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 R_{2 }→ R_{2 }– R_{1}, R_{3} → R_{3} - aR_{1})

Applying R_{3} → R_{3} - \(\frac{a(a-1)}{b-1}\)R_{2}

\(=\begin{bmatrix}1&1&1&3\\0&b-1&b^2-1&4\\0&0&a(a-1)(a-b)&\frac{-4a^2-3ab+7a+b-1}{b-1}\end{bmatrix}\)

For infinitely many solution, we have a_{33} = 0 & a_{34} = 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

⇒ -4a^{2} - 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 a_{22} = 0 & a_{23} = 0 but a_{24} = 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

-4a^{2} - 3a^{2} + 7a + a - 1 = 0

⇒ -7a^{2} + 8a - 1 = 0

⇒ -7a^{2} + 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) = (\(\frac17,\frac17\))