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Use product  $$\left[\begin{array}{ccc} {1} & {1} & {2} \\ {0} & {2} & {3} \\ {3} & {2} & {4} \end{array}\right]\left[\begin{array}{ccc} {2} & {0} & {1} \\ {9} & {2} & {3} \\ {6} & {1} & {2} \end{array}\right]$$to solve the system of equation
x – y + 2z = 1, 2y – 3z = 1, 3x – 2y + 4z = 2
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## QuestionMathsClass 12

Use product  $$\left[\begin{array}{ccc} {1} & {1} & {2} \\ {0} & {2} & {3} \\ {3} & {2} & {4} \end{array}\right]\left[\begin{array}{ccc} {2} & {0} & {1} \\ {9} & {2} & {3} \\ {6} & {1} & {2} \end{array}\right]$$to solve the system of equation
x – y + 2z = 1, 2y – 3z = 1, 3x – 2y + 4z = 2

Consider the product $$\left[ {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ 0&2&{ - 3} \\ 3&{ - 2}&4 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { - 2}&0&1 \\ 9&2&{ - 3} \\ 6&1&{ - 2} \end{array}} \right]$$
$$\left[\begin{array}{ccc} {-2-9+12} & {0-2+2} & {1+3-4} \\ {0+18-18} & {0+4-3} & {0-6+6} \\ {-6-18+24} & {0-4+4} & {3+6-8} \end{array}\right]$$ $$= \left[ {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right]$$
Hence $${\left[ {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ 0&2&{ - 3} \\ 3&{ - 2}&4 \end{array}} \right]^{ - 1}} = \left[ {\begin{array}{*{20}{c}} { - 2}&0&1 \\ 9&2&{ - 3} \\ 6&1&{ - 2} \end{array}} \right]$$
Now, given system of equations can be written, in matrix form, as follows
$$\left[ {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ 0&2&{ - 3} \\ 3&{ - 2}&4 \end{array}} \right] \left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1 \\ 1 \\ 2 \end{array}} \right]$$
or $$\begin{array}{l} {x} \\ {y} \\ {z} \end{array}=\left[\begin{array}{rrr} {1} & {-1} & {2} \\ {0} & {2} & {-3} \\ {3} & {-2} & {4} \end{array}\right]^{-1}\left[\begin{array}{l} {1} \\ {1} \\ {2} \end{array}\right]$$ = $$\left[\begin{array}{rrr} {2} & {0} & {1} \\ {9} & {2} & {3} \\ {6} & {1} & {2} \end{array}\right]\left[\begin{array}{l} {1} \\ {1} \\ {2} \end{array}\right]$$
$$\left[\begin{array}{c} {-2+0+2} \\ {9+2-6} \\ {6+1-4} \end{array}\right]=\left[\begin{array}{c} {0} \\ {5} \\ {3} \end{array}\right]$$
Hence, x = 0, y = 5, z = 3