Question

If $straight A space equals space open square brackets table row 2 cell space space space minus 3 end cell cell space space space space 5 end cell row 3 cell space space space space 2 end cell cell negative 4 end cell row 1 cell space space space 1 end cell cell negative 2 end cell end table close square brackets comma$ find A ^–1 .
Using A^–1, solve the following system of linear equations.
2x – 3y + 5z = 11
3x + 2y – 4z = – 5
x + y – 2z = – 3

Solution

Here, $straight A space equals space open square brackets table row 2 cell space space minus 3 end cell cell space space space 5 end cell row 3 cell space space space space 2 end cell cell space minus 4 end cell row 1 cell space space space 1 end cell cell space minus 2 end cell end table close square brackets$
$therefore space space space space space open vertical bar straight A close vertical bar space equals space open vertical bar table row 2 cell space space minus 3 end cell cell space space space 5 end cell row 3 cell space space space space 2 end cell cell negative 4 end cell row 1 cell space space space 1 end cell cell negative 2 end cell end table close vertical bar space equals space 2 open vertical bar table row 2 cell space space minus 4 end cell row 1 cell space space minus 2 end cell end table close vertical bar minus left parenthesis negative 3 right parenthesis space open vertical bar table row 3 cell space space space minus 4 end cell row 1 cell space space space minus 2 end cell end table close vertical bar plus 5 space open vertical bar table row 3 cell space space space 2 end cell row 1 cell space space 1 end cell end table close vertical bar space space space space space space space space space space space space space space space space space equals 2 left parenthesis negative 4 plus 4 right parenthesis plus 3 left parenthesis negative 6 plus 4 right parenthesis plus 5 left parenthesis 3 minus 2 right parenthesis space equals space 0 minus 6 plus 5 space equals space minus 1 not equal to 0 therefore space space space space space space space straight A to the power of negative 1 end exponent space exists.$
Co-factors of the elements of first row of | A | are
$open vertical bar table row 2 cell space space space minus 4 end cell row 1 cell space space space minus 2 end cell end table close vertical bar comma space space space space minus open vertical bar table row 3 cell space space minus 4 end cell row 1 cell space space minus 2 end cell end table close vertical bar comma space open vertical bar table row 3 cell space space space space space 2 end cell row 1 cell space space space space 1 end cell end table close vertical bar$
i.e., 0, 2, 1 respectively
Co-factors of the elements of second row of | A | are
$negative open vertical bar table row cell negative 3 end cell cell space space space 5 end cell row 1 cell negative 2 end cell end table close vertical bar comma space space minus open vertical bar table row 2 cell space space space space space 5 end cell row 1 cell space minus 2 end cell end table close vertical bar comma space space minus open vertical bar table row 2 cell space space minus 3 end cell row 1 cell space space space space space 1 end cell end table close vertical bar$
i.e., –1 –9 , –5 respectively.
Co-factors of the elements of third row of | A | are
$open vertical bar table row cell negative 3 end cell cell space space space space space space space 5 end cell row 2 cell space space space minus 4 end cell end table close vertical bar comma space space space space minus open vertical bar table row 2 cell space space space space space 5 end cell row 3 cell space space minus 4 end cell end table close vertical bar comma space space space open vertical bar table row 2 cell space space space minus 3 end cell row 3 cell space space space space space 2 end cell end table close vertical bar$
i.e., 2, 23, 13 respectively.
$therefore space space adj space straight A space equals space open square brackets table row 0 cell space space space space 2 end cell cell space space 1 end cell row cell negative 1 end cell cell negative 9 end cell cell space minus 5 end cell row 2 cell space space 23 end cell cell space space 13 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row 0 cell space space minus 1 end cell cell space space 2 end cell row 2 cell space minus 9 end cell cell space 23 end cell row 1 cell space minus 5 end cell cell space 13 end cell end table close square brackets straight A to the power of negative 1 end exponent space equals space fraction numerator adj space straight A over denominator open vertical bar straight A close vertical bar end fraction space equals space minus open square brackets table row 0 cell space space minus 1 end cell cell space space 2 end cell row 2 cell space minus 9 end cell cell space 23 end cell row 1 cell negative 5 end cell 13 end table close square brackets space equals open square brackets table row 0 cell space space 1 end cell cell space space minus 2 end cell row cell negative 2 end cell cell space space 9 end cell cell space minus 23 end cell row cell negative 1 end cell cell space space 5 end cell cell space minus 13 end cell end table close square brackets$

The given equations are
2x – 3y + 5z = 11
3x + 2y – 4z = – 5
x + y – 2z = – 3
These equations can be written as
$open square brackets table row 2 cell space space space minus 3 end cell cell space space space space 5 end cell row 3 cell space space space space 2 end cell cell negative 4 end cell row 1 cell space space 1 end cell cell negative 2 end cell end table close square brackets space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space open square brackets table row cell space 11 end cell row cell negative 5 end cell row cell negative 3 end cell end table close square brackets$
or $AX space equals space straight B space space where space straight A space equals space open square brackets table row 2 cell space space minus 3 end cell cell space space space space 5 end cell row 3 cell space space space space 2 end cell cell negative 4 end cell row 1 cell space space 1 end cell cell negative 2 end cell end table close square brackets comma space space space straight X space equals open square brackets table row straight x row straight y row straight z end table close square brackets comma space space space straight B space equals open square brackets table row cell space space 11 end cell row cell negative 5 end cell row cell negative 3 end cell end table close square brackets$
Now $AX space equals space straight B space space space space space rightwards double arrow space space space space space straight X space equals space straight A to the power of negative 1 end exponent straight B$
$therefore space space space space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space open square brackets table row 0 cell space space space 1 end cell cell space space minus 2 end cell row cell negative 2 end cell cell space space space 9 space end cell cell space minus 23 end cell row cell negative 1 end cell cell space space 5 end cell cell negative 13 end cell end table close square brackets space open square brackets table row cell space space 11 end cell row cell negative 5 end cell row cell negative 3 end cell end table close square brackets$
$rightwards double arrow space space space space space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space open square brackets table row cell 0 minus 5 plus 6 end cell row cell negative 22 minus 45 plus 69 end cell row cell negative 11 minus 25 plus 39 end cell end table close square brackets space space space space rightwards double arrow space space space space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space open square brackets table row 1 row 2 row 3 end table close square brackets therefore space space space space space space space space straight x space equals space 1 comma space space space straight y space equals 2 comma space space space space straight z space equals space 3$

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