Find X and Y , if X+Y=left[begin{array}{ll}7 & 0 2 & 5end{array}right] and X-Y=left[begin{ar

Hindi

3 and 4 .Determinants and Matrices

medium

Find $X$ and $Y$, if

$X+Y=\left[\begin{array}{ll}7 & 0 \\ 2 & 5\end{array}\right]$ and $X-Y=\left[\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right]$

$X = \left[ {\begin{array}{*{20}{l}}
5&0 \\
1&4
\end{array}} \right]$, $Y=\left[\begin{array}{ll}2 & 0 \\ 1 & 1\end{array}\right]$

Solution

$X+Y=\left[\begin{array}{ll}7 & 0 \\ 2 & 5\end{array}\right]$ ………… $(1)$

$X-Y=\left[\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right]$ ………… $(2)$

Adding equations $(1)$ and $(2)$, we get:

$2 X=$ $\left[ {\begin{array}{*{20}{l}}
7&0 \\
2&5
\end{array}} \right] + \left[ {\begin{array}{*{20}{l}}
3&0 \\
0&3
\end{array}} \right]$ $ = \left[ {\begin{array}{*{20}{l}}
{7 + 3}&{0 + 0} \\
{2 + 0}&{5 + 3}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{10}&0 \\
2&8
\end{array}} \right]$

$\therefore \,\,X = \frac{1}{2}\left[ {\begin{array}{*{20}{l}}
{10}&0 \\
2&8
\end{array}} \right] = \left[ {\begin{array}{*{20}{l}}
5&0 \\
1&4
\end{array}} \right]$

Now, $X+Y=\left[\begin{array}{ll}7 & 0 \\ 2 & 5\end{array}\right]$

$\Rightarrow\left[\begin{array}{ll}5 & 0 \\ 1 & 4\end{array}\right]+Y=\left[\begin{array}{ll}7 & 0 \\ 2 & 5\end{array}\right]$

$\Rightarrow Y=\left[\begin{array}{ll}7 & 0 \\ 2 & 5\end{array}\right]-\left[\begin{array}{ll}5 & 0 \\ 1 & 4\end{array}\right]$

$\Rightarrow Y=\left[\begin{array}{ll}7-5 & 0-0 \\ 2-1 & 5-4\end{array}\right]$

$\therefore Y=\left[\begin{array}{ll}2 & 0 \\ 1 & 1\end{array}\right]$

Standard 12

Mathematics

If $A = \left[ {\begin{array}{*{20}{c}}4&2\\{ – 1}&1\end{array}} \right]$and $I$ is the identity matrix of order $2$, then $(A – 2I)(A – 3I) = $

easy

Let $A$ be a square matrix such that ${a_{ij}} \in \left\{ { – 1,0,1} \right\}\forall\,\, i,j$ and it has only one non-zero entry in each row as well as in each column, then

hard

If the matrix $\left[ {\begin{array}{*{20}{c}}1&3&{\lambda + 2}\\2&4&8\\3&5&{10}\end{array}} \right]$ is singular, then $\lambda = $

easy

Let the determinant of a $3 \times 3$ matrix A be 6 then $B$ is a matrix defined by $B = 5{A^2}$. Then det of $B$ is

hard

$A = \left[ {\begin{array}{{20}{c}}4&6&{ – 1}\\3&0&2\\1&{ – 2}&5\end{array}} \right]$,$B = \left[ {\begin{array}{{20}{c}}2&4\\0&1\\{ – 1}&2\end{array}} \right],\,\,C = \left[ {\begin{array}{*{20}{c}}3\\1\\2\end{array}} \right]$, then the expression which is not defined is

easy