For what value of k is the matrix $\begin{bmatrix} 2\cos 2\theta & 2\cos 2\theta & 6 \ 1 -2 \sin^2\theta & 2 \cos^2\theta -1 & 3 \ k & 2k & 1 \end{bmatrix}$ singular?

Question

For what value of k is the matrix $\begin{bmatrix} 2\cos 2\theta & 2\cos 2\theta & 6 \ 1 -2 \sin^2\theta & 2 \cos^2\theta -1 & 3 \ k & 2k & 1 \end{bmatrix}$  singular?

Prepp Team · Accepted Answer

Understanding Singular Matrices

A matrix is considered singular if its determinant is equal to zero. If the determinant is non-zero, the matrix is called non-singular or invertible.

Simplifying the Matrix Elements

The given matrix is:

$ A = \begin{bmatrix} 2\cos 2\theta & 2\cos 2\theta & 6 \\ 1 -2 \sin^2\theta & 2 \cos^2\theta -1 & 3 \\ k & 2k & 1 \end{bmatrix} $

We can simplify the elements in the second row using standard trigonometric identities:

$1 - 2\sin^2\theta = \cos 2\theta$
$2 \cos^2\theta - 1 = \cos 2\theta$

Substituting these identities, the matrix simplifies to:

$ A = \begin{bmatrix} 2\cos 2\theta & 2\cos 2\theta & 6 \\ \cos 2\theta & \cos 2\theta & 3 \\ k & 2k & 1 \end{bmatrix} $

Analyzing for Linear Dependence in Rows

A powerful property of determinants is that if the rows (or columns) of a matrix are linearly dependent, the determinant is zero, making the matrix singular. Let's look closely at the rows of the simplified matrix:

Row 1 ($R_1$): $[2\cos 2\theta \quad 2\cos 2\theta \quad 6]$
Row 2 ($R_2$): $[\cos 2\theta \quad \cos 2\theta \quad 3]$
Row 3 ($R_3$): $[k \quad 2k \quad 1]$

Let's compare the elements of Row 1 and Row 2:

First element: $2\cos 2\theta$ in $R_1$ and $\cos 2\theta$ in $R_2$. $2\cos 2\theta = 2 \times \cos 2\theta$.
Second element: $2\cos 2\theta$ in $R_1$ and $\cos 2\theta$ in $R_2$. $2\cos 2\theta = 2 \times \cos 2\theta$.
Third element: $6$ in $R_1$ and $3$ in $R_2$. $6 = 2 \times 3$.

We can observe that each element in Row 1 is exactly twice the corresponding element in Row 2. This means that $R_1$ is a scalar multiple of $R_2$ ($R_1 = 2R_2$).

Determinant and Singularity Condition

Since Row 1 and Row 2 of the matrix are linearly dependent ($R_1$ is a scalar multiple of $R_2$), the determinant of the matrix is guaranteed to be zero. This is a fundamental property of determinants.

$|A| = 0$

This condition ($|A| = 0$) holds true regardless of the values in the third row, which contains the variable $k$. The linear dependency between the first two rows alone makes the determinant zero.

Conclusion for the Value of k

Because the determinant of the matrix is always zero, irrespective of the value of $k$, the matrix is singular for any real value of $k$.

Revision Table: Key Matrix Properties for Singularity

Property	Explanation related to Singularity
Singular Matrix	A matrix whose determinant is zero.
Non-Singular Matrix	A matrix whose determinant is non-zero. It has an inverse.
Linear Dependence	If one row (or column) is a linear combination of other rows (or columns), the determinant is zero, and the matrix is singular.
Determinant	A scalar value calculated from a square matrix; indicates properties like invertibility and linear dependence.

Additional Information: Why Linear Dependence Means Determinant is Zero

The determinant of a matrix is closely related to the volume of the parallelepiped spanned by its row or column vectors. If the vectors are linearly dependent, they lie in a lower-dimensional space, and the volume of the parallelepiped they span is zero. Hence, the determinant is zero.

For example, if $R_1 = c \times R_2$, the row vectors do not span a full 3D space; they are confined to a plane (or a line or a point), which has zero volume in 3D. This is why the determinant is zero.

In systems of linear equations $Ax = \mathbf{b}$, a singular matrix A implies that there is either no unique solution or infinitely many solutions, depending on $\mathbf{b}$. For $Ax = \mathbf{0}$, a singular matrix guarantees non-trivial solutions.

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