(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

The number of real values of $t$ such that the system of homogeneous equations
$\begin{aligned}
t x+(t+1) y+(t-1) z &=0 \\
(t+1) x+t y+(t+2) z &=0 \\
(t-1) x+(t+2) y+t z &=0
\end{aligned}$
has non-trivial solutions is

DifficultAP EAMCET 2013

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What is the area of the triangle with vertices $(4, 4)$,$(3, -2)$,and $(3, -16)$?

Easy

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Prove that $\left|\begin{array}{ccc}b+c & a & a \\ b & c+a & b \\ c & c & a+b\end{array}\right|=4abc$

Medium

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If $f(x) = \begin{vmatrix} 1 & x & x+1 \\ 2x & x(x-1) & (x+1)x \\ 3x(x-1) & x(x-1)(x-2) & (x+1)x(x-1) \end{vmatrix}$, then $f(100)$ is equal to:

MediumWBJEE 2015

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If the matrix $\begin{bmatrix} 0 & 1 & -2 \\ -1 & 0 & 3 \\ \lambda & -3 & 0 \end{bmatrix}$ is singular,then $\lambda = $

Easy

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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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The number of real values of $t$ such that the system of homogeneous equations$\begin{aligned}t x+(t+1) y+(t-1) z &=0 \\(t+1) x+t y+(t+2) z &=0 \\(t-1) x+(t+2) y+t z &=0\end{aligned}$has non-trivial solutions is

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The number of real values of $t$ such that the system of homogeneous equations
$\begin{aligned}
t x+(t+1) y+(t-1) z &=0 \\
(t+1) x+t y+(t+2) z &=0 \\
(t-1) x+(t+2) y+t z &=0
\end{aligned}$
has non-trivial solutions is