Given system of equations
\(x + y + az = 2 \)…(i)
\(3x + y + z = 4\) …(ii)
\(x + 2z = 1\) …(iii)
Solving (i), (ii) and (iii), we get
\(x = 1,\) \(y = 1\) , \(z = 0\) (and for unique solution \(a ≠–3\))
Now, \((α, 1), (1, α)\) and \((1, –1)\) are collinear
\(∴\) \(\begin{vmatrix} \alpha&1 & 1\\1&\alpha&1\\1&-1 &1 \end{vmatrix}=0\)
⇒ \(α(α + 1) – 1(0) + 1(–1 – α) = 0\)
\(⇒ α^2 – 1 = 0\)
\(∴ α = ±1\)
\(∴\) Sum of absolute values of \(α = 1 + 1 = 2\)
Hence, the correct option is (C): \(2\)
Electrolysis of 600 mL aqueous solution of NaCl for 5 min changes the pH of the solution to 12. The current in Amperes used for the given electrolysis is ….. (Nearest integer).
If the system of equations \[ x + 2y - 3z = 2, \quad 2x + \lambda y + 5z = 5, \quad 14x + 3y + \mu z = 33 \] has infinitely many solutions, then \( \lambda + \mu \) is equal to:}