Determinants JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Determinants, free to read — no sign-in needed. The full chapter has 22 questions; sign in to attempt the remaining 17 in the exam simulator.

1. Q1JEE Advanced Adv 2024 (Paper 1)
   Let $S=\left\{A=\left(\begin{array}{lll}0 & 1 & c \\ 1 & a & d \\ 1 & b & e\end{array}\right): a, b, c, d, e \in\{0,1\}\right.$ and $\left.|A| \in\{-1,1\}\right\}$, where $|A|$ denotes the determinant of $A$. Then the number of elements in $S$ is
   Show answer & solution

   Answer: 16

   $|A|=-(e-d)+c(b-a)= \pm 1$ Case (i) : $c=0 \Rightarrow(e, d)=(1,0),(0,1) \rightarrow 2$ ways b and a can be each 2 ways $\Rightarrow$ Total $=8$ ways Case (ii) : $c \neq 0 \Rightarrow c=1$ $\Rightarrow d-e+b-a= \pm 1$ $\left.\begin{array}{llll}1 & 1 & 1 & 0 \\ 1 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{array}\right\} \rightarrow 4 \times 2=8$ ways Total $=16$ ways
2. Q2JEE Advanced Adv 2021 (Paper 1)
   Let $\alpha, \beta$ and $\gamma$ be real numbers such that the system of linear equations $$\begin{array}{c} x+2 y+3 z=\alpha \\ 4 x+5 y+6 z=\beta \\ 7 x+8 y+9 z=\gamma-1 \end{array}$$ is consistent. Let $|M|$ represent the determinant of the matrix $$M=\left[\begin{array}{ccc} \alpha & 2 & \gamma \\ \beta & 1 & 0 \\ -1 & 0 & 1 \end{array}\right]$$ Let $P$ be the plane containing all those $(\alpha, \beta, \gamma)$ for which the above system of linear equations is consistent, and $D$ be the square of the distance of the point $(0,1,0)$ from the plane $P$. The value of $D$ is
   Show answer & solution

   Answer: 1.5

   $x+2y+3z=\alpha$ $4x+5y+6z=\beta$ $7x+8y+9z=\gamma -1$ The system of linear equations is consistent it means it has unique solution or infinite solutions Here, $\Delta =\left|\begin{matrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9\end{matrix}\right|=0$ Hence, there are infinitely many solutions. If equations have infinitely many solutions, then the equations are linearly connected, i.e., ${L}_{1}+\lambda {L}_{2}={L}_{3}$ $\Rightarrow \left(x+2y+3z-\alpha \right)+\lambda (4x+5y+6z-\beta )=7x+8y+9z-\gamma +1$ $\dfrac{1+4\lambda }{7}=\dfrac{2+5\lambda }{8}=\dfrac{3+6\lambda }{9}=\dfrac{\alpha +\lambda \beta }{\gamma -1}$ $\dfrac{1+4\lambda }{7}=\dfrac{2+5\lambda }{8}\Rightarrow \lambda =-2$ Also,$\dfrac{1+4\lambda }{7}=\dfrac{\alpha +\lambda B}{\gamma -1}$ $\Rightarrow -1=\dfrac{\alpha -2\beta }{r-1}$ $\Rightarrow \alpha -2\beta +\gamma =1$ Now, $P$ is the plane containing the points $\left(\alpha ,\beta ,\gamma \right)$ So, the equation of the plane is $x-2y+z=1$ (replacing $\alpha ,\beta ,\gamma$ by $x,y,z$) We know, the distance of a point $\left({x}_{1},{y}_{1},{z}_{1}\right)$ from plane $ax+by+cz+d=0$ is $\left|\dfrac{a{x}_{1}+b{y}_{1}+c{z}_{1}+d}{\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}}\right|$ So, $D={\left(\dfrac{0\times 1-2\times 1+0\times 1-1}{\sqrt{{1}^{2}+(-2{)}^{2}+{1}^{2}}}\right)}^{2}=\dfrac{9}{6}=1.50$
3. Q3JEE Advanced Adv 2021 (Paper 1)
   Let $\alpha, \beta$ and $\gamma$ be real numbers such that the system of linear equations $$\begin{array}{c} x+2 y+3 z=\alpha \\ 4 x+5 y+6 z=\beta \\ 7 x+8 y+9 z=\gamma-1 \end{array}$$ is consistent. Let $|M|$ represent the determinant of the matrix $$M=\left[\begin{array}{ccc} \alpha & 2 & \gamma \\ \beta & 1 & 0 \\ -1 & 0 & 1 \end{array}\right]$$ Let $P$ be the plane containing all those $(\alpha, \beta, \gamma)$ for which the above system of linear equations is consistent, and $D$ be the square of the distance of the point $(0,1,0)$ from the plane $P$. The value of $|M|$ is
   Show answer & solution

   Answer: 1

   $x+2y+3z=\alpha$ $4x+5y+6z=\beta$ $7x+8y+9z=\gamma -1$ The system of linear equations is consistent it means it has unique solution or infinite solutions Here, $\Delta =\left|\begin{matrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9\end{matrix}\right|=0$ Hence, there are infinitely many solutions. If equations have infinitely many solutions, then the equations are linearly connected, i.e., ${L}_{1}+\lambda {L}_{2}={L}_{3}$ $\Rightarrow \left(x+2y+3z-\alpha \right)+\lambda (4x+5y+6z-\beta )=7x+8y+9z-\gamma +1$ $\dfrac{1+4\lambda }{7}=\dfrac{2+5\lambda }{8}=\dfrac{3+6\lambda }{9}=\dfrac{\alpha +\lambda \beta }{\gamma -1}$ $\dfrac{1+4\lambda }{7}=\dfrac{2+5\lambda }{8}\Rightarrow \lambda =-2$ Also,$\dfrac{1+4\lambda }{7}=\dfrac{\alpha +\lambda B}{\gamma -1}$ $\Rightarrow -1=\dfrac{\alpha -2\beta }{r-1}$ $\Rightarrow \alpha -2\beta +\gamma =1$ Now, $\left|M\right|=\left|\begin{matrix}\alpha & 2 & \gamma \\ \beta & 1 & 0 \\ -1 & 0 & 1\end{matrix}\right|$ $=\alpha +2(-\beta )+\gamma (1)=\alpha -2\beta +\gamma =1$
4. Q4JEE Advanced Adv 2019 (Paper 2)
   Let $x\in R$ and let $P=\left[\begin{matrix}1 & 1 & 1 \\ 0 & 2 & 2 \\ 0 & 0 & 3\end{matrix}\right],Q=\left[\begin{matrix}2 & x & x \\ 0 & 4 & 0 \\ x & x & 6\end{matrix}\right]$ and $R=PQ{P}^{-1}.$ Then which of the following options is/are correct?
   1. A.For $x=1,$ there exists a unit vector $\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$ for which $R\left[\begin{matrix}\alpha \\ \beta \\ \gamma \end{matrix}\right]=\left[\begin{matrix}0 \\ 0 \\ 0\end{matrix}\right]$
   2. B.There exists a real number $x$ such that $PQ=QP$
   3. C.$\det ⁡R=\det ⁡\left[\begin{matrix}2 & x & x \\ 0 & 4 & 0 \\ x & x & 5\end{matrix}\right]+8,$ for all $x\in R$
   4. D.For $x=0,$ if $R\left[\begin{matrix}1 \\ a \\ b\end{matrix}\right]=6\left[\begin{matrix}1 \\ a \\ b\end{matrix}\right],$ then $a+b=5$
   Show answer & solution

   Answer: C,D

   $\begin{matrix}P=\left[\begin{matrix}1 & 1 & 1 \\ 0 & 2 & 2 \\ 0 & 0 & 3\end{matrix}\right] & Q=\left[\begin{matrix}2 & x & x \\ 0 & 4 & 0 \\ x & x & 6\end{matrix}\right]\end{matrix}$ Option $(C)$ Now $R=PQ{P}^{-1}$ $\det ⁡\left(R\right)=\det ⁡\left(R\right)\det ⁡\left(Q\right)\det ⁡\left({P}^{-1}\right)$ $\begin{matrix}\det ⁡\left(R\right)=\det ⁡\left(Q\right) & \left(\det ⁡\left({P}^{-1}\right)=\dfrac{1}{\det ⁡\left(P\right)}\right)\end{matrix}$ $\det ⁡\left(R\right)=\det ⁡\left[\begin{matrix}2 & x & x \\ 0 & 4 & 0 \\ x & x & 6\end{matrix}\right]$ $\det ⁡\left(R\right)=48-4{x}^{2}$ now $\det ⁡\left[\begin{matrix}2 & x & x \\ 0 & 4 & 0 \\ x & x & 5\end{matrix}\right]=40-4{x}^{2}$ $\det ⁡R=\det ⁡\left[\begin{matrix}2 & x & x \\ 0 & 4 & 0 \\ x & x & 5\end{matrix}\right]+8$ Option $\left(A\right)$ $R\left[\begin{matrix}\alpha \\ \beta \\ \gamma \end{matrix}\right]=\left[\begin{matrix}0 \\ 0 \\ 0\end{matrix}\right]$ must have not trivial solution So $\det ⁡\left(R\right)=0$ $48-4{x}^{2}=0\Rightarrow x=\pm 2\sqrt{3}$ Option $\left(D\right)$ $R\left[\begin{matrix}1 \\ a \\ b\end{matrix}\right]=\left[\begin{matrix}6 \\ 6a \\ 6b\end{matrix}\right]$ $PQ{P}^{-1}\left[\begin{matrix}1 \\ a \\ b\end{matrix}\right]=\left[\begin{matrix}6 \\ 6a \\ 6b\end{matrix}\right]$ …(i) ${P}^{-1}=\dfrac{1}{6}\left[\begin{matrix}6 & -3 & 0 \\ 0 & 3 & -2 \\ 0 & 0 & 2\end{matrix}\right]$ Putting $P,Q,{P}^{-1}$ in equation (i) $\dfrac{1}{6}\left[\begin{matrix}12 & 6 & 4 \\ 0 & 24 & 8 \\ 0 & 0 & 36\end{matrix}\right]\left[\begin{matrix}1 \\ a \\ b\end{matrix}\right]=\left[\begin{matrix}6 \\ 6a \\ 6b\end{matrix}\right]$ $\Rightarrow 12+6a+4b=36$ $24a+8b=36a$ $a=2$ and $b=3$ $a+b=5$ Option $\left(B\right)$ $PQ=QP$ $PQ=QP\Rightarrow PQ{P}^{-1}=Q$ $R=Q$ Not possible for any value of $x.$
5. Q5JEE Advanced Adv 2018 (Paper 2)
   Let P be a matrix of order $3\times 3$ such that all the entries in P are from the set $\left\{-1,0,1\right\}$ . Then, the maximum possible value of the determinant of P is _______.
   Show answer & solution

   Answer: 4

   $∆=\left|\begin{matrix}{a}_{1} & {a}_{2} & {a}_{3} \\ {b}_{1} & {b}_{2} & {b}_{3} \\ {c}_{1} & {c}_{2} & {c}_{3}\end{matrix}\right|=\begin{matrix}\left({a}_{1}{b}_{2}{c}_{3}+{a}_{2}{b}_{3}{c}_{1}+{a}_{3}{b}_{1}{c}_{2}\right)_{⏟} \\ x\end{matrix}-\begin{matrix}\left({a}_{3}{b}_{2}{c}_{1}+{a}_{2}{b}_{1}{c}_{3}+{a}_{1}{b}_{3}{c}_{2}\right)_{⏟} \\ y\end{matrix}$ Now if $x\leq 3$ and $y\geq -3$ The $∆$ can be maximum 6 but it is not possible because $x=3\Rightarrow$ each term of $x=1$ And $y=-3\Rightarrow$ each term of $y=-1$ $\Rightarrow$ $∏_{i=1}^{3}{a}_{i}{b}_{i}{c}_{i}=1$ and $∏_{i=1}^{3}{a}_{i}{b}_{i}{c}_{i}=-1$, which is a contradiction. Now for value to be $5$ one the terms must be zero but that will make two terms zero which means answer cannot be $5$. Next possibility is $4$ which can be obtained as $\left|\begin{matrix}1 & 1 & 1 \\ -1 & 1 & 1 \\ 1 & -1 & 1\end{matrix}\right|=1\left(1+1\right)-1\left(-1-1\right)+1\left(1-1\right)=4$