Ncert Solutions For Class 12 Maths Ex 4.1

Ncert Solutions For Class 12 Maths Chapter 4 Ex 4.1


Q1: Evaluate the determinants.

\(\begin{vmatrix} 2 & 4\\ -5 & -1 \end{vmatrix}\)

Ans:

\(\begin{vmatrix} 2 & 4\\ -5 & -1 \end{vmatrix}\) = 2 ( -1 ) – 4 ( -5 ) = – 2 + 20 = 18

 

Q2: Evaluate the determinant.

(i) \(\begin{vmatrix} cos \Theta & -sin \Theta \\ sin \Theta & cos \Theta \end{vmatrix}\\\)

(ii) \(\\\begin{vmatrix} x^{ 2 } – x + 1 & x – 1 \\ x + 1 & x + 1 \end{vmatrix}\)

Ans:

(i) (cos Ɵ )(cos Ɵ ) – ( -sin Ɵ ) (sin Ɵ ) = cos2 Ɵ + sin2 Ɵ = 1

(ii) = ( x2 – x + 1 ) ( x + 1 ) – ( x – 1 ) ( x + 1 )

= x3 – x2 + x + x2 – x + 1 – ( x2 – 1 )

= x3 + 1 – x2 + 1

= x3 – x2 + 2

 

 

Q3: If A = \(\begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}\) , then show that |2A| = 4 |A|

Ans: The given matrix is A = \(\begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}\)

Therefore, 2A = \(2 \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} 2 & 4 \\ 8 & 4 \end{bmatrix}\)

Therefore, L.H.S. = |2A| = \(\begin{vmatrix} 2 & 4 \\ 8 & 4 \end{vmatrix}\) = 2 × 4 – 4 × 8 = 8 – 32 = -24

Now, |A| = \(\begin{vmatrix} 1 & 2 \\ 4 & 2 \end{vmatrix}\) =

1 × 2 – 2 × 4 = 2 – 8 = -6

Therefore, R.H.S. = 4|A| = 4 × ( -6) = -24

Therefore, L.H.S. = R.H.S.

Q.4: If A = \(\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{bmatrix}\) , then show that |3A| = 27|A|.

Ans: 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

Therefore, 27|A| = 27 (4) = 108 …(i)

Now, 3A = \(3 \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{bmatrix} = \begin{bmatrix} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{bmatrix}\)

Therefore, |3A| = \(3 \begin{vmatrix} 3 & 6 \\ 0 & 12 \end{vmatrix} – 0 \begin{vmatrix} 0 & 3 \\ 0 & 12 \end{vmatrix} + 0 \begin{vmatrix} 0 & 3 \\ 3 & 6 \end{vmatrix}\)

= 3 ( 36 – 0 ) = 3 ( 36 ) = 108 …(ii)

From equations (i) and (ii), we have:

|3A| = 27|A|

Hence, the given result is proved.

Q5: Evaluate the determinants

(i) \(\begin{vmatrix} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{vmatrix}\\\)

(ii) \(\begin{vmatrix} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{vmatrix}\\\)

(iii) \(\begin{vmatrix} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{vmatrix}\\\)

(iv) \(\begin{bmatrix} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{bmatrix}\)

Ans:

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

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

|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 \(\begin{bmatrix} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{bmatrix}\)

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 = \(\begin{bmatrix} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{bmatrix}\)

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 = \(\begin{bmatrix} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{bmatrix}\)

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

Q6: If A = \(\begin{bmatrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{bmatrix}\) , find |A|.

Ans:

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

Q7: Find values of x, if

(i) \(\begin{vmatrix} 2 & 4 \\ 2 & 1 \end{vmatrix} = \begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}\\\)

(ii) \(\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = \begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}\)

Ans:

(i) \(\begin{vmatrix} 2 & 4 \\ 2 & 1 \end{vmatrix} = \begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}\)

\(\boldsymbol{\Rightarrow }\) 2 × 1 – 5 × 4 = 2x × x – 6 × 4

\(\boldsymbol{\Rightarrow }\) 2 – 20 = 2x2 – 24

\(\boldsymbol{\Rightarrow }\) 2x2 = 6

\(\boldsymbol{\Rightarrow }\) x2 = 3

\(\boldsymbol{\Rightarrow }\) x = ± \(\sqrt{ 3 }\)

(ii) \(\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = \begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}\)

\(\boldsymbol{\Rightarrow }\) 2 × 5 – 3 × 4 = x × 5 – 3 × 2x

\(\boldsymbol{\Rightarrow }\) 10 -12 = 5x – 6x

\(\boldsymbol{\Rightarrow }\) -2 = -x

\(\boldsymbol{\Rightarrow }\) x = 2

Q8: If \(\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix} = \begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix}\) , then x is equal to

(A) 6

(B) ± 6

(C) – 6

(D) 0

Ans:

\(\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix} = \begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix}\\\)

\(\\\boldsymbol{\Rightarrow }\) x2 – 36 = 36 – 36

\(\boldsymbol{\Rightarrow }\) x2 – 36 = 0

\(\boldsymbol{\Rightarrow }\) x = ± 6

Hence, the correct answer is B