Animated Solution for Mathematics - Matrices and Determinants: If the system of equations $11x+y+\lambda z=-5$, $2x+3y+5z=3$, $8x-19y-39z=\mu$ has infinitely many solutions, then ${\lambda }^4-\mu$ is equal to :

Imagine you are standing in a 3D space, looking at three giant, flat sheets of glass—these are our planes, represented by the equations $11x+y+\lambda z=-5$, $2x+3y+5z=3$, and $8x-19y-39z=\mu$.

Usually, three planes intersect at a single point, like the corner of a room. However, the problem states that they have infinitely many solutions.

This implies that these planes are dependent, intersecting along a common line or coinciding entirely. This is the beauty of linear algebra—it is the geometry of the invisible.

To unlock this mystery, we turn to Cramer's Rule. The main determinant, $D$, acts as our gatekeeper.

If $D eq 0$, the system has a unique solution. Since we are told the system is dependent, the gatekeeper must step aside, meaning $D$ must be zero.

We construct this determinant using the coefficients of $x$, $y$, and $z$:

Expanding this along the first row is our first tactical move. We take $11$ multiplied by the minor determinant $\left|\begin{array}{cc}3& 5\\ -19& -39\end{array}\right|$, subtract $1$ times the minor $\left|\begin{array}{cc}2& 5\\ 8& -39\end{array}\right|$, and add $\lambda$ times the minor $\left|\begin{array}{cc}2& 3\\ 8& -19\end{array}\right|$.

This is where the JEE tests your discipline. One small sign error can cause the entire structure to collapse.

Let us calculate carefully:

Inside the brackets, we find:

Solving this, we find $\lambda =-2$. We have successfully navigated the first hurdle.

Is $D=0$ enough? Not quite. If $D=0$, the system could still be inconsistent (no solution).

To guarantee infinitely many solutions, we must ensure that the augmented determinants $D_1$, $D_2$, and $D_3$ are also zero. We focus on $D_1$, where we replace the first column of $D$ with the constants from the right-hand side: $-5$, $3$, and $\mu$.

Expanding this along the first row:

Simplifying the terms:

Solving for $\mu$, we get $\mu =-31$. The logic holds, and the path is clear.

We have our values: $\lambda =-2$ and $\mu =-31$. The problem asks for the value of ${\lambda }^4-\mu$.

Let us compute this final step with confidence:

The final answer is 47. You have navigated the geometry, respected the determinants, and executed the algebra with precision.