(A) To find the determinant of matrix $A$,we expand along the first row: $|A| = \begin{vmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{vmatrix}$ $|A

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(A) To find the determinant of matrix $A$,we expand along the first row: $|A| = \begin{vmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{vmatrix}$ $|A| = 1 \begin{vmatrix} 1 & 0 \\ 2 & 1 \end{vmatrix} - 0 \begin{vmatrix} 2 & 0 \\ 3 & 1 \end{vmatrix} + 1 \begin{vmatrix} 2 & 1 \\ 3 & 2 \end{vmatrix}$ $|A| = 1(1 \times 1 - 0 \times 2) - 0 + 1(2 \times 2 - 1 \times 3)$ $|A| = 1(1 - 0) + 1(4 - 3)$ $|A| = 1(1) + 1(1) = 1 + 1 = 2$ Thus,$\det A = 2$.

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 $A = \begin{bmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix}$,then $\det A$ =

A
$2$
B
$3$
C
$4$
D
$5$

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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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Medium

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DifficultJEE MAIN 2024

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EasyKCET 2015

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If $A = \begin{bmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix}$,then $\det A$ =

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Find the equation of the line joining $(1, 2)$ and $(3, 6)$ using determinants.

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The value of $a$ for which the system of equations has a non-zero solution is
$a^{3} x+(a+1)^{3} y+(a+2)^{3} z=0$
$a x+(a+1) y+(a+2) z=0$
$x+y+z=0$

If the system of equations
$x+(\sqrt{2} \sin \alpha) y+(\sqrt{2} \cos \alpha) z=0$
$x+(\cos \alpha) y+(\sin \alpha) z=0$
$x+(\sin \alpha) y-(\cos \alpha) z=0$
has a non-trivial solution,then $\alpha \in \left(0, \frac{\pi}{2}\right)$ is equal to :