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TS EAMCET 9-Sep-2020 Shift 1 Solved Paper

Section: Mathematics

Question : 6 of 160

Marks: +1, -0

The equations

x+y+z=3,

x+2 y+2 z=6

x+a y+3 z=b

have

Solution:

We have,

x+y+z=3

x+2 y+2 z=6

x+a y+3 z=b

Above equations can be written as

A X=B

Where,

A=\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & a & 3 \end{bmatrix}, X=\begin{bmatrix} x \\ y \\ z \end{bmatrix}, B=\begin{bmatrix} 3 \\ 6 \\ b \end{bmatrix}

Now,

|A|=\begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & a & 3 \end{vmatrix}

=1[6-2 a]-1[3-2]+1[a-2]

=6-2 a-1+a-2=3-a

For unique solution

|A| \neq 0 \Rightarrow 3-a \neq 0 \Rightarrow a \neq 3

\therefore

Given system have unique solution when

a \neq 3, b

can have any value.

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