The matrix form of given linear equations is Ax = b
where A = \(\begin{bmatrix}1&1&1\\1&b&b^2\\a&a^2&a^3\end{bmatrix}\), x = \(\begin{bmatrix}x\\y\\z\end{bmatrix}\), b = \(\begin{bmatrix}3\\7\\1\end{bmatrix}\)
Its argumented matrix is [A : B]
\(=\begin{bmatrix}1&1&1&3\\1&b&b^2&7\\a&a^2&a^3&1\end{bmatrix}\)
\(=\begin{bmatrix}1&1&1&3\\0&b-1&b^2-1&4\\0&a^2-a&a^3-a&1-3a\end{bmatrix}\)(By applying R2 → R2 – R1, R3 → R3 - aR1)
Applying R3 → R3 - \(\frac{a(a-1)}{b-1}\)R2
\(=\begin{bmatrix}1&1&1&3\\0&b-1&b^2-1&4\\0&0&a(a-1)(a-b)&\frac{-4a^2-3ab+7a+b-1}{b-1}\end{bmatrix}\)
For infinitely many solution, we have a33 = 0 & a34 = 0
for argumented matrix.
\(\therefore\) a(a - 1)(a - b) = 0
⇒ a = 0 or a = 1 or a = b----(1)
& \(\frac{-4a^2-3ab+7a+b-1}{b-1}\) = 0
⇒ -4a2 - 3ab + 7a + b - 1 = 0---(2)
Case I: If a \(\neq\) 0
Then from (2), we obtain
b - 1 = 0 ⇒ b = 1
But if b = 1 then a22 = 0 & a23 = 0 but a24 = 0
Then given system has no solution.
\(\therefore\) This case is not required.
Case II: If a = 1, then from (2) we obtain
-4 - 3b + 7 + b - 1 = 0
⇒ -2b + 2 = 0
⇒ b = 2/2 = 1
Then system has no solution.
\(\therefore\) This case is not required.
Case III: If a = b then from (2), we obtain
-4a2 - 3a2 + 7a + a - 1 = 0
⇒ -7a2 + 8a - 1 = 0
⇒ -7a2 + 7a + a - 1 = 0
⇒ -7a(a - 1) + 1(a - 1) = 0
⇒ (-7a + 1)(a - 1) = 0
⇒ -7a + 1 = 0 or a - 1 = 0
⇒ a = 1/7 or a = 1
⇒ b = a = 1/7 or b = a = 1
b = 1 is not possible for infinite many solution.
\(\therefore\) b = a = 1/7
Hence, ordered pair (a, b) = (\(\frac17,\frac17\))