} \\
\end{pmatrix} 1 0 0 1 1 2 1 2 α − 1 } \right|\]
\[ = \left( {\alpha - 1} \right) - 4 = \left( {\alpha - 5} \right)\]
for infinite solutions \[D = 0 \Rightarrow \alpha = 5\]
\[{D_x} = 0 \Rightarrow \left| { ( 5 1 1 9 2 3 β 3 5 ) \begin{pmatrix} 5 & 1 & 1 \\
9 & 2 & 3 \\
\beta & 3 & 5 \\
\end{pmatrix} 5 9 β 1 2 3 1 3 5 } \right| = 0\]
\[ \Rightarrow \left| { ( 0 0 1 − 1 − 1 3 β − 15 − 2 5 ) \begin{pmatrix} 0 & 0 & 1 \\
{ - 1} & { - 1} & 3 \\
{\beta - 15} & { - 2} & 5 \\
\end{pmatrix} 0 − 1 β − 15 0 − 1 − 2 1 3 5 } \right| = 0\]
\[ \Rightarrow 2 + \beta - 15 = 0 \Rightarrow \beta - 13 = 0\]
on \[\beta = 13\] we get D y = D z = 0 {D_y} = {D_z} = 0 D y = D z = 0
\[\alpha = 5,\beta = 13\] Answer the question to unlock