If A = bmatrix 2 & 1 & -1 \\ 0 & 1 & 2 \\ 2 & -1 & bmatrix is a singular matrix, then the value of is Option 1: 0 Option 2: 3 Option 3: 5 Option 4: -5

Answer: -5. Solution: A matrix is singular when its determinant equals zero. For the given matrix: det(A) = vmatrix 2 & 1 & -1 \\ 0 & 1 & 2 \\ 2 & -1 & vmatrix Expanding along the first row: det(A) = 2 vmatrix 1 & 2 \\ -1 & vmatrix - 1 vmatrix 0 & 2 \\ 2 & vmatrix + (-1) vmatrix 0 & 1 \\ 2 & -1 vmatrix For the first minor: 2 vmatrix 1 & 2 \\ -1 & vmatrix = 2[(1)() - (2)(-1)] = 2[ + 2] = 2 + 4 For the second minor: -1 vmatrix 0 & 2 \\ 2 & vmatrix = -1[(0)() - (2)(2)] = -1[-4] = 4 For the third minor: -1 vmatrix 0 & 1 \\ 2 & -1 vmatrix = -1[(0)(-1) - (1)(2)] = -1[-2] = 2 Combining all terms: det(A) = 2 + 4 + 4 + 2 det(A) = 2 + 10 Since the matrix is singular: 2 + 10 = 0 2 = -10 = -5 Therefore, the value of is -5.

Algebra

Matrices & Determinants

Medium

common

If $A = \begin{bmatrix} 2 & 1 & -1 \\ 0 & 1 & 2 \\ 2 & -1 & \lambda \end{bmatrix}$ is a singular matrix, then the value of $\lambda$ is

-5

✅ Correct Option: 4

A matrix is singular when its determinant equals zero.

For the given matrix:

$\text{det}(A) = \begin{vmatrix} 2 & 1 & -1 \\ 0 & 1 & 2 \\ 2 & -1 & \lambda \end{vmatrix}$

Expanding along the first row:

$\text{det}(A) = 2 \begin{vmatrix} 1 & 2 \\ -1 & \lambda \end{vmatrix} - 1 \begin{vmatrix} 0 & 2 \\ 2 & \lambda \end{vmatrix} + (-1) \begin{vmatrix} 0 & 1 \\ 2 & -1 \end{vmatrix}$

For the first minor:

$2 \begin{vmatrix} 1 & 2 \\ -1 & \lambda \end{vmatrix} = 2[(1)(\lambda) - (2)(-1)]$

$= 2[\lambda + 2]$

$= 2\lambda + 4$

For the second minor:

$-1 \begin{vmatrix} 0 & 2 \\ 2 & \lambda \end{vmatrix} = -1[(0)(\lambda) - (2)(2)]$

$= -1[-4]$

$= 4$

For the third minor:

$-1 \begin{vmatrix} 0 & 1 \\ 2 & -1 \end{vmatrix} = -1[(0)(-1) - (1)(2)]$

$= -1[-2]$

$= 2$

Combining all terms:

$\text{det}(A) = 2\lambda + 4 + 4 + 2$

$\text{det}(A) = 2\lambda + 10$

Since the matrix is singular:

$2\lambda + 10 = 0$

$2\lambda = -10$

$\lambda = -5$

Therefore, the value of $\lambda$ is $-5$ .

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If $A = \begin{bmatrix} 2 & 1 & -1 \\ 0 & 1 & 2 \\ 2 & -1 & \lambda \end{bmatrix}$ is a singular matrix, then the value of $\lambda$ is