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

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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, the value of $\det(A)$ is $2$.

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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If $A = \begin{bmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix}$, then $\det(A)$ is equal to

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