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Singular Matrix Solved Problem | Maths

*Solution for \( x \)* Since the given matrix *\( A \)* is singular, its determinant must be **zero**: \[ \det(A) = 0 \] #### *Matrix Given:* \[ A = \begin{bmatrix} 3 - x & 2 & 2 \\ 2 & 4 - x & 1 \\ -2 & -4 & -1 - x \end{bmatrix} \] Now, we compute *\( \det(A) \)* and solve for \( x \). Let's calculate: \[ \det(A) = (3-x) \begin{vmatrix} 4-x & 1 \\ -4 & -1-x \end{vmatrix} 2 \begin{vmatrix} 2 & 1 \\ -2 & -1-x \end{vmatrix} 2 \begin{vmatrix} 2 & 4-x \\ -2 & -4 \end{vmatrix} \] Now, solving for \( x \): \[ \begin{vmatrix} 4-x & 1 \\ -4 & -1-x \end{vmatrix} = (4-x)(-1-x) - (1)(-4) = (-4 -4x + x - x^2) + 4 = -x^2 -3x \] \[ \begin{vmatrix} 2 & 1 \\ -2 & -1-x \end{vmatrix} = (2)(-1-x) - (1)(-2) = (-2 - 2x +2) = -2x \] \[ \begin{vmatrix} 2 & 4-x \\ -2 & -4 \end{vmatrix} = (2)(-4) - (4-x)(-2) = -8 + (8 -2x) = -2x \] Thus, the determinant equation simplifies to: \[ (3-x)(-x^2 -3x) - 2(-2x) + 2(-2x) = 0 \] \[ (3-x)(-x^2 -3x) + 4x - 4x = 0 \] \[ -3x^2 -9x + x^3 + 3x^2 = 0 \] \[ x^3 -9x = 0 \] Factoring: \[ x(x^2 - 9) = 0 \] \[ x(x-3)(x+3) = 0 \] Thus, the possible values for \( x \) are: \[ x = 0, 3, -3 \] *Final Answer:* \[ x = 0, 3, -3 \] --- 1. *Solving for \( x \) in a Singular Matrix* 2. *Find \( x \) for a Determinant of Zero* 3. *Matrix Singularity Condition: Solve for \( x \)* 4. *Determinant-Based Equation for Finding \( x \)* *Best Description:* In this video, we solve for \( x \) in a given *3×3 matrix* under the condition that the matrix is *singular* (i.e., determinant is zero). We compute the determinant, set it to zero, and solve for \( x \). This concept is fundamental in **linear algebra**, **matrix theory**, and **engineering applications**. #LinearAlgebra #MatrixDeterminant #SingularMatrix #MathTutorial #EngineeringMath #DeterminantZero #Algebra #Mathematics