(B) Given the matrix $A = \begin{bmatrix} \lambda & i \\ i & -\lambda \end{bmatrix}$.For the inverse $A^{-1}$ to not exist,the determinant of the matr

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(B) Given the matrix $A = \begin{bmatrix} \lambda & i \\ i & -\lambda \end{bmatrix}$.For the inverse $A^{-1}$ to not exist,the determinant of the matrix must be zero,i.e.,$|A| = 0$.Calculating the determinant:$|A| = (\lambda)(-\lambda) - (i)(i) = -\lambda^2 - i^2$.Since $i = \sqrt{-1}$,we have $i^2 = -1$.Substituting this into the determinant equation:$|A| = -\lambda^2 - (-1) = -\lambda^2 + 1$.Setting the determinant to zero:$-\lambda^2 + 1 = 0 \Rightarrow \lambda^2 = 1$.Therefore,$\lambda = \pm 1$.

Class 12 Mathematics 3-4. Matrices and Determinants Expansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line

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If $A = \begin{bmatrix} \lambda & i \\ i & -\lambda \end{bmatrix}$ and $A^{-1}$ does not exist,then $\lambda = $ (where $i = \sqrt{-1}$)

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A
$\pm 2$
B
$\pm 1$
C
$0$
D
$\pm 3$

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3-4. Matrices and Determinants

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Expansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line

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If $A = \begin{bmatrix} \lambda & i \\ i & -\lambda \end{bmatrix}$ and $A^{-1}$ does not exist,then $\lambda = $ (where $i = \sqrt{-1}$)

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Let $N = \left| \begin{array}{ccc} 28 & 25 & 38 \\ 42 & 38 & 65 \\ 56 & 47 & 83 \end{array} \right|$. Then,the number of ways in which $N$ can be expressed as a product of two relatively prime divisors is:

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