Let P=left[begin{array}{ccc}1 & 0 & 0 4 & 1 & 0 16 & 4 & 1end{array}right] a

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3 and 4 .Determinants and Matrices

hard

Let $P=\left[\begin{array}{ccc}1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1\end{array}\right]$ and $I$ be the identity matrix of order $3$ . If $\left.Q=q_{i j}\right]$ is a matrix such that $P^{50}-Q=I$, then $\frac{q_{31}+q_{32}}{q_{21}}$ equals

$52$

$103$

$201$

$205$

(IIT-2016)

Solution

$P=\left[\begin{array}{ccc}1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1\end{array}\right] $

$ P^2=\left[\begin{array}{ccc}1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1\end{array}\right]\left[\begin{array}{ccc}1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0 \\ 8 & 1 & 0 \\ 48 & 8 & 1\end{array}\right] $

$ P ^3=\left[\begin{array}{ccc}1 & 0 & 0 \\ 8 & 1 & 0 \\ 48 & 8 & 1\end{array}\right]\left[\begin{array}{ccc}1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0 \\ 12 & 1 & 0 \\ 56 & 12 & 1\end{array}\right]$

$P^4=\left[\begin{array}{ccc}1 & 0 & 0 \\ 8 & 1 & 0 \\ 48 & 8 & 1\end{array}\right]\left[\begin{array}{ccc}1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0 \\ 16 & 1 & 0 \\ 160 & 16 & 1\end{array}\right] $

$\therefore P ^{50}=\left[\begin{array}{ccc}1 & 0 & 0 \\ 4 \times 50 & 1 & 0 \\ p _{31} & 4 \times 50 & 1\end{array}\right] Q =\left[\begin{array}{lll} q _{11} & q _{12} & q _{13} \\ q _{21} & q _{22} & q _{23} \\ q _{31} & q _{32} & q _{33}\end{array}\right]$

$ P^{50}-Q=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$

$\Rightarrow 4 \times 50- q _{21}= o$

$p _{31}- q _{31}= o$

$4 \times 50- q _{32}= q$

On observing,

$p _{31^1}=16 \times 1$

$p _{31^2}=16 \times(1+2)$

$p _{31^3}=16 \times(1+2+3)$

$p _{3150}=16 \times(1+2+3+\ldots .50)$

$=16 \times \frac{50 \times 51}{2}$

$\therefore \frac{ q _{31}+ q _{32}}{ q _{21}}=\frac{ p _{31}+ p _{32}}{ p _{21}}$

$=\frac{16 \times 25 \times 51+4 \times 50}{4 \times 50}$

$=103$

Standard 12

Mathematics

The parameter on which the value of the determinant $\left| {\,\begin{array}{*{20}{c}}1&a&{{a^2}}\\{\cos (p – d)x}&{\cos px}&{\cos (p + d)x}\\{\sin (p – d)x}&{\sin px}&{\sin (p + d)x}\end{array}\,} \right|$ does not depend upon

medium

(IIT-1997)

If $\omega $ is the cube root of unity, then $\left| {\begin{array}{*{20}{c}}1&\omega &{{\omega ^2}}\\\omega &{{\omega ^2}}&1\\{{\omega ^2}}&1&\omega \end{array}} \right|$=

easy

Let the numbers $2, b, c$ be in an $A.P$ and $A = \left[ {\begin{array}{*{20}{c}}
1&1&1 \\
2&b&c \\
4&{{b^2}}&{{c^2}}
\end{array}} \right]$. If $det(A) \in [2,16]$ then $c$ lies in the interval

hard

(JEE MAIN-2019)

Let $P=\left[\begin{array}{ccc}3 & -1 & -2 \\ 2 & 0 & \alpha \\ 3 & -5 & 0\end{array}\right]$, where $\alpha \in \mathbb{R}$. Suppose $Q=\left[q_{i j}\right]$ is a matrix such that $P Q=k I$, where $k \in \mathbb{R}, k \neq 0$ and $I$ is the identity matrix of order $3$ . If $q_{23}=-\frac{k}{8}$ and $\operatorname{det}(Q)=\frac{k^2}{2}$, then

($A$) $\quad \alpha=0, k=8$

($B$) $4 \alpha-k+8=0$

($C$) $\operatorname{det}(P \operatorname{adj}(Q))=2^9$

($D$) $\operatorname{det}(Q \operatorname{adj}(P))=2^{13}$

normal

(IIT-2016)

If ${a^{ – 1}} + {b^{ – 1}} + {c^{ – 1}} = 0$ such that $\left| {\,\begin{array}{*{20}{c}}{1 + a}&1&1\\1&{1 + b}&1\\1&1&{1 + c}\end{array}\,} \right| = \lambda $, then the value of $\lambda $is

hard

Let $P=\left[\begin{array}{ccc}1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1\end{array}\right]$ and $I$ be the identity matrix of order $3$ . If $\left.Q=q_{i j}\right]$ is a matrix such that $P^{50}-Q=I$, then $\frac{q_{31}+q_{32}}{q_{21}}$ equals

Solution

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Let the numbers $2, b, c$ be in an $A.P$ and $A = \left[ {\begin{array}{*{20}{c}}
1&1&1 \\
2&b&c \\
4&{{b^2}}&{{c^2}}
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