Compute A –1 for the following matrix Hence, solve the system of equations: x + 2y + 5z = 10 x – y – z = – 2 2x + 3y – 2z = – 1 - Wired Faculty

Question

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Compute A^–1 for the following matrix
$straight A space equals space open square brackets table row 1 cell space space space space space 2 end cell cell space space space 5 end cell row 1 cell space minus 1 end cell cell space minus 1 end cell row 2 cell space space space 3 end cell cell space minus 2 end cell end table close square brackets.$
Hence, solve the system of equations:
x + 2y + 5z = 10
x – y – z = – 2
2x + 3y – 2z = – 1

Solution

straight A space equals space open square brackets table row 1 cell space space space space space space 2 end cell cell space space space space 5 end cell row 1 cell space space minus 1 end cell cell space minus 1 end cell row 2 cell space space space space 3 end cell cell space minus 2 end cell end table close square brackets therefore space space space open vertical bar straight A close vertical bar space equals space open vertical bar table row 1 cell space space space space space 2 end cell cell space space space space space space 5 end cell row 1 cell space space minus 1 end cell cell space minus 1 end cell row 2 cell space space space space space 3 end cell cell space space minus 2 end cell end table close vertical bar space equals space 1 space open vertical bar table row cell negative 1 end cell cell space space space minus 1 end cell row 3 cell space space space minus 2 end cell end table close vertical bar minus 2 open vertical bar table row 1 cell space space minus 1 end cell row 2 cell space space minus 2 end cell end table close vertical bar plus 5 space open vertical bar table row 1 cell space space minus 1 end cell row 2 cell space space space space 3 end cell end table close vertical bar space space space space space space space space space space space space space space equals 1 left parenthesis 2 plus 3 right parenthesis minus 2 left parenthesis negative 2 plus 2 right parenthesis space plus space 5 left parenthesis 3 plus 2 right parenthesis space equals space 5 minus 0 plus 25 space equals space 30 space not equal to space 0 therefore space space space space space space straight A to the power of negative 1 end exponent space exists.

Co-factors of the elements of first row of | A | are

open vertical bar table row cell negative 1 end cell cell space space space minus 1 end cell row cell space 3 end cell cell space space minus 2 end cell end table close vertical bar comma space space space minus open vertical bar table row 1 cell space space space minus 1 end cell row 2 cell space space space minus 2 end cell end table close vertical bar comma space space space open vertical bar table row 1 cell space space minus 1 end cell row 2 cell space space space space 3 end cell end table close vertical bar

i.e., 5, 0, 5 respectively
Co-factors of the elements of second row of | A | are

negative open vertical bar table row 2 cell space space space space space space 5 end cell row 3 cell space space minus 2 end cell end table close vertical bar comma space space space open vertical bar table row 1 cell space space space space space 5 end cell row 2 cell space space minus 2 end cell end table close vertical bar comma space space minus open vertical bar table row 1 cell space space space space space 2 end cell row 2 cell space space space space 3 end cell end table close vertical bar

i.e., 19, – 12, 1 respectively
Co-factors of the elements of third row of | A | are
$open vertical bar table row cell space space 2 end cell cell space space space space space 5 end cell row cell negative 1 end cell cell space minus 1 end cell end table close vertical bar comma space space space space space minus open vertical bar table row 1 cell space space space space space 5 end cell row 1 cell space minus 1 end cell end table close vertical bar comma space space space open vertical bar table row 1 cell space space space space space 2 end cell row 1 cell space space minus 1 end cell end table close vertical bar$
i.e., 3, 6, – 3 respectively
$therefore space space space adj space straight A space equals space open square brackets table row 5 cell space space space space 0 end cell cell space space space space 5 end cell row 19 cell space minus 12 end cell cell space space space 1 end cell row 3 cell space space space 6 end cell cell space minus 3 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row 5 cell space space space space space 9 end cell cell space space space 3 end cell row 0 cell space space minus 12 end cell cell space space space 6 end cell row 5 cell space space space 1 end cell cell negative 3 end cell end table close square brackets space space space space space space space space straight A to the power of negative 1 end exponent space equals space fraction numerator adj. space straight A over denominator open vertical bar straight A close vertical bar end fraction space equals space 1 over 30 open square brackets table row 5 cell space space space space space 9 end cell cell space space space 3 end cell row 0 cell space space minus 12 end cell cell space space space 6 end cell row 5 cell space space space 1 end cell cell negative 3 end cell end table close square brackets$

The given equations are
x + 2y + 5z = 10
x – y – z = – 2
2x + 3y – 2z = – 1
These equations can be written as
$open square brackets table row 1 cell space space space space space space 2 end cell cell space space space space space 5 end cell row 1 cell space space minus 1 end cell cell space space minus 1 end cell row 2 cell space space space space space space 3 end cell cell space space minus 2 end cell end table close square brackets space space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space open square brackets table row cell space space 10 end cell row cell negative 2 end cell row cell negative 1 end cell end table close square brackets$
or $AX space equals space straight B space space where space straight A space equals space open square brackets table row 1 cell space space space space space 2 end cell cell space space space space space space 5 end cell row 1 cell space space minus 1 end cell cell space space space minus 1 end cell row 2 cell space space space space 3 end cell cell space space space minus 2 end cell end table close square brackets comma space space space straight X space equals space open square brackets table row straight x row straight y row straight z end table close square brackets comma space space space space space straight B space equals open square brackets table row cell space space 10 end cell row cell negative 2 end cell row cell negative 1 end cell end table close square brackets$
$Now comma space AX space equals space straight B space space space space rightwards double arrow space space space space straight X space equals space straight A to the power of negative 1 end exponent straight B space space space rightwards double arrow space space space space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space 1 over 30 open square brackets table row 5 cell space space space 19 end cell cell space space space space space 3 end cell row 0 cell space minus 12 end cell cell space space space space space 6 end cell row 5 cell space space space 1 end cell cell space space minus 3 end cell end table close square brackets space open square brackets table row 10 row cell negative 2 end cell row cell negative 1 end cell end table close square brackets to the power of apostrophe rightwards double arrow space space space space space space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space 1 over 30 open square brackets table row cell 50 minus 38 minus 3 end cell row cell 0 plus 24 minus 6 end cell row cell 50 minus 2 plus 3 end cell end table close square brackets space space space space rightwards double arrow space space space space space space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space 1 over 30 open square brackets table row 9 row 18 row 51 end table close square brackets rightwards double arrow space space space space space space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space open square brackets table row cell 3 over 10 end cell row cell 3 over 5 end cell row cell 17 over 10 end cell end table close square brackets space space space rightwards double arrow space space space space straight x space equals space 3 over 10 comma space space straight y space equals space 3 over 5 comma space space straight z space equals 17 over 10$

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