Question

Consider the system of equations $2x + P^2y + 6z = 8, x + 2y + 2qz = 5$ and $x + y + 3z = 4$

• A. Given system has unique solution for $P \neq \pm \sqrt{2}$ and $q = \frac{3}{2}$
• B. Given system has no solution for $P = \pm \sqrt{2}$ and $q = \frac{3}{2}$
• C. Given system has infinite solution for $P = \pm \sqrt{2}$ and $q \in R$
• D. None of these

Answer 1

Answer: (C)

Given system has infinite solution for $P = \pm \sqrt{2}$ and $q \in \mathbb{R}$

Answer 2

For $P=\pm\sqrt{2}$, equation (1) becomes identical to equation (3), reducing the system to 2 equations in 3 unknowns which always have infinitely many solutions regardless of $q$.

Step 1: Write the coefficient matrix

$$A=\begin{pmatrix} 2 & P^2 & 6 \\ 1 & 2 & 2q \\ 1 & 1 & 3 \end{pmatrix}.$$

The determinant is

$$\det(A)=2\begin{vmatrix}2&2q\\1&3\end{vmatrix}-P^2\begin{vmatrix}1&2q\\1&3\end{vmatrix}+6\begin{vmatrix}1&2\\1&1\end{vmatrix}.$$

Calculate the 2×2 determinants:

$$\begin{aligned} \det\begin{pmatrix}2&2q\\1&3\end{pmatrix}&=2\cdot3-1\cdot2q=6-2q,\\[1mm] \det\begin{pmatrix}1&2q\\1&3\end{pmatrix}&=1\cdot3-1\cdot2q=3-2q,\\[1mm] \det\begin{pmatrix}1&2\\1&1\end{pmatrix}&=1\cdot1-1\cdot2= -1. \end{aligned}$$

Thus,

$$\det(A)=2(6-2q)-P^2(3-2q)+6(-1)=12-4q- P^2(3-2q)-6,$$ $$\det(A)=6-4q- P^2(3-2q).$$

Rewriting,

$$\det(A)=6-3P^2+(2P^2-4)q.$$

Notice that if $P^2\neq2$ (i.e. $P\neq\pm\sqrt{2}$), then

$$\det(A)=0 \quad\text{if}\quad (2P^2-4)q=3P^2-6.$$

Since

$$3P^2-6=3(P^2-2) \quad \text{and} \quad 2P^2-4=2(P^2-2),$$

we get

$$q=\frac{3(P^2-2)}{2(P^2-2)}=\frac{3}{2}.$$

Thus, for $P^2\neq2$, the determinant is zero when $q=\frac{3}{2}$ and nonzero when $q\neq\frac{3}{2}$. Hence, a unique solution exists for $P\neq \pm\sqrt{2}$ and $q\neq\frac{3}{2}$.

Step 2: Consider the case $P^2=2$ (i.e. $P=\pm\sqrt{2}$).

For $P=\pm\sqrt{2}$, equation (1) becomes

$$2x+2y+6z=8 \quad\Longrightarrow \quad x+y+3z=4,$$

which is exactly the same as equation (3). Thus, the system reduces to

$$\begin{aligned} x+y+3z&=4,\\[1mm] x+2y+2qz&=5. \end{aligned}$$

This is a system of 2 equations with 3 unknowns. They are consistent (subtracting, we get $y+(2q-3)z=1$ which always has a solution with a free parameter), so the system has infinitely many solutions for any $q$.