Let $P = left[ {begin{array}{*{20}{c}} 1&0&0 3&1&0

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3 and 4 .Determinants and Matrices

hard

Let $P = \left[ {\begin{array}{*{20}{c}}
1&0&0 \\
3&1&0 \\
9&3&1
\end{array}} \right]$ and $Q = [q_{ij}]$ be two $3\times3$ matrices such that $Q -P^5 = I_3$. Then $\frac{{{q_{21}} + {q_{31}}}}{{{q_{32}}}}$ is equal to

$10$

$135$

$15$

$9$

(JEE MAIN-2019)

Solution

${P^2} = \left[ {\begin{array}{*{20}{c}}
1&0&0\\
6&1&0\\
{24}&6&1
\end{array}} \right]{P^3} = \left[ {\begin{array}{*{20}{c}}
1&0&0\\
9&1&0\\
{54}&9&1
\end{array}} \right].\therefore {P^5} = \left[ {\begin{array}{*{20}{c}}
1&0&0\\
{15}&1&0\\
{135}&{15}&1
\end{array}} \right]$

$Q = \left[ {\begin{array}{*{20}{c}}
1&0&0\\
{15}&1&0\\
{135}&{15}&1
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
1&0&0\\
0&1&0\\
0&0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
2&0&0\\
{15}&2&0\\
{135}&{15}&2
\end{array}} \right]$

So $\frac{{{q_{21}} + {q_{31}}}}{{{q_{32}}}} = \frac{{15 + 135}}{{15}} = 10$

Standard 12

Mathematics

Let $A=\left[\begin{array}{rr}2 & -1 \\ 3 & 4\end{array}\right], B=\left[\begin{array}{ll}5 & 2 \\ 7 & 4\end{array}\right], C=\left[\begin{array}{ll}2 & 5 \\ 3 & 8\end{array}\right] .$ Find a matrix $D$ such that $CD-A B=O$.

medium

If $A=\left[\begin{array}{lll}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right]$ and $M=A+A^{2}+A^{3}+\ldots .+A^{20}$, then the sum of all the elements of the matrix $\mathrm{M}$ is equal to $…..$

hard

(JEE MAIN-2021)

If $A =$ $\left( {\begin{array}{*{20}{c}}1&a\\0&1\end{array}} \right)$ , then $A^n$ (where $n \in N$) equals

normal

If $A = \left[ {\begin{array}{{20}{c}}\alpha &0\\1&1\end{array}} \right]$ and $B = \left[ {\begin{array}{{20}{c}}1&0\\5&1\end{array}} \right]$, then value of $\alpha $for which ${A^2} = B$, is

easy

(IIT-2003)

Let $A = \left[ {\begin{array}{*{20}{c}}
1&2&3\\
2&2&{ – 1}\\
3&0&k
\end{array}} \right]$ and $f(x) = {x^3} – 2{x^2} – \alpha x + \beta = 0$ . If $A$ satisfies $f(x)=0$ ,then

normal

Let $P = \left[ {\begin{array}{*{20}{c}} 1&0&0 \\ 3&1&0 \\ 9&3&1 \end{array}} \right]$ and $Q = [q_{ij}]$ be two $3\times3$ matrices such that $Q -P^5 = I_3$. Then $\frac{{{q_{21}} + {q_{31}}}}{{{q_{32}}}}$ is equal to

Solution

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