Use product [ ccc 1 & 1 & 2 0 & 2 & 3 3 & 2 & 4 ] [ ccc 2 & 0 & 1 9 & 2 & 3 6 & 1 & 2 ]to solve the system of equations x – y + 2z = 1, 2y – 3z = 1, 3x – 2y + 4z = 2

Consider the product [ * 20 c 1& - 1 &2 0&2& - 3 3& - 2 &4 ] [ * 20 c - 2 &0&1 9&2& - 3 6&1& - 2 ] = [ 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 ] = [ * 20 c 1&0&0 0&1&0 0&0&1 ] Hence [ * 20 c 1& - 1 &2 0&2& - 3 3& - 2 &4 ]^ - 1 = [ * 20 c - 2 &0&1 9&2& - 3 6&1& - 2 ] Now, given system of equations can be written, in matrix form, as follows [ * 20 c 1& - 1 &2 0&2& - 3 3& - 2 &4 ] [ * 20 c x y z ] = [ * 20 c 1 1 2 ] or l x y z = [ rrr 1 & -1 & 2 0 & 2 & -3 3 & -2 & 4 ]^ -1 [ l 1 1 2 ] = [ rrr 2 & 0 & 1 9 & 2 & 3 6 & 1 & 2 ] [ l 1 1 2 ] = [ c -2+0+2 9+2-6 6+1-4 ]= [ c 0 5 3 ] Hence, x = 0, y = 5, z = 3

Use product [ ccc 1 & 1 & 2 0 & 2 & 3 3 & 2 & 4 ] [ ccc 2 & 0 & 1…

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Question

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 equations
x – y + 2z = 1, 2y – 3z = 1, 3x – 2y + 4z = 2

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Answer

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