If $left[ {begin{array}{*{20}{c}} 2&1 1&2 end{array}} right] Alef

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3 and 4 .Determinants and Matrices

normal

If $\left[ {\begin{array}{{20}{c}}
2&1 \\
1&2
\end{array}} \right]$ $A\left[ {\begin{array}{{20}{c}}
{ - 3}&2 \\
5&{ - 3}
\end{array}} \right] = {I_2}$ then $A =$

$\left[ {\begin{array}{*{20}{c}}
1&1 \\
1&0
\end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}}
1&1 \\
0&1
\end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}}
1&0 \\
1&1
\end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}}
0&1 \\
1&1
\end{array}} \right]$

Solution

$\left[\begin{array}{ll}{2} & {1} \\ {3} & {2}\end{array}\right] {A}\left[\begin{array}{cc}{-3} & {2} \\ {5} & {-3}\end{array}\right]=\left[\begin{array}{ll}{1} & {0} \\ {0} & {1}\end{array}\right]$

$\left[\begin{array}{ll}{2} & {1} \\ {3} & {2}\end{array}\right] \mathrm{A}=\left[\begin{array}{ll}{1} & {0} \\ {0} & {1}\end{array}\right]\left[\begin{array}{ll}{-3} & {2} \\ {5} & {-3}\end{array}\right]^{-1}$

$A=\left[\begin{array}{ll}{2} & {1} \\ {3} & {2}\end{array}\right]^{-1}\left[\begin{array}{ll}{1} & {0} \\ {0} & {1}\end{array}\right]\left[\begin{array}{cc}{-3} & {2} \\ {5} & {-3}\end{array}\right]^{-1}$

$=\frac{1}{1}\left[\begin{array}{cc}{2} & {1} \\ {3} & {2}\end{array}\right] \frac{1}{-1}\left[\begin{array}{cc}{-3} & {-2} \\ {-5} & {-3}\end{array}\right]$

$=\left[\begin{array}{cc}{2} & {-1} \\ {-2} & {2}\end{array}\right]\left[\begin{array}{cc}{3} & {2} \\ {5} & {3}\end{array}\right]$

$ \Rightarrow\left[\begin{array}{cc}{6-5} & {4-3} \\ {-9+10} & {-6+6}\end{array}\right]=\left[\begin{array}{cc}{1} & {1} \\ {1} & {0}\end{array}\right]$

Standard 12

Mathematics

Find $x$ and $y,$ if $2\left[\begin{array}{ll}1 & 3 \\ 0 & x\end{array}\right]+\left[\begin{array}{ll}y & 0 \\ 1 & 2\end{array}\right]=\left[\begin{array}{ll}5 & 6 \\ 1 & 8\end{array}\right]$

medium

Let $A=\left[a_{i j}\right]$ be a square matrix of order $3$ such that $a_{i j}=2^{j-i}$, for all $i, j=1,2,3$. Then, the matrix $A ^{2}+ A ^{3}+\ldots+ A ^{10}$ is equal to

hard

(JEE MAIN-2022)

Square matrix ${[{a_{ij}}]_{n \times n}}$ will be an upper triangular matrix, if

easy

If $A = \left[ {\begin{array}{*{20}{c}}\lambda &1\\{ – 1}&{ – \lambda }\end{array}} \right]$, then for what value of $\lambda ,\,{A^2} = O$

easy

The solution $(s)$ of the equation $\left| {\begin{array}{*{20}{c}}x&a&b\\ a&x&a\\b&b&x\end{array}} \right|$ $= 0$ is/are :