If the system of equations x + 5y + 6z = 4 2x + 3y + 4z = 7 x + 6y + az = b has infinitely…

Question

If the system of equations $x + 5y + 6z = 4$ $2x + 3y + 4z = 7$ $x + 6y + az = b$ has infinitely many solutions, then the point $(a, b)$ lies on the line

Accepted Answer

The given system of equations is:

$x+5y+6z=4$

$2x+3y+4z=7$

$x+6y+az=b$

Find the determinant of the coefficient matrix ($\Delta$)

The coefficient matrix is formed by the coefficients of $x$, $y$, and $z$.

$\Delta=\begin{vmatrix} 1 \quad  5 \quad  6 \ 2 \quad  3 \quad  4 \ 1 \quad  6 \quad  a \end{vmatrix}$

Let us expand this determinant along the first row:

$\Delta=1(3\cdot a-6\cdot 4)-5(2\cdot a-1\cdot 4)+6(2\cdot 6-1\cdot 3)$

$\Delta=1(3a-24)-5(2a-4)+6(12-3)$

$\Delta=3a-24-10a+20+6(9)$

$\Delta=-7a-4+54$

$\Delta=-7a+50$

For the system to have infinitely many solutions, we equate $\Delta$ to $0$:

$-7a+50=0\, \implies \, 7a=50\, \implies \, a=\frac{50}{7}$

Find the determinant ${\Delta}_{z}$

To find ${\Delta}_{z}$, we replace the $z$-column (third column) with the constant terms from the right-hand side of the equations.

${\Delta}_{z}=\begin{vmatrix} 1 \quad  5 \quad  4 \ 2 \quad  3 \quad  7 \ 1 \quad  6 \quad  b \end{vmatrix}$

We again expand along the first row:

${\Delta}_{z}=1(3\cdot b-6\cdot 7)-5(2\cdot b-1\cdot 7)+4(2\cdot 6-1\cdot 3)$

${\Delta}_{z}=1(3b-42)-5(2b-7)+4(12-3)$

${\Delta}_{z}=3b-42-10b+35+4(9)$

${\Delta}_{z}=-7b-7+36$

${\Delta}_{z}=-7b+29$

Equating ${\Delta}_{z}$ to zero for infinitely many solutions, we get:

$-7b+29=0\, \implies \, 7b=29\, \implies \, b=\frac{29}{7}$

Check the options for the point $(a,b)$

We have obtained the point coordinates as $(a,b)=(\frac{50}{7},\frac{29}{7})$.

Let us evaluate the difference $a-b$ to find out which linear profile equation matches our outcome:

$a-b=\frac{50}{7}-\frac{29}{7}$

$a-b=\frac{50-29}{7}$

$a-b=\frac{21}{7}=3$

Writing this relation in terms of $x$ and $y$ (where $x=a$ and $y=b$), it perfectly satisfies the line equation:

$x-y=3$

Thus, the point $(a,b)$ lies on the line given in Option B.

Solutions · 1