Question 16.8.1: For all values of p, find the solutions of the system px + y......

The coefficient matrix has determinant

According to Theorem 16.8.1, the system has a unique solution if $1-p\neq0$ —that is, if $p\neq1.$ In this case, the determinants in (16.8.2) are

$D_{j}=\left | \begin{matrix} a_{11} & … & a_{1,j-1} & b_{1} & a_{1,j+1} & … & a_{1n} \\ a_{21} & … & a_{2,j-1} & b_{2} & a_{2+j+1} & … & a_{2n} \\ . & & . & . & . & & . \\ : & & : & : & : & & : \\ a_{n1}&…& a_{n, j−1}& b_{n}& a_{n, j+1} & …& a_{nn}\end{matrix} \right |$(16.8.2)

whose numerical values are $D_{1} = 2, D_{2} = 1 − 3p,$ and $D_{3} = −1 − 3p.$ Then, for $p \neq 1,$ Eq. (16.8.4) yields

$x_{1}={\frac{D_{1}}{|\mathbf{A}|}}\,,\,\,\,x_{2}={\frac{D_{2}}{|\mathbf{A}|}}\,,\cdot\cdot\cdot\,,\,\,\,x_{n}={\frac{D_{n}}{|\mathbf{A}|}}$      (16.8.4)

On the other hand, in case p = 1, the first equation becomes x + y = 1. Yet adding the last two of the original equations implies that x + y = 3. There is no solution to these two contradictory equations in case p = 1.$^{9}$

$^{9}$ It might be instructive to solve this problem by using Gaussian elimination, starting by interchanging the first two equations.