Matrices & Determinants · JEE Main previous year questions

2023 31 questions

1. Q152023 · 25 Jan 2023 · Shift 2

   Let $A=\left[\begin{array}{cc}\frac{1}{\sqrt{10}}& \frac{3}{\sqrt{10}}\\ \frac{-3}{\sqrt{10}}& \frac{1}{\sqrt{10}}\end{array}\right]$ and $B=\left[\begin{array}{cc}1& -i\\ 0& 1\end{array}\right]$, where $i=\sqrt{-1}$. If $\mathrm{M}={\mathrm{A}}^{\mathrm{T}}\mathrm{BA}$, then the inverse of the matrix $\mathrm{A}{\mathrm{M}}^{2023}{\mathrm{A}}^{\mathrm{T}}$ is

   1. A.$\left[\begin{array}{cc}1& -2023i\\ 0& 1\end{array}\right]$
   2. B.$\left[\begin{array}{cc}1& 0\\ 2023i& 1\end{array}\right]$
   3. C.$\left[\begin{array}{cc}1& 2023i\\ 0& 1\end{array}\right]$
   4. D.$\left[\begin{array}{cc}1& 0\\ -2023i& 1\end{array}\right]$
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   Answer: C

2. Q102023 · 24 Jan 2023 · Shift 1

   If A and B are two non-zero $n\times n$ matrices such that $A^2+B=A^{2}B$, then :

   1. A.$A^{2}B=I$
   2. B.$A^2=I$ or $B=I$
   3. C.$A^{2}B=B{A}^2$
   4. D.$AB=I$
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   Answer: C

3. Q202023 · 24 Jan 2023 · Shift 1

   Let $\alpha$ be a root of the equation $(a-c)x^2+(b-a)x+(c-b)=0$ where a, b, c are distinct real numbers such that the matrix $\left[\begin{array}{ccc}{\alpha }^2& \alpha & 1\\ 1& 1& 1\\ a& b& c\end{array}\right]$ is singular. Then, the value of $\frac{(a-c{)}^2}{(b-a)(c-b)}+\frac{(b-a{)}^2}{(a-c)(c-b)}+\frac{(c-b{)}^2}{(a-c)(b-a)}$ is

   1. A.3
   2. B.6
   3. C.12
   4. D.9
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   Answer: A

4. Q52023 · 24 Jan 2023 · Shift 2

   If the system of equations $x+2y+3z=3$ $4x+3y-4z=4$ $8x+4y-\lambda z=9+\mu$ has infinitely many solutions, then the ordered pair $(\lambda ,\mu )$ is equal to :

   1. A.$\left(\frac{72}{5},\frac{21}{5}\right)$
   2. B.$\left(-\frac{72}{5},-\frac{21}{5}\right)$
   3. C.$\left(-\frac{72}{5},\frac{21}{5}\right)$
   4. D.$\left(\frac{72}{5},-\frac{21}{5}\right)$
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   Answer: D

5. Q102023 · 24 Jan 2023 · Shift 2

   The number of square matrices of order 5 with entries from the set {0, 1}, such that the sum of all the elements in each row is 1 and the sum of all the elements in each column is also 1, is :

   1. A.125
   2. B.150
   3. C.225
   4. D.120
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   Answer: D

6. Q162023 · 24 Jan 2023 · Shift 2

   Let A be a $3\times 3$ matrix such that $|\mathrm{adj}(\mathrm{adj}(\mathrm{adj}\mathrm{A}))|=12^4$. Then $|{\mathrm{A}}^{-1}\mathrm{adj}\mathrm{A}|$ is equal to

   1. A.12
   2. B.$2\sqrt{3}$
   3. C.1
   4. D.$\sqrt{6}$
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   Answer: B