Question

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.

Answer 1

Hint: 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.
 \[\therefore 2B = 2\left[ {\begin{array}{*{20}{c}}
  { - 3}&2 \\
  4&0
\end{array}} \right]\]
    \[ = \left[ {\begin{array}{*{20}{c}}
  {2 \times - 3}&{2 \times 2} \\
  {2 \times 4}&{2 \times 0}
\end{array}} \right]\]
    \[ = \left[ {\begin{array}{*{20}{c}}
  { - 6}&4 \\
  8&0
\end{array}} \right]\]
Now, add the matrices 2B and C.
 \[\therefore 2B + C = \left[ {\begin{array}{*{20}{c}}
  { - 6}&4 \\
  8&0
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&2
\end{array}} \right]\]
      \[
   = \left[ {\begin{array}{*{20}{c}}
  { - 6 + 1}&{4 + 0} \\
  {8 + 0}&{0 + 2}
\end{array}} \right] \\
   = \left[ {\begin{array}{*{20}{c}}
  { - 5}&4 \\
  8&2
\end{array}} \right] \\
 \]
Now,
 \[A + X = 2B + C\]
 \[
  \therefore \left[ {\begin{array}{*{20}{c}}
  2&{ - 1} \\
  2&0
\end{array}} \right] + X = \left[ {\begin{array}{*{20}{c}}
  { - 5}&4 \\
  8&2
\end{array}} \right] \\
  \therefore X = \left[ {\begin{array}{*{20}{c}}
  { - 5}&4 \\
  8&2
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  2&{ - 1} \\
  2&0
\end{array}} \right] \\
  \therefore X = \left[ {\begin{array}{*{20}{c}}
  { - 5 - 2}&{4 - \left( { - 1} \right)} \\
  {8 - 2}&{2 - 0}
\end{array}} \right] \\
  \therefore X = \left[ {\begin{array}{*{20}{c}}
  { - 7}&5 \\
  6&2
\end{array}} \right] \\
 \]
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.
 \[\therefore 3\left[ {\begin{array}{*{20}{c}}
  3&2 \\
  0&{ - 3}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {3 \times 3}&{3 \times 2} \\
  {3 \times 0}&{3 \times - 3}
\end{array}} \right]\]
   \[ = \left[ {\begin{array}{*{20}{c}}
  9&6 \\
  0&{ - 9}
\end{array}} \right]\]
Now, we have
       \[\left[ {\begin{array}{*{20}{c}}
  1&4 \\
  { - 2}&3
\end{array}} \right] + 2M = \left[ {\begin{array}{*{20}{c}}
  9&6 \\
  0&{ - 9}
\end{array}} \right]\]
 \[\therefore 2M = \left[ {\begin{array}{*{20}{c}}
  9&6 \\
  0&{ - 9}
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  1&4 \\
  { - 2}&3
\end{array}} \right]\]
 \[\therefore 2M = \left[ {\begin{array}{*{20}{c}}
  {9 - 1}&{6 - 4} \\
  {0 - \left( { - 2} \right)}&{ - 9 - 3}
\end{array}} \right]\]
 \[
  \therefore 2M = \left[ {\begin{array}{*{20}{c}}
  8&2 \\
  2&{ - 12}
\end{array}} \right] \\
  \therefore M = \dfrac{1}{2}\left[ {\begin{array}{*{20}{c}}
  8&2 \\
  2&{ - 12}
\end{array}} \right] \\
  \therefore M = \left[ {\begin{array}{*{20}{c}}
  {\dfrac{8}{2}}&{\dfrac{2}{2}} \\
  {\dfrac{2}{2}}&{\dfrac{{ - 12}}{2}}
\end{array}} \right] \\
  \therefore M = \left[ {\begin{array}{*{20}{c}}
  4&1 \\
  1&{ - 6}
\end{array}} \right] \\
 \]
Thus, \[M = \left[ {\begin{array}{*{20}{c}}
  4&1 \\
  1&{ - 6}
\end{array}} \right]\] .

Note: Whenever it is said to multiply any matrix by any number, we have to multiply each and every element of the matrix by that particular number. While, in Determinant if it is said to multiply the determinant by any number, we have to multiply any single row or any single column.