Question

For what values of k, the system of linear equations
x+y+2=6
x + 2y + 3Z=10
x + 2y + kz=10
has infinitely many solutions?​

Answer 1

Answer:

K-4=0

Step-by-step explanation:

When we write the given equation in matrix form and then convert to echelon form.

So for checking that on which value it will be infinitely many solutions so makes the 3rd row is equal to zero.

Note.

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Answer 2

$\large\underline{\sf{Solution-}}$

Given equations are

• x+y+z=6

• x + 2y + 3Z=10

• x + 2y + kz=10

We use rank method to find the value of k.

Here,

$\begin{gathered}\sf A=\left[\begin{array}{ccc}1&1&1\\1&2&3\\1&2&k\end{array}\right]\end{gathered}$

$\begin{gathered}\sf B=\left[\begin{array}{c}6\\10\\10\end{array}\right]\end{gathered}$

The augmented matrix [A : B] is

$\begin{gathered}\sf =\left[\begin{array}{cccc}1&1&1&6\\1&2&3&10\\1&2&k&10\end{array}\right]\end{gathered}$

$\rm :\longmapsto\:OP \: R_2 \: \to \: R_2 - R_1$

$\begin{gathered}\sf =\left[\begin{array}{cccc}1&1&1&6\\0&1&2&4\\1&2&k&10\end{array}\right]\end{gathered}$

$\rm :\longmapsto\:OP \: R_3 \: \to \: R_3 - R_1$

$\begin{gathered}\sf =\left[\begin{array}{cccc}1&1&1&6\\0&1&2&4\\0&1&k - 1&4\end{array}\right]\end{gathered}$

$\rm :\longmapsto\:OP \: R_3 \: \to \: R_3 - R_2$

$\begin{gathered}\sf =\left[\begin{array}{cccc}1&1&1&6\\0&1&2&4\\0&0&k - 3&0\end{array}\right]\end{gathered}$

Since system of equation have Infinitely many solutions

$\rm :\implies\: \rho \: [A : B] = \rho \: (A) < n$

$\rm :\implies\:k - 3 = 0 \: so \: that \: \rho \: [A : B] = \rho \: (A) = 2$

$\bf\implies \:k = 3$