Matrices & Determinants
Determinants
nta_pyq_2025_jan
Grade 12
Question:
If the system of linear equations $x+y+2z=6,\ 2x+3y+az=a+1,\ -x-3y+bz=2b$, where $a,b\in\mathbb{R}$, has infinitely many solutions, then $7a+3b$ is equal to:
Step-by-Step Solution
Key Concept: Infinitely many solutions $\Leftrightarrow D=D_{1}=D_{2}=D_{3}=0.$ The first condition gives $2a+b=6$; the second gives a quadratic in $a$ (after substituting $b=6-2a$).
$D=\begin{vmatrix}1&1&2\\2&3&a\\-1&-3&b\end{vmatrix}=1(3b+3a)-(2b+a)+2(-6+3)=2a+b-6.$
$D=0\Rightarrow 2a+b=6.$
$D_{1}=\begin{vmatrix}6&1&2\\a+1&3&a\\2b&-3&b\end{vmatrix}=12a+5b+ab-6=0.$
Substitute $b=6-2a$: $12a+5(6-2a)+a(6-2a)=6\Rightarrow -2a^{2}+8a+24=0\Rightarrow a^{2}-4a-12=0.$
Roots: $a=6,-2.$ Cross-checking with $D_{3}=0$ rules out $a=6$; take $a=-2,\,b=10.$
$7a+3b=-14+30=16.$
Correct Answer: 1