JEE MAIN Matrices Previous Year Questions (PYQs) – Page 1 of 15
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If the system of equations
$x+\big(\sqrt{2}\sin\alpha\big)y+\big(\sqrt{2}\cos\alpha\big)z=0$
$x+(\cos\alpha)y+(\sin\alpha)z=0$
$x+(\sin\alpha)y-(\cos\alpha)z=0$
has a non-trivial solution, then $\alpha\in\left(0,\frac{\pi}{2}\right)$ is equal to:
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Solution
If $P = \begin{bmatrix} 1 & 0 \\ \tfrac{1}{2} & 1 \end{bmatrix}$, then $P^{50}$ is :
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Solution
Let $A=\begin{bmatrix}1\\1\\1\end{bmatrix}$ and
$B=\begin{bmatrix}
9^{2} & -10^{2} & 11^{2}\\
12^{2} & 13^{2} & -14^{2}\\
-15^{2} & 16^{2} & 17^{2}
\end{bmatrix}$,
then the value of $A'BA$ is:
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Solution
If $A$ is a $3\times 3$ non-singular matrix such that $AA' = A'A$ and
$B = A^{-1}A'$, then $BB'$ equals :
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Solution
Let the sum of the focal distances of the point $P(4,3)$ on the hyperbola $H:\ \dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$ be $8\sqrt{\dfrac{5}{3}}$. If for $H$, the length of the latus rectum is $l$ and the product of the focal distances of the point $P$ is $m$, then $9l^{2}+6m$ is equal to:
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Solution
If $\alpha,\beta \ne 0$, and $f(n) = \alpha^{n} + \beta^{n}$ and
$\begin{vmatrix}
3 & 1 + f(1) & 1 + f(2) \\
1+f(1) & 1 + f(2) & 1 + f(3) \\
1+f(2) & 1 + f(3) & 1 + f(4)
\end{vmatrix}
= K(1-\alpha)^{2}(1-\beta)^{2}(\alpha-\beta)^{2}$, then $K$ is equal to :
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Solution
For the system of linear equations $\alpha x+y+z=1,\quad x+\alpha y+z=1,\quad x+y+\alpha z=\beta$, which one of the following statements is **NOT** correct?
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Solution
The values of $\lambda$ and $\mu$ for which the system of linear equations
\[
\begin{aligned}
x + y + z &= 2,\\
x + 2y + 3z &= 5,\\
x + 3y + \lambda z &= \mu
\end{aligned}
\]
has infinitely many solutions are, respectively:
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Solution
Let m and M be respectively the minimum and maximum values of
\[
\left|
\begin{array}{ccc}
\cos^{2}x & 1+\sin^{2}x & \sin 2x\\
1+\cos^{2}x & \sin^{2}x & \sin 2x\\
\cos^{2}x & \sin^{2}x & 1+\sin 2x
\end{array}
\right|.
\]
Then the ordered pair (m, M) is equal to :
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Solution
Let $m$ and $M$ be respectively the minimum and maximum values of
\[
\left|\begin{matrix}
\cos^2 x & 1+\sin^2 x & \sin 2x\\
1+\cos^2 x & \sin^2 x & \sin 2x\\
\cos^2 x & \sin^2 x & 1+\sin 2x
\end{matrix}\right|.
\]
Then the ordered pair $(m, M)$ is equal to :
Solution:
Apply $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$
\[
\Delta =
\left|\begin{matrix}
\cos^2 x & 1+\sin^2 x & \sin 2x\\
1 & -1 & 0\\
0 & -1 & 1
\end{matrix}\right|
\]
Expanding along the first row,
\[
\Delta = \cos^2 x(-1) - (1+\sin^2 x)(1) - \sin 2x(1)
\]
\[
\Delta = -(\cos^2 x + \sin^2 x + 1 + \sin 2x)
\]
\[
\Delta = -2 - \sin 2x
\]
Since $\sin 2x \in [-1,1]$,
\[
\Delta \in [-3, -1]
\]
Hence, $m = -3$, $M = -1$
Therefore, the ordered pair is $\boxed{(-3, -1)}$.
A value of $\theta \in \left( {0,{\pi \over 3}} \right)$, for which
$\left| {\matrix{ {1 + {{\cos }^2}\theta } & {{{\sin }^2}\theta } & {4\cos 6\theta } \cr {{{\cos }^2}\theta } & {1 + {{\sin }^2}\theta } & {4\cos 6\theta } \cr {{{\cos }^2}\theta } & {{{\sin }^2}\theta } & {1 + 4\cos 6\theta } \cr } } \right| = 0$, is :
$\left| {\matrix{ {1 + {{\cos }^2}\theta } & {{{\sin }^2}\theta } & {4\cos 6\theta } \cr {{{\cos }^2}\theta } & {1 + {{\sin }^2}\theta } & {4\cos 6\theta } \cr {{{\cos }^2}\theta } & {{{\sin }^2}\theta } & {1 + 4\cos 6\theta } \cr } } \right| = 0$, is :
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Solution
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