Class 12 Mathematics - Chapter Determinants NCERT Solutions | Evaluate the determinants (i) \(\begin{v
Evaluate the determinants
(i) \(\begin{vmatrix}3 & -1 & -2\\0 & 1 & 2\\0 & 0 & 4\end{vmatrix}\) (iii) \(\begin{vmatrix}3 & -4 & 5\\1 & 1 & -2\\2 & 3 & 1\end{vmatrix}\)
(ii) \(\begin{vmatrix}0 & 1 & 2\\-1 & 0 & -3\\-2 & 3 & 0\end{vmatrix}\)(iv) \(\begin{vmatrix}2 & -1 & -2\\0 & 2 & -1\\3 & -5 & 0\end{vmatrix}\)
(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
More Questions From Class 12 Mathematics - Chapter Determinants
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