Question
Question: i.Given \[A = \left[ {\begin{array}{*{20}{c}} 2&{ - 1} \\\ 2&0 \end{array}} \right]\] , ...
i.Given A = \left[ {\begin{array}{*{20}{c}}
2&{ - 1} \\\
2&0
\end{array}} \right] , B = \left[ {\begin{array}{*{20}{c}}
{ - 3}&2 \\\
4&0
\end{array}} \right] and C = \left[ {\begin{array}{*{20}{c}}
1&0 \\\
0&2
\end{array}} \right] . Find the matrix X such that A+X=2B+C .
ii.If \left[ {\begin{array}{*{20}{c}}
1&4 \\\
{ - 2}&3
\end{array}} \right] + 2M = 3\left[ {\begin{array}{*{20}{c}}
3&2 \\\
0&{ - 3}
\end{array}} \right] , find the matrix M.
Solution
i. Firstly, multiply each element of the matrix B by 2. After that, add the matrices 2B and C. Now, to find the matrix X, finally subtract matrix A from the matrix 2B+C.
ii. For the first step multiply each of the element of the matrix \left[ {\begin{array}{*{20}{c}}
3&2 \\\
0&{ - 3}
\end{array}} \right] by 3. Now, subtract the matrix \left[ {\begin{array}{*{20}{c}}
1&4 \\\
{ - 2}&3
\end{array}} \right] from matrix 3\left[ {\begin{array}{*{20}{c}}
3&2 \\\
0&{ - 3}
\end{array}} \right]. This will give a matrix, let’s say matrix Q. For the final answer, divide the matrix Q by 2 to get the answer of matrix M.
Complete step-by-step answer:
The given equation of matrices to find the matrix X, is A+X=2B+C .
Firstly, multiply each element of the matrix B by 2.
Now,
A+X=2B+C
Thus, X = \left[ {\begin{array}{*{20}{c}}
{ - 7}&5 \\\
6&2
\end{array}} \right] .
The given equation to find the matrix M is \left[ {\begin{array}{*{20}{c}}
1&4 \\\
{ - 2}&3
\end{array}} \right] + 2M = 3\left[ {\begin{array}{*{20}{c}}
3&2 \\\
0&{ - 3}
\end{array}} \right] .
First of all, multiply each element of the matrix \left[ {\begin{array}{*{20}{c}}
3&2 \\\
0&{ - 3}
\end{array}} \right] by 3.
\therefore 2M = \left[ {\begin{array}{{20}{c}}
8&2 \\
2&{ - 12}
\end{array}} \right] \\
\therefore M = \dfrac{1}{2}\left[ {\begin{array}{{20}{c}}
8&2 \\
2&{ - 12}
\end{array}} \right] \\
\therefore M = \left[ {\begin{array}{{20}{c}}
{\dfrac{8}{2}}&{\dfrac{2}{2}} \\
{\dfrac{2}{2}}&{\dfrac{{ - 12}}{2}}
\end{array}} \right] \\
\therefore M = \left[ {\begin{array}{{20}{c}}
4&1 \\
1&{ - 6}
\end{array}} \right] \\