Matrices and Determinants JEE Main PYQs (2020–2026)

Solved Matrices and Determinants PYQs

3 recent questions with answers. The full set of 289 is free to practise.

1. JEE Main 2026EasySystem of Linear Equations using Cramer's Rule
   Q1

   If the system of linear equations :
   
   $$\begin{array}{rl}& x+y+z=6\\ & x+2y+5z=10\\ & 2x+3y+\lambda z=\mu \end{array}$$
   
   has infinitely many solutions, then the value of $\lambda +\mu$ equals:

   • A12
   • B16
   • C22
   • D28

   Show answer & solution

   Correct answer: (C)

   To find the condition for infinitely many solutions, we first calculate the determinant $D$ of the coefficients of $x,y,z$ :
   
   $$D=\left|\begin{array}{ccc}1& 1& 1\\ 1& 2& 5\\ 2& 3& \lambda \end{array}\right|=\lambda -6$$
   
   For infinitely many solutions, $D=0$ .
   
   , $\lambda -6=0\Rightarrow \lambda =6.$
   
   Now, let us find $D_1$ , the determinant obtained by replacing the first column with the constants:
   
   $$D_1=\left|\begin{array}{ccc}6& 1& 1\\ 10& 2& 5\\ \mu & 3& \lambda \end{array}\right|=2\lambda +3\mu -60$$
   
   Next, the determinant $D_2$ is obtained by replacing the second column with the constants:
   
   $$D_2=\left|\begin{array}{ccc}1& 6& 1\\ 1& 10& 5\\ 2& \mu & \lambda \end{array}\right|=4(\lambda -\mu +10)$$
   
   Finally, $D_3$ is obtained by replacing the third column with the constants:
   
   $$D_3=\left|\begin{array}{ccc}1& 1& 6\\ 1& 2& 10\\ 2& 3& \mu \end{array}\right|=\mu -16$$
   
   For the system to have infinitely many solutions, all the determinants must be zero:
   
   $$D=D_1=D_2=D_3=0$$
   
   Substitute $\lambda =6$ :
   
   From $D_3=0$ ,
   
   $$\mu -16=0\Rightarrow \mu =16.$$
   
   Thus,
   
   $$\lambda +\mu =6+16=22$$
2. JEE Main 2026MediumAdjoint of a Matrix & Inverse of a Matrix
   Q2

   Let $A=\left[\begin{array}{ccc}\alpha & 1& 2\\ 2& 3& 0\\ 0& 4& 5\end{array}\right]$ and $B=\left[\begin{array}{ccc}1& 0& 0\\ 0& -5\alpha & 0\\ 0& 4\alpha & -2\alpha \end{array}\right]+adj(A)$ . If $det(B)=66$ , then $det(adj(A))$ equals :

   • A289
   • B361
   • C441
   • D529

   Show answer & solution

   Correct answer: (C)

   $$A=\left[\begin{array}{ccc}\alpha & 1& 2\\ 2& 3& 0\\ 0& 4& 5\end{array}\right]$$
   
   $$\Rightarrow |\mathbf{A}|=15\alpha -10+16=15\alpha +6$$
   
   Now adj $A=\left[\begin{array}{ccc}15& 3& -6\\ -10& 5\alpha & 4\\ 8& -4\alpha & 3\alpha -2\end{array}\right]$
   
   $$\therefore B=\left[\begin{array}{ccc}1& 0& 0\\ 0& -5\alpha & 0\\ 0& 4\alpha & -2\alpha \end{array}\right]+adj(A)$$
   
   $$B=\left[\begin{array}{ccc}16& 3& -6\\ -10& 0& 4\\ 8& 0& \alpha -2\end{array}\right]$$
   
   $$\begin{array}{rlr}& & |B|=6(5\alpha +6)=66\text{ (given in question) }\\ & \Rightarrow \alpha =1& \\ & \Rightarrow |A|=21& \end{array}$$
   
   $$\mathbf{A}\cdot Adj\left(\mathbf{A}\right)=\left|\mathbf{A}\right|\mathbf{I}$$
   
   $$\Rightarrow \left|\mathbf{A}\cdot Adj\left(\mathbf{A}\right)\right|=\parallel \mathbf{A}\left|\mathbf{I}\right|$$
   
   $$=\left(|\mathbf{A}{|}^3\right)|\mathbf{I}|=|\mathbf{A}{|}^3$$
   
   $$\Rightarrow |\mathbf{A}||Adj(\mathbf{A})|=|\mathbf{A}{|}^3$$
   
   $$\Rightarrow |Adj(\mathbf{A})|=|\mathbf{A}{|}^2=(21{)}^2=441$$
3. JEE Main 2026HardEvaluation and Properties of Determinants
   Q3

   Let $A=\left[\begin{array}{ccc}1& 0& 0\\ 3& 1& 0\\ 9& 3& 1\end{array}\right]$ and $B=\left[b_{ij}\right],1\le i,j\le 3$ . If $B=A^{99}-I$ , then the value of $\frac{b_{31}-b_{21}}{b_{32}}$ is :

   • A99
   • B199
   • C149
   • D159

   Show answer & solution

   Correct answer: (C)

   $$\begin{array}{rl}& A=P+I\\ & A=\left[\begin{array}{ccc}0& 0& 0\\ 3& 0& 0\\ 9& 3& 0\end{array}\right]+\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\\ & A=P+I\text{ where }{P}^3=0\\ & A^{99}=I+99P{+}^{99}{C}_2{P}^2\end{array}$$
   
   $$\begin{array}{rl}& A^{99}-I=99\left[\begin{array}{ccc}0& 0& 0\\ 3& 0& 0\\ 9& 3& 0\end{array}\right]{+}^{99}{C}_2\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 9& 0& 0\end{array}\right]\\ & =\left[\begin{array}{ccc}0& 0& 0\\ 99\times 3& 0& 0\\ 44550& 99\times 3& 0\end{array}\right]\end{array}$$
   
   $$b_{31}=44550,b_{21}=99\times 3,b_{32}=99\times 3$$
   
   $\therefore \frac{b_{31}-b_{21}}{b_{32}}$ = $\frac{44550-297}{297}=149$

Topic-wise question count

Where JEE Main has asked from Matrices and Determinants, most frequent first.

• System of Linear Equations using Cramer's Rule64
• Powers of Matrices, Characteristic Equation & Matrix Equations58
• Evaluation and Properties of Determinants54
• Adjoint of a Matrix & Inverse of a Matrix35
• System of Linear Equations using Matrix Method33
• Types of Matrices, Operations & Transpose32
• Symmetric, Skew-Symmetric, Orthogonal & Nilpotent Matrices12

Year-wise Matrices and Determinants questions in JEE Main

Number of questions from this chapter in each year's paper(s).