Animated Solution for Mathematics - Matrices and Determinants: Let the system of linear equations $x+y+kz=2$, $2x+3y-z=1$, $3x+4y+2z=k$ have infinitely many solutions. Then the system $(k+1)x+(2k-1)y=7$, $(2k+1)x+(k+5)y=10$ has :

To determine the condition under which the system of planes possesses infinitely many solutions, we must ensure that the determinant of the coefficient matrix is zero. This condition implies that the normal vectors of the planes are linearly dependent, causing the system to collapse into a shared intersection.

We define the coefficient matrix $A$ and set its determinant $\Delta$ to zero:

Expanding the determinant along the first row, we calculate the volume of the parallelepiped formed by the normal vectors. Setting this volume to zero ensures the planes are coplanar:

Simplifying the terms inside the expansion:

Solving this linear equation yields the critical value $k=3$.

With the parameter $k$ identified, we substitute $k=3$ into the second system of equations provided. This transforms the problem into a standard two-dimensional linear system:

To solve for $x$ and $y$, we employ the method of elimination. We multiply the first equation by $7$ and the second by $4$ to align the coefficients of $x$:

Subtracting the second equation from the first eliminates $x$ entirely:

Substituting $y=3$ back into the equation $4x+5y=7$:

The unique solution to the system is $(x,y)=(-2,3)$. Evaluating the sum of these coordinates, we find:

The final result confirms that the unique solution satisfies the condition $x+y=1$.