Animated Solution for Mathematics - Matrices and Determinants: If the system of linear equations: $x+y+2z=6$ $2x+3y+az=a+1$ $-x-3y+bz=2b$ where $a,b\in R,$ has infinitely many solutions, then $7a+3b$ is equal to:

We are presented with a system of three linear equations: $x+y+2z=6$ $2x+3y+az=a+1$ $-x-3y+bz=2b$

The problem states that this system has infinitely many solutions. Geometrically, this implies that the three planes intersect along a common line rather than at a single point.

According to Cramer's Rule, for a system to have infinitely many solutions, the main determinant $D$ must be zero, and the auxiliary determinants $D_x,D_y,$ and $D_z$ must also be zero. We construct $D$ using the coefficients of our variables:

Expanding along the first row: $1(3b-(-3a))-1(2b-(-a))+2(-6-(-3))=0$

Simplifying the expression: $1(3b+3a)-1(2b+a)+2(-3)=0$ $3b+3a-2b-a-6=0$

This yields our first relation: $2a+b=6$

To find the values of $a$ and $b$, we evaluate $D_z$ by replacing the third column with the constants from the right-hand side:

Expanding along the first row: $1(6b-(-3(a+1)))-1(4b-(-1(a+1)))+6(-6-(-3))=0$

Expanding the terms: $1(6b+3a+3)-1(4b+a+1)+6(-3)=0$ $6b+3a+3-4b-a-1-18=0$

Simplifying leads to: $2a+2b-16=0$ $a+b=8$

We now solve the system of two linear equations: 1) $2a+b=6$ 2) $a+b=8$

Subtracting the second equation from the first: $(2a+b)-(a+b)=6-8$ $a=-2$

Substituting $a=-2$ into $a+b=8$: $-2+b=8$ $b=10$

Finally, we calculate the value of $7a+3b$: $7(-2)+3(10)=-14+30$ Result = 16