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JEE Main 20244 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
4 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Matrices

JEE Main 2024 · 4 April · Evening Shift · Mathematics · Matrices

Let A be a 2 × 2 symmetric matrix such that A [ array l 1 1 array...

Let AA be a 2×22 \times 2 symmetric matrix such that AA\left[11\begin{array}{l}1 \\ 1\end{array}\right]=\left[37\begin{array}{l}3 \\ 7\end{array}\right] and the determinant of AA be 1 . If A−1=αA+βIA^{-1}=\alpha A+\beta I, where II is an identity matrix of order 2×22 \times 2, then α+β\alpha+\beta equals _______

Correct answer

0

Step-by-step solution

 Let A=[abbd][abbd][11]=[37],ad−b2=1a+b=3,b+d=7,(3−b)(7−b)−b2=121−10b=1→b=2,a=1,d=5A=[1225],A−1=[5−2−21]A−1=αA+βI[5−2−21]=[α+β2α2α5α+β]α=−1,β=6→α+β=5\begin{aligned} & \text{ Let } A=\left[\begin{array}{ll}a & b \\ b & d\end{array}\right] \\ & {\left[\begin{array}{ll}a & b \\ b & d\end{array}\right]\left[\begin{array}{l}1 \\ 1\end{array}\right]=\left[\begin{array}{l}3 \\ 7\end{array}\right], a d-b^2=1} \\ & a+b=3, b+d=7,(3-b)(7-b)-b^2=1 \\ & 21-10 b=1 \rightarrow b=2, a=1, d=5 \\ & A=\left[\begin{array}{ll}1 & 2 \\ 2 & 5\end{array}\right], A^{-1}=\left[\begin{array}{cc}5 & -2 \\ -2 & 1\end{array}\right] \\ & A^{-1}=\alpha A+\beta I \\ & {\left[\begin{array}{cc}5 & -2 \\ -2 & 1\end{array}\right]=\left[\begin{array}{cc}\alpha+\beta & 2 \alpha \\ 2 \alpha & 5 \alpha+\beta\end{array}\right]} \\ & \alpha=-1, \beta=6 \rightarrow \alpha+\beta=5\end{aligned}
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