Question
Let $A=\left[\begin{array}{cc}2 & -1 \\ 3 & 4\end{array}\right], B=\left[\begin{array}{ll}5 & 2 \\ 7 & 4\end{array}\right], C=\left[\begin{array}{ll}2 & 5 \\ 3 & 8\end{array}\right]$. Find a matrix D such that $\mathrm{CD}-\mathrm{AB}=0$
✓
Answer
Let $D = \left[ {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right]$
Given that CD - AB = O
$\left[ {\begin{array}{*{20}{c}} 2&5 \\ 3&8 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 2&{ - 1} \\ 3&4 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 5&2 \\ 7&4 \end{array}} \right] = 0$
$\left[ {\begin{array}{*{20}{c}} {2a + 5c}&{2b + 5d} \\ {3a + 8c}&{3b + 8d} \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 3&0 \\ {43}&{22} \end{array}} \right] = 0$
$\left[ {\begin{array}{*{20}{c}} {2a + 5c - 3}&{2b + 5d} \\ {3a + 8c - 43}&{3b + 8d - 22} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&0 \\ 0&0 \end{array}} \right]$
2a + 5c – 3 = 0........(1)
2b + 5d = 0.........(2)
3a + 8c – 43 = 0..........(3)
3b + 8d – 22 = 0..........(4)
Solving (1) and (3), we get, a = -191, c = 77.
Solving (2) and (4), we get, b = -110, d = 44
$D = \left[ {\begin{array}{*{20}{c}} { - 191}&{ - 110} \\ {77}&{44} \end{array}} \right]$
Given that CD - AB = O
$\left[ {\begin{array}{*{20}{c}} 2&5 \\ 3&8 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 2&{ - 1} \\ 3&4 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 5&2 \\ 7&4 \end{array}} \right] = 0$
$\left[ {\begin{array}{*{20}{c}} {2a + 5c}&{2b + 5d} \\ {3a + 8c}&{3b + 8d} \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 3&0 \\ {43}&{22} \end{array}} \right] = 0$
$\left[ {\begin{array}{*{20}{c}} {2a + 5c - 3}&{2b + 5d} \\ {3a + 8c - 43}&{3b + 8d - 22} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&0 \\ 0&0 \end{array}} \right]$
2a + 5c – 3 = 0........(1)
2b + 5d = 0.........(2)
3a + 8c – 43 = 0..........(3)
3b + 8d – 22 = 0..........(4)
Solving (1) and (3), we get, a = -191, c = 77.
Solving (2) and (4), we get, b = -110, d = 44
$D = \left[ {\begin{array}{*{20}{c}} { - 191}&{ - 110} \\ {77}&{44} \end{array}} \right]$
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