Animated Solution for Mathematics - Matrices and Determinants: Let $\lambda$ be a real number for which the system of linear equations $x+y+z=6$, $4x+\lambda y-\lambda z=-2$, $3x+2y-4z=-5$ has infinitely many solutions. Then $\lambda$ is a root of the quadratic equation.

Imagine you are standing in a three-dimensional space. Each of the equations in our system, $x+y+z=6$, $4x+\lambda y-\lambda z=-2$, and $3x+2y-4z=-5$, represents a plane.

Normally, three planes intersect at a single point, giving us a unique solution. But the problem tells us something fascinating: this system has infinitely many solutions.

Geometrically, this means these three planes do not meet at a single point; instead, they intersect along a common line or are coincident. This is a state of perfect harmony, a dependency where the equations are essentially telling the same story in different ways.

In the language of linear algebra, this dependency is captured by the determinant of the coefficient matrix, $D$. If $D eq 0$, the system is independent and has a unique solution.

But when $D=0$, the system is singular, and the planes are no longer independent. This is our golden ticket. To find the value of $\lambda$ that forces this dependency, we must set the determinant of the coefficient matrix to zero:

Now, let's break this down. We expand along the first row to keep our calculations clean. Remember the sign convention for determinants: plus, minus, plus.

For the first element, $1$, we hide its row and column, leaving us with the minor $\left|\begin{array}{cc}\lambda & -\lambda \\ 2& -4\end{array}\right|$. The calculation is $1\cdot ((\lambda )(-4)-(-\lambda )(2))=-4\lambda +2\lambda =-2\lambda$.

For the second element, $1$, we apply the negative sign: $-1\cdot \left|\begin{array}{cc}4& -\lambda \\ 3& -4\end{array}\right|$. This gives us $-1\cdot ((4)(-4)-(-\lambda )(3))=-1\cdot (-16+3\lambda )=16-3\lambda$.

Finally, for the third element, $1$, we have $+1\cdot \left|\begin{array}{cc}4& \lambda \\ 3& 2\end{array}\right|$. This results in $1\cdot ((4)(2)-(\lambda )(3))=8-3\lambda$.

Combining these, we get the equation:

Grouping the terms, we have $(-2\lambda -3\lambda -3\lambda )+(16+8)=0$, which simplifies to $-8\lambda +24=0$.

Solving for $\lambda$, we find $\lambda =3$. It is a moment of pure satisfaction when the algebra collapses into such a clean, integer result.

The question asks us to identify which quadratic equation has $\lambda =3$ as a root. Testing the options, we look at ${\lambda }^2-\lambda -6=0$.

Substituting $\lambda =3$ gives $(3{)}^2-3-6=9-3-6=0$. The condition is perfectly satisfied.

We have navigated the geometry, mastered the determinant, and arrived at the truth. Keep this logic in your toolkit—whenever you see "infinitely many solutions," let $D=0$ be your first instinct.