Question

The system of equations $2x-3y+6z-5t=3$, $y-4z+t=1$, $4x-5y+8z-9t=k$ has

• A. no solution if $k\ne 7$
• B. no solution if $k=0$
• C. infinite solutions if $k\ne 7$
• D. infinite solutions if $k=7$

Answer 1

Answer: (A), (D)

The correct options are
A no solution if $k\ne 7$
D infinite solutions if $k=7$

The augmented matrix $C=\left[AB\right]$

$=\left[\begin{array}{ccccc}2& -3& 6& -5& 3\\ 0& 1& -4& 1& 1\\ 4& -5& 8& -9& k\end{array}\right]$

Applying $R_3\to R_3-2R_1$

$=\left[\begin{array}{ccccc}2& -3& 6& -5& 3\\ 0& 1& -4& 1& 1\\ 0& 1& -4& 1& k-6\end{array}\right]$

Applying $R_3\to R_3-R_2$

$=\left[\begin{array}{ccccc}2& -3& 6& -5& 3\\ 0& 1& -4& 1& 1\\ 0& 0& 0& 0& k-7\end{array}\right]$

(i) There is no solution if $\rho \left(A\right)\ne \rho \left(C\right)$

Here, $\rho \left(A\right)=2.\rho \left(C\right)=3$ if $k-7\ne 0\Rightarrow k\ne 7$.

(ii) There are infinite solutions if $\rho \left(A\right)=\rho \left(C\right)<4$.

Already, $\rho \left(A\right)=2.\rho \left(C\right)=2$ if $k-7=0\Rightarrow k=7$