GO Classes DPP | GATE CS | Linear Algebra | System of Linear Equations

Answer 1

The given system is

$x+y+z=1$

$x+2y+4z=a$

$x+4y+10z=a^2$

Its augmented matrix is

$\left[\begin{array}{ccc|c}
1&1&1&1\\
1&2&4&a\\
1&4&10&a^2
\end{array}\right]$

Apply the row operations

$R_2 \to R_2-R_1,\qquad R_3 \to R_3-R_1$

Then we get

$\left[\begin{array}{ccc|c}
1&1&1&1\\
0&1&3&a-1\\
0&3&9&a^2-1
\end{array}\right]$

Now apply

$R_3 \to R_3-3R_2$

So,

$\left[\begin{array}{ccc|c}
1&1&1&1\\
0&1&3&a-1\\
0&0&0&a^2-1-3(a-1)
\end{array}\right]$

Simplifying the last entry,

$a^2-1-3a+3=a^2-3a+2=(a-1)(a-2)$

Hence,

$\left[\begin{array}{ccc|c}
1&1&1&1\\
0&1&3&a-1\\
0&0&0&(a-1)(a-2)
\end{array}\right]$

For the system to have a solution, the last row must not be of the form

$[0\ \ 0\ \ 0\ |\ \text{non-zero}]$

So we need

$(a-1)(a-2)=0$

Therefore,

$a=1$ or $a=2$

Hence, the correct option is $D$.

Answer 2

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Answer 3

For (a = 1), the solution is ({1, 0, 0}), which represents the column number one.

For (a = 2), the solution is ({0, 1, 0}), which represents the column number second.