(A) matrix $A$ is non-singular if its determinant $|A| \ne 0$.Given $A = \left[ {\begin{array}{*{20}{c}}2&\lambda &{ - 4}\\{ - 1}&3&4\\1&{ - 2}&{ - 3}

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(A) matrix $A$ is non-singular if its determinant $|A| \ne 0$.Given $A = \left[ {\begin{array}{*{20}{c}}2&\lambda &{ - 4}\\{ - 1}&3&4\\1&{ - 2}&{ - 3}\end{array}} \right]$.Calculating the determinant $|A|$:$|A| = 2[3(-3) - 4(-2)] - \lambda[(-1)(-3) - 4(1)] + (-4)[(-1)(-2) - 3(1)]$$|A| = 2[-9 + 8] - \lambda[3 - 4] - 4[2 - 3]$$|A| = 2(-1) - \lambda(-1) - 4(-1)$$|A| = -2 + \lambda + 4$$|A| = \lambda + 2$For the matrix to be non-singular,$|A| \ne 0$,so $\lambda + 2 \ne 0$,which implies $\lambda \ne -2$.

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

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The matrix $\left[ {\begin{array}{*{20}{c}}2&\lambda &{ - 4}\\{ - 1}&3&4\\1&{ - 2}&{ - 3}\end{array}} \right]$ is non-singular,if

A
$\lambda \ne - 2$
B
$\lambda \ne 2$
C
$\lambda \ne 3$
D
$\lambda \ne - 3$

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3 and 4 .Determinants and Matrices

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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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The matrix $\left[ {\begin{array}{*{20}{c}}2&\lambda &{ - 4}\\{ - 1}&3&4\\1&{ - 2}&{ - 3}\end{array}} \right]$ is non-singular,if

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