JEE MAIN Matrices Previous Year Questions (PYQs) – Page 1 of 17
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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
For the matrices
$A = \begin{bmatrix} 3 & -4 \ 1 & -1 \end{bmatrix}, \quad B = \begin{bmatrix} -29 & 49 \ -13 & 18 \end{bmatrix}$
if $(A^{15} + B)\begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}$, then among the following which one is true?
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Solution
Here
$A^n = \begin{bmatrix} 2n+1 & -4n \ n & -2n+1 \end{bmatrix}$
$\Rightarrow A^{15} = \begin{bmatrix} 31 & -60 \ 15 & -29 \end{bmatrix}$
$A^{15} + B = \begin{bmatrix} 2 & -11 \ 2 & -11 \end{bmatrix}$
Now
$\begin{bmatrix} 2 & -11 \ 2 & -11 \end{bmatrix}\begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}$
$\Rightarrow 2x - 11y = 0$
$\Rightarrow x = 11,; y = 2$
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
Given sum of focal distances:
$SP_1 + SP_2 = 2ex = 8\sqrt{\frac{5}{3}}$
At $P(4,3)$ ⇒ $2e \cdot 4 = 8\sqrt{\frac{5}{3}}$
⇒ $e=\sqrt{\frac{5}{3}}$
Now, $b^2 = a^2(e^2 - 1) = a^2\left(\frac{5}{3}-1\right)=\frac{2}{3}a^2$
Using point $(4,3)$: $\frac{16}{a^2}-\frac{9}{b^2}=1$
Substitute $b^2=\frac{2}{3}a^2$:
$\frac{16}{a^2}-\frac{9}{(2/3)a^2}=1$
⇒ $\frac{16}{a^2}-\frac{27}{2a^2}=1$
⇒ $\frac{5}{2a^2}=1$
⇒ $a^2=\frac{5}{2},\ b^2=\frac{5}{3}$
Latus rectum: $l=\frac{2b^2}{a}$ ⇒ $l^2=\frac{4b^4}{a^2}$
⇒ $l^2=\frac{4(25/9)}{5/2}=\frac{40}{9}$
⇒ $9l^2=40$
Product of focal distances: $m=(ex+a)(ex-a)=e^2x^2-a^2$
⇒ $m=\frac{5}{3}\cdot16-\frac{5}{2}=\frac{145}{6}$
⇒ $6m=145$
$9l^2+6m=40+145=185$
$\boxed{185}$
At $P(4,3)$ ⇒ $2e \cdot 4 = 8\sqrt{\frac{5}{3}}$
⇒ $e=\sqrt{\frac{5}{3}}$
Now, $b^2 = a^2(e^2 - 1) = a^2\left(\frac{5}{3}-1\right)=\frac{2}{3}a^2$
Using point $(4,3)$: $\frac{16}{a^2}-\frac{9}{b^2}=1$
Substitute $b^2=\frac{2}{3}a^2$:
$\frac{16}{a^2}-\frac{9}{(2/3)a^2}=1$
⇒ $\frac{16}{a^2}-\frac{27}{2a^2}=1$
⇒ $\frac{5}{2a^2}=1$
⇒ $a^2=\frac{5}{2},\ b^2=\frac{5}{3}$
Latus rectum: $l=\frac{2b^2}{a}$ ⇒ $l^2=\frac{4b^4}{a^2}$
⇒ $l^2=\frac{4(25/9)}{5/2}=\frac{40}{9}$
⇒ $9l^2=40$
Product of focal distances: $m=(ex+a)(ex-a)=e^2x^2-a^2$
⇒ $m=\frac{5}{3}\cdot16-\frac{5}{2}=\frac{145}{6}$
⇒ $6m=145$
$9l^2+6m=40+145=185$
$\boxed{185}$
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
The system of equations
$3x + y + 4z = 3$
$2x + ay - z = -3$
$x + 2y + z = 4$
has no solution, then the value of $\alpha$ is equal to:
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Solution
for no solution $\Delta = 0$
$\begin{vmatrix} 3 & 1 & 4 \ 2 & \alpha & -1 \ 1 & 2 & 1 \end{vmatrix} = 0$
$\Rightarrow 3(\alpha + 2) + 1(-1 - 2) + 4(4 - \alpha) = 0$
$\Rightarrow 3\alpha + 6 - 3 + 16 - 4\alpha = 0$
$\Rightarrow 19 - \alpha = 0 \Rightarrow \alpha = 19$
$\Delta_x = \begin{vmatrix} 3 & 1 & 4 \ -3 & \alpha & -1 \ 4 & 2 & 1 \end{vmatrix}$
$\neq 0$
$\therefore$ no solution for $\alpha = 19$
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
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