CBSE Class 12 Maths Chapter 4: Determinants NCERT Solutions

Solution:

To find the value of the determinant \(\begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix}\), we use the formula for a 2x2 determinant, which is \(ad - bc\) for a matrix \(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\).

Here, \(a=2\), \(b=4\), \(c=-5\), and \(d=-1\).

Determinant = \((2) \times (-1) - (4) \times (-5)\)

= \(-2 - (-20)\)

= \(-2 + 20\)

= \(18\)

Thus, the value of the determinant is 18.

Solution:

Part (i):

We need to evaluate the determinant \(\begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{vmatrix}\).

Using the formula \(ad - bc\) for a 2x2 determinant:

Determinant = \((\cos \theta) \times (\cos \theta) - (-\sin \theta) \times (\sin \theta)\)

= \(\cos^2 \theta - (-\sin^2 \theta)\)

= \(\cos^2 \theta + \sin^2 \theta\)

Using the trigonometric identity \(\cos^2 \theta + \sin^2 \theta = 1\), the value of the determinant is 1.

Part (ii):

We need to evaluate the determinant \(\begin{vmatrix} x^2 - x + 1 & x - 1 \\ x + 1 & x + 1 \end{vmatrix}\).

Using the formula \(ad - bc\):

Determinant = \((x^2 - x + 1) \times (x + 1) - (x - 1) \times (x + 1)\)

First, expand \((x^2 - x + 1)(x + 1)\):

= \(x(x^2 - x + 1) + 1(x^2 - x + 1)\)

= \(x^3 - x^2 + x + x^2 - x + 1\)

= \(x^3 + 1\)

Next, expand \((x - 1)(x + 1)\), which is a difference of squares:

= \(x^2 - 1\)

Now, subtract the second expansion from the first:

Determinant = \((x^3 + 1) - (x^2 - 1)\)

= \(x^3 + 1 - x^2 + 1\)

= \(x^3 - x^2 + 2\)

Thus, the value of the determinant is \(x^3 - x^2 + 2\).

Solution:

Given the matrix

A = \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix} . First, let's find the value of \(|A|\).

\(|A| = \begin{vmatrix} 1 & 2 \\ 4 & 2 \end{vmatrix} = (1 \times 2) - (2 \times 4) = 2 - 8 = -6\).

Now, let's find the matrix \(2A\).

\(2A = 2 \times \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} 2 \times 1 & 2 \times 2 \\ 2 \times 4 & 2 \times 2 \end{bmatrix} = \begin{bmatrix} 2 & 4 \\ 8 & 4 \end{bmatrix}\).

Next, let's find the determinant of \(2A\), denoted as \(|2A|\).

\(|2A| = \begin{vmatrix} 2 & 4 \\ 8 & 4 \end{vmatrix} = (2 \times 4) - (4 \times 8) = 8 - 32 = -24\).

Now, let's find the value of \(4|A|\).

\(4|A| = 4 \times (-6) = -24\).

Comparing the values, we see that \(|2A| = -24\) and \(4|A| = -24\).

Therefore, we have shown that \(|2A| = 4|A|\).

Solution:

Given the matrix

A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{bmatrix} . First, let's calculate the determinant of A, \(|A|\).

We can expand along the first row:

\(|A| = 1 \times \begin{vmatrix} 1 & 2 \\ 0 & 4 \end{vmatrix} - 0 \times \begin{vmatrix} 0 & 2 \\ 0 & 4 \end{vmatrix} + 1 \times \begin{vmatrix} 0 & 1 \\ 0 & 0 \end{vmatrix}\)

= \(1 \times (1 \times 4 - 2 \times 0) - 0 + 1 \times (0 \times 0 - 1 \times 0)\)

= \(1 \times (4 - 0) - 0 + 1 \times (0 - 0)\)

= \(1 \times 4 = 4\).

So, \(|A| = 4\).

Now, let's find the matrix \(3A\).

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

Now, let's calculate the determinant of \(3A\), \(|3A|\).

Expanding along the first row:

\(|3A| = 3 \times \begin{vmatrix} 3 & 6 \\ 0 & 12 \end{vmatrix} - 0 \times \begin{vmatrix} 0 & 6 \\ 0 & 12 \end{vmatrix} + 3 \times \begin{vmatrix} 0 & 3 \\ 0 & 0 \end{vmatrix}\)

= \(3 \times (3 \times 12 - 6 \times 0) - 0 + 3 \times (0 \times 0 - 3 \times 0)\)

= \(3 \times (36 - 0) - 0 + 3 \times (0 - 0)\)

= \(3 \times 36 = 108\).

So, \(|3A| = 108\).

Finally, let's calculate \(27|A|\).

\(27|A| = 27 \times 4 = 108\).

Since \(|3A| = 108\) and \(27|A| = 108\), we have shown that \(|3A| = 27|A|\).

Solution:

Part (i):

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

Expanding along the first row:

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

= \(3 \times (0 \times 0 - (-1) \times (-5)) + 1 \times (0 \times 0 - (-1) \times 3) - 2 \times (0 \times (-5) - 0 \times 3)\)

= \(3 \times (0 - 5) + 1 \times (0 + 3) - 2 \times (0 - 0)\)

= \(3 \times (-5) + 1 \times 3 - 2 \times 0\)

= \(-15 + 3 - 0\)

= \(-12\)

Part (ii):

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

Expanding along the first row:

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

= \(3 \times (1 \times 1 - (-2) \times 3) + 4 \times (1 \times 1 - (-2) \times 2) + 5 \times (1 \times 3 - 1 \times 2)\)

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

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

= \(21 + 20 + 5\)

= \(46\)

Part (iii):

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

Expanding along the first row:

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

= \(0 - 1 \times ((-1) \times 0 - (-3) \times (-2)) + 2 \times ((-1) \times 3 - 0 \times (-2))\)

= \(-1 \times (0 - 6) + 2 \times (-3 - 0)\)

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

= \(6 - 6\)

= \(0\)

Part (iv):

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

Expanding along the first row:

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

= \(2 \times (2 \times 0 - (-1) \times (-5)) + 1 \times (0 \times 0 - (-1) \times 3) - 2 \times (0 \times (-5) - 2 \times 3)\)

= \(2 \times (0 - 5) + 1 \times (0 + 3) - 2 \times (0 - 6)\)

= \(2 \times (-5) + 1 \times 3 - 2 \times (-6)\)

= \(-10 + 3 + 12\)

= \(5\)

Solution:

We need to find the determinant of the matrix

A = \begin{bmatrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{bmatrix} . Expanding along the first row:

\(|A| = 1 \times \begin{vmatrix} 1 & -3 \\ 4 & -9 \end{vmatrix} - 1 \times \begin{vmatrix} 2 & -3 \\ 5 & -9 \end{vmatrix} + (-2) \times \begin{vmatrix} 2 & 1 \\ 5 & 4 \end{vmatrix}\)

= \(1 \times (1 \times (-9) - (-3) \times 4) - 1 \times (2 \times (-9) - (-3) \times 5) - 2 \times (2 \times 4 - 1 \times 5)\)

= \(1 \times (-9 + 12) - 1 \times (-18 + 15) - 2 \times (8 - 5)\)

= \(1 \times 3 - 1 \times (-3) - 2 \times 3\)

= \(3 + 3 - 6\)

= \(0\)

Thus, the value of the determinant \(|A|\) is 0.

Solution:

Part (i):

We are given the equation

\begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix} = \begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix} . First, calculate the determinant on the left side:

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

Next, calculate the determinant on the right side:

\(\begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix} = (2x \times x) - (4 \times 6) = 2x^2 - 24\).

Now, set the two determinants equal to each other:

\(-18 = 2x^2 - 24\)

Add 24 to both sides:

\(-18 + 24 = 2x^2\)

\(6 = 2x^2\)

Divide by 2:

\(x^2 = 3\)

Take the square root of both sides:

\(x = \pm \sqrt{3}\)

Part (ii):

We are given the equation

\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = \begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix} . Calculate the determinant on the left side:

\(\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = (2 \times 5) - (3 \times 4) = 10 - 12 = -2\).

Calculate the determinant on the right side:

\(\begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix} = (x \times 5) - (3 \times 2x) = 5x - 6x = -x\).

Set the two determinants equal:

\(-2 = -x\)

Multiply by -1:

\(x = 2\)