IIT JEE Main Maths -Unit 3 -Matrices- Exam Style Questions- New Syllabus
▶ Answer/Explanation
Ans. (3)
Sol. $\Delta = \begin{vmatrix} 1 & 1 & 2 \\ 2 & 3 & a \\ -1 & -3 & b \end{vmatrix} = 0$
$\Rightarrow 2a + b – 6 = 0$ $\cdots$(1)
$\Delta_1 = \begin{vmatrix} 1 & 1 & 6 \\ 2 & 3 & a + 1 \\ -1 & -3 & 2b \end{vmatrix} = 0$
$\Rightarrow a + b – 8 = 0$ $\cdots$(2)
Solving (1) + (2) $a = -2$, $b = 10$
$\Rightarrow 7a + 3b = 16$
Sol. $\Delta = \begin{vmatrix} 1 & 1 & 2 \\ 2 & 3 & a \\ -1 & -3 & b \end{vmatrix} = 0$
$\Rightarrow 2a + b – 6 = 0$ $\cdots$(1)
$\Delta_1 = \begin{vmatrix} 1 & 1 & 6 \\ 2 & 3 & a + 1 \\ -1 & -3 & 2b \end{vmatrix} = 0$
$\Rightarrow a + b – 8 = 0$ $\cdots$(2)
Solving (1) + (2) $a = -2$, $b = 10$
$\Rightarrow 7a + 3b = 16$
▶️ Answer/Explanation
Ans. (2)
\( (\lambda – 1)x + (\lambda – 4)y + \lambda z = 5 \)
\( \lambda x + (\lambda – 1)y + (\lambda – 4)z = 7 \)
\( (\lambda + 1)x + (\lambda + 2)y – (\lambda + 2)z = 9 \)
For infinitely many solutions,
\( D =
\begin{vmatrix}
\lambda – 1 & \lambda – 4 & \lambda \\
\lambda & \lambda – 1 & \lambda – 4 \\
\lambda + 1 & \lambda + 2 & -(\lambda + 2)
\end{vmatrix} = 0 \)
\( (\lambda – 3)(2\lambda + 1) = 0 \)
\( D_x =
\begin{vmatrix}
5 & \lambda – 4 & \lambda \\
7 & \lambda – 1 & \lambda – 4 \\
9 & \lambda + 2 & -(\lambda + 2)
\end{vmatrix} = 0 \)
\( 2(3 – \lambda)(23 – 2\lambda) = 0 \)
\( \lambda = 3 \)
\( \therefore \lambda^2 + \lambda = 9 + 3 = 12 \)