# If the system of linear equations $\begin{array}{ll} & x+y+z=3 \\ & x+b y+b^{2} z=7 \\ \text { and } \quad & a x+a^{2} y+a^{3} z=1 \end{array}$

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If the system of linear equations $\begin{array}{ll} & x+y+z=3 \\ & x+b y+b^{2} z=7 \\ \text { and } \quad & a x+a^{2} y+a^{3} z=1 \end{array}$ has infinitely many solutions, then the ordered pair $(a, b)$ is

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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→ R– 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) = ($\frac17,\frac17$)