Class 12 Mathematics Chapter 4: Determinants - NCERT Solutions

\[ \begin{vmatrix} \mathbf{2} & \mathbf{4} \\ \mathbf{-5} &  \mathbf{-1} \end{vmatrix} \]

= 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