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Welcome to the Chapter 4 - Determinants, Class 12 Mathematics NCERT Solutions page. Here, we provide detailed question answers for Chapter 4 - Determinants. The page is designed to help students gain a thorough understanding of the concepts related to natural resources, their classification, and sustainable development.
Our solutions explain each answer in a simple and comprehensive way, making it easier for students to grasp key topics Determinants and excel in their exams. By going through these Determinants question answers, you can strengthen your foundation and improve your performance in Class 12 Mathematics. Whether you’re revising or preparing for tests, this chapter-wise guide will serve as an invaluable resource.
\[ \begin{vmatrix} \mathbf{2} & \mathbf{4} \\ \mathbf{-5} & \mathbf{-1} \end{vmatrix} \]
= 2(−1) − 4(−5)
= − 2 + 20
= 18
(i) \begin{vmatrix} \mathbf{Cosθ} & \mathbf{−sin θ} \\ \mathbf{sin θ} & \mathbf{cos θ} \end{vmatrix}
= (cos θ)(cos θ) − (−sin θ)(sin θ)
= cos2 θ+ sin2 θ
= 1
(ii) \begin{vmatrix} \mathbf{x^2 − x + 1} & \mathbf{x − 1} \\ \mathbf{x + 1} & \mathbf{x + 1} \end{vmatrix}
= (x2 − x + 1)(x + 1) − (x − 1)(x + 1)
= x3 − x2 + x + x2 − x + 1 − (x2 − 1)
= x3 + 1 − x2 + 1
= x3 − x2 + 2
The given matrix is
\(u=\begin{bmatrix}1 & 2\\4 & 2\end{bmatrix}\)
So 2A = 2\(\begin{bmatrix}1 & 2\\4 & 2\end{bmatrix}\)
\(= \begin{bmatrix}2 & 4\\8 & 4\end{bmatrix}\)
so L.H.S. = |2A| \(= \begin{bmatrix}2 & 4\\8 & 4\end{bmatrix}\)
= 2 x 4 - 4 x 8
= 8 - 32
= -24
The given matrix is
A=\(\begin{bmatrix}1 & 0 & 1\\0 & 1 & 2\\0 & 0 & 4\end{bmatrix}\)
It can be observed that in the first column, two entries are zero. Thus, we expand along the first column (C1) for easier calculation.
| A| = 1\(\begin{vmatrix}1 & 2\\0 & 4\end{vmatrix}\) - 0\(\begin{vmatrix}0 & 1\\0 & 4\end{vmatrix}\) + 0\(\begin{vmatrix}0 &1\\1 & 2\end{vmatrix}\) = 1(4 – 0) – 0 + 0 = 4
So 27 |A| = 27 (4) = 108 ……. (i)
Now 3A = 3\(\begin{bmatrix}1 & 0 & 1\\0 & 1 & 2\\0 & 0 & 4\end{bmatrix}\)=\(\begin{bmatrix}3 & 0 & 3\\0 & 3 & 6\\0 & 0 & 12\end{bmatrix}\)
So |3A| = 3\(\begin{vmatrix}3 & 6\\0 & 12\end{vmatrix}\) - 0\(\begin{vmatrix}0 & 3\\0 & 12\end{vmatrix}\) + 0\(\begin{vmatrix}0 &3\\0 & 6\end{vmatrix}\)
= 3 (36 – 0) = 3(36) 108 ……….. (ii)
From equations (i) and (ii), we have:
|3A| = 27|A|
(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
Let A = \(\begin{bmatrix}1 & 1 & -2\\2 & 1 & -3\\5 & 4 & -9\end{bmatrix}\)
By expanding along the first row, we have:
|A| = 1\(\begin{vmatrix}1 & -3\\4 & -9\end{vmatrix}\) - 1\(\begin{vmatrix}2 & -3\\5 & -9\end{vmatrix}\) - 2\(\begin{vmatrix}2 & 1\\5 & 4\end{vmatrix}\)
= 1(-9 + 12) – 1(-18 + 15) -2(8 – 5)
= 1(3) – 1 (-3) – 2(3)
= 3 + 3 – 6
= 6 – 6
= 0
(i) \(\begin{vmatrix}2 & 4\\2 & 1\end{vmatrix}\) = \(\begin{vmatrix}2x & 4\\6 & x\end{vmatrix}\)
⇒2 x 1 – 5 x 4 = 2x x x – 6 x 4)
⇒ 2- 20 = 2x2 – 24
⇒2x2 = 6
⇒ x2 = 3
⇒ x = ±√3
(ii) \(\begin{vmatrix}2 & 3\\4 & 5\end{vmatrix}\) = \(\begin{vmatrix}x & 3\\2x & 5\end{vmatrix}\)
⇒ 2 x 5 – 3 x 4 = x x 5 – 3 x 2x
⇒10 – 12 = 5x – 6x
⇒ -2 = -x
⇒ x = 2
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