GO Classes CS/DA 2025 | Weekly Quiz 4 | Linear Algebra | Question: 5

Answer 1

The given system of equations are:

$x + 3y = 2$

$3x – hy = k$

Here $h$ and $k$ are real numbers..

Let’s write this in $Augmented$ matrix form:

$\left(\begin{array}{cc|c} 1 & 3 & 2 \\ 3 & -h & k \end{array}\right)$

Let’s now perform elementary row operations on it:

$R_{2} → R_{2} – 3R_{1}$ then we get –

$\left(\begin{array}{cc|c} 1 & 3 & 2 \\ 0 & -h-9 & k – 6 \end{array}\right)$

Now, if the all entries of last row i.e. $R_{2}$  are $zero$ then the system have infinitely many solutions.

$\therefore $     $–  h  –  9 = 0 => h = – 9$  and  $k – 6 = 0 => k = 6$

Hence, the values of $h$ & $k$ are $-9$ and $6$ respectively.

${\color{Green} Ans: C. h = – 9}$ ${\color{Green}and}$ ${\color{Green} k = 6}$ 

Answer 2

here  Even without Using Augmented Matrix, you can still easily solve this..

If one of the Equation is a Multiple of the Other, then it confirms infinite solution. 

Here if we multiply 3 to the 1st equation x + 3y = 2, then we will get this :- 

3x + 9y = 6 . By comparing this equation with the 2nd equation,  we can easily conclude that h = -9 and k=6.

Answer 3

$$
\begin{gathered}
x+3 y=2 \\
3 x-h y=k \\
{[A \mid b]=\left[\begin{array}{ll|l}
1 & 3 & 2 \\
3 & -h & k
\end{array}\right]} \\
{\left[\begin{array}{cc|c}
1 & 3 & 2 \\
0 & -h-9 & k-6
\end{array}\right] .}
\end{gathered}
$$
system has infinite solutions if we have a zero row.
$$
\begin{aligned}
\therefore \quad-h-9=0, & k-6=0 \\
\Rightarrow h=-9, & k=6
\end{aligned}
$$