Evaluate the determinants (i) (egin{v | Class 12 Mathematics Chapter Determinants, Determinants NCERT Solutions

(i) Let A = \(\begin{vmatrix}3 & -1 & -2\\0 & 1 & 2\\0 & 0 & 4\end{vmatrix}\)

It can be observed that in the second row, two entries are zero. Thus, we expand along the second row for easier calculation.

|A|  = -0\(\begin{vmatrix}-1 & -2\\-5 & 0\end{vmatrix}\) + 0\(\begin{vmatrix}3 & -2\\3 & 0\end{vmatrix}\) – (-1)\(\begin{vmatrix}3 &-1\\3 &-5\end{vmatrix}\) = (-15 + 3) = -12

(ii) Let A = \(\begin{vmatrix}0 & 1 & 2\\-1 & 0  & -3\\-2 & 3 & 0\end{vmatrix}\)

By expanding along the first row, we have:

|A|  = 3\(\begin{vmatrix}1 & -2\\3 & 1\end{vmatrix}\) + 4\(\begin{vmatrix}1 & -2\\2 & 1\end{vmatrix}\)  + 5\(\begin{vmatrix}1 &1\\2 &3\end{vmatrix}\)

= 3 (1+6) + 4(1+4) + 5(3-2)

= 3 (7) + 4 (5) + 5 (1)

= 21 + 20 + 5

= 46

(iii) Let  A = \(\begin{vmatrix}3 & -4 & 5\\1 & 1 & -2\\2 & 3 & 1\end{vmatrix}\)

By expanding along the first row, we have:

|A|  = 0\(\begin{vmatrix}0 & -3\\3 & 0\end{vmatrix}\) - 1\(\begin{vmatrix}-1 & -3\\-2 & 0\end{vmatrix}\)  + 2\(\begin{vmatrix}-1 & 0\\-2 &3\end{vmatrix}\)

= 0 – 1(0 – 6) + 2 (-3 - 0)

= -1 (-6) + 2(-3)

= 6 – 6

= 0

(iv) Let  A = \(\begin{vmatrix}2 & -1 & -2\\0 & 2 & -1\\3 & -5 & 0\end{vmatrix}\)

By expanding along the first column, we have:

|A|  = 2\(\begin{vmatrix}2 & -1\\-5 & 0\end{vmatrix}\) - 0\(\begin{vmatrix}-1 & -2\\-5 & 0\end{vmatrix}\)  + 3\(\begin{vmatrix}-1 & -2\\2 & -1\end{vmatrix}\)

= 2(0 – 5) – 0 + 3(1 + 4)

= -10 + 15 = 5