NCERT Exemplar Solutions
Class 12 - Mathematics - Chapter 4: DETERMINANTS
Short Answer (S.A.)

Question. 1

Using the properties of determinants, evaluate:

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

Answer:

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

Question. 2

Using the properties of determinants, evaluate:

\(\begin{vmatrix} a + x & y & z \\ x & a + y & z \\ x & y & a + z \end{vmatrix}\)

Answer:

\(a^2 (a + x + y + z)\)

Question. 3

Using the properties of determinants, evaluate:

\(\begin{vmatrix} 0 & x y^2 & x z^2 \\ x^2 y & 0 & y z^2 \\ x^2 z & z y^2 & 0 \end{vmatrix}\)

Answer:

\(2 x^3 y^3 z^3\)

Question. 4

Using the properties of determinants, evaluate:

\(\begin{vmatrix} 3x & -x + y & -x + z \\ x - y & 3y & z - y \\ x - z & y - z & 3z \end{vmatrix}\)

Answer:

\(3 (x + y + z)(xy + yz + zx)\)

Question. 5

Using the properties of determinants, evaluate:

\(\begin{vmatrix} x + 4 & x & x \\ x & x + 4 & x \\ x & x & x + 4 \end{vmatrix}\)

Answer:

\(16(3x + 4)\)

Question. 6

Using the properties of determinants, evaluate:

\(\begin{vmatrix} a - b - c & 2a & 2a \\ 2b & b - c - a & 2b \\ 2c & 2c & c - a - b \end{vmatrix}\)

Answer:

\((a + b + c)^3\)

Question. 7

Using the proprties of determinants, prove that:

\(\begin{vmatrix} y^2 z^2 & y z & y + z \\ z^2 x^2 & z x & z + x \\ x^2 y^2 & x y & x + y \end{vmatrix} = 0\)

Question. 8

Using the proprties of determinants, prove that:

\(\begin{vmatrix} y + z & z & y \\ z & z + x & x \\ y & x & x + y \end{vmatrix} = 4xyz\)

Question. 9

Using the proprties of determinants, prove that:

\(\begin{vmatrix} a^2 + 2a & 2a + 1 & 1 \\ 2a + 1 & a + 2 & 1 \\ 3 & 3 & 1 \end{vmatrix} = (a - 1)^3\)

Question. 10

If \(A + B + C = 0\), then prove that

\(\begin{vmatrix} 1 & \cos C & \cos B \\ \cos C & 1 & \cos A \\ \cos B & \cos A & 1 \end{vmatrix} = 0\)

Question. 11

If the co-ordinates of the vertices of an equilateral triangle with sides of length \(a\) are \((x_1, y_1), (x_2, y_2), (x_3, y_3)\), then show that

\(\left|\begin{matrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{matrix}\right|^{2} = \dfrac{3a^{4}}{4}\)

Question. 12

Find the value of \(\theta\) satisfying

\(\begin{vmatrix} 1 & 1 & \sin 3\theta \\ -4 & 3 & \cos 2\theta \\ 7 & -7 & -2 \end{vmatrix} = 0\)

Answer:

\(\theta = n\pi\) or \(\theta = n\pi + (-1)^n \dfrac{\pi}{6}\)

Question. 13

If

\(\begin{vmatrix} 4 - x & 4 + x & 4 + x \\ 4 + x & 4 - x & 4 + x \\ 4 + x & 4 + x & 4 - x \end{vmatrix} = 0\), then find values of \(x\).

Answer:

\(x = 0, -12\)

Question. 14

If \(a_1, a_2, a_3, \ldots, a_r\) are in G.P., then prove that the determinant

\(\begin{vmatrix} a_{r+1} & a_{r+5} & a_{r+9} \\ a_{r+7} & a_{r+11} & a_{r+15} \\ a_{r+11} & a_{r+17} & a_{r+21} \end{vmatrix}\)

is independent of \(r\).

Question. 15

Show that the points \((a + 5, a - 4), (a - 2, a + 3)\) and \((a, a)\) do not lie on a straight line for any value of \(a\).

Question. 16

Show that the triangle \(ABC\) is an isosceles triangle if the determinant

\(\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 + \cos A & 1 + \cos B & 1 + \cos C \\ \cos^{2} A + \cos A & \cos^{2} B + \cos B & \cos^{2} C + \cos C \end{vmatrix} = 0\).

Question. 17

Find \(A^{-1}\) if \(A = \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix}\) and show that

\(A^{-1} = \dfrac{A^{2} - 3I}{2}\).

NCERT Exemplar Solutions Class 12 – Mathematics – Chapter 4: DETERMINANTS – Short Answer (S.A.) | Detailed Answers