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Maths-

General

Easy

Question

The element in the first row and third column of the inverse of the matrix $open square brackets table row 1 2 cell negative 3 end cell row 0 1 2 row 0 0 1 end table close square brackets$ is

– 2
0
1
7

The correct answer is: 7

Let $A equals open square brackets table row 1 2 cell negative 3 end cell row 0 1 2 row 0 0 1 end table close square brackets$ Þ |A| = 1
$a d j left parenthesis A right parenthesis equals open square brackets table row 1 2 cell negative 1 end cell row cell negative 2 end cell 1 1 row 7 cell negative 2 end cell 1 end table close square brackets to the power of T end exponent$ .
Hence, $A to the power of negative 1 end exponent equals fraction numerator a d j left parenthesis A right parenthesis over denominator vertical line A vertical line end fraction$
Þ $A to the power of negative 1 end exponent equals open square brackets table row 1 cell negative 2 end cell 7 row 2 1 cell negative 2 end cell row cell negative 1 end cell 1 1 end table close square brackets$ . Hence, element $z equals 3$ .

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