System of linear equations

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If the system of linear equations $x + y + z = a, x - y + b z = 2,$ $2 x + 3 y - z = 1$ has infinitely many solutions, then $b - 5 a =$

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Solution

Augmented matrix $= [\begin{matrix} 1 & 1 & 1 & a \\ 1 & - 1 & b & 2 \\ 2 & 3 & - 1 & 1 \end{matrix}]$
$\overset{\begin{array}{l} R_{2} \to R_{2} - R_{1} \\ R_{3} \to R_{3} - 2 R_{1} \end{array}}{~} [\begin{matrix} 1 & 1 & 1 & a \\ 0 & - 2 & b - 1 & 2 - a \\ 0 & 1 & - 3 & 1 - 2 a \end{matrix}]$
$\overset{R_{3} \to 2 R_{3} + R_{2}}{~} [\begin{matrix} 1 & 1 & 1 & a \\ 0 & - 2 & b - 1 & 2 - a \\ 0 & 1 & b - 7 & 4 - 5 a \end{matrix}]$
Given system has infinitely many solutions
$\Rightarrow b - 7 = 0, 4 - 5 a = 0$
$b = 7, a = \frac{4}{5}$
$b - 5 a = 7 - 5 \times \frac{4}{5} = 7 - 4 = 3$

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System of linear equations

Remember concepts with our Masterclasses.

If the system of linear equations x+y+z=a, x−y+bz=2,2x+3y−z=1 has infinitely many solutions, then b−5a=

Augmented matrix =111a1-1b223-11 ~R2→R2−R1R3→R3 −2R1 111a0-2b-12-a01-31-2a ~R3→2R3 +R2 111a0-2b-12-a01b-74-5aGiven system has infinitely many solutions⇒b−7=0, 4−5a=0b=7, a=45b−5a= 7−5×45 = 7−4=3

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If the system of linear equations $x + y + z = a, x - y + b z = 2,$ $2 x + 3 y - z = 1$ has infinitely many solutions, then $b - 5 a =$