NCERT Solutions for Class 12 Maths Chapter 4 - Determinants

$$\textbf{Q. Prove that the determinant} \begin{vmatrix} x & sin\space\theta & cos\space\theta \\ -sin\space\theta & -x & 1 \\ cos\space\theta & 1 & x \end{vmatrix}\textbf{is independent of}\space \theta. $$

$$\textbf{Ans. } \text{Let} \space \space \Delta = \begin{vmatrix} x & sin\space\theta & cos\space\theta \\ -sin\space\theta & -x & 1 \\ cos\space\theta & 1 & x \end{vmatrix} \\ = x \begin{vmatrix} -x & 1 \\ 1 & x \end{vmatrix} – sin\space \theta\begin{vmatrix} -sin\space \theta & 1 \\ cos\space \theta & x \end{vmatrix} + cos\space \theta \begin{vmatrix} -sin\space \theta & -x \\ cos\space \theta & 1 \end{vmatrix} \\ = x(-x^2-1)-sin\space \theta(-x\space sin\space \theta- cos\space \theta) + cos \space (-sin\space \theta + x\space cos\space \theta) \\ =-x^3-x+xsin^2\space \theta + sin\space \theta \space cos\space \theta-cos\space \theta sin\space \theta + x\space cos^2 \space \theta \\ =- x^3-x+x(sin^2\space \theta + cos^2\space \theta ) \\ =-x^3-x + x \\ =-x^3 \\ \text{Hence,}\space \Delta\space \text{is independent of }\space \theta.$$

$$\textbf{Q. Write minors and cofactors of elements of determinant } \begin{vmatrix} a & c \\ b & d \end{vmatrix}$$

$$ \textbf{Ans. } \text{Given,} \begin{vmatrix} a & c \\ b & d \end{vmatrix} \\ \text{Minors} \space \space M_{11} = d \space \text{and} \space M_{12} = b \\ M_{21} = c \space \text{and} \space M_{22} = a \\ \text{Cofactors are} \\ C_{11} = (-1)^{1+1}d = d \\ C_{12} = (-1)^{1+2}b = -b \\ C_{21} = (-1)^{2+1}c = -c \\ C_{22} = (-1)^{2+2}a = a.$$

$$\textbf{Q. If area of a triangle is 35 sq. units with vertices (2,- 6), (5, 4) and (k,4), then find the the values of k.}$$

$$ \textbf{Ans. } \text{Given,}\\ \frac{1}{2}\begin{vmatrix}2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1\end{vmatrix}= \pm \text{ 35}\\ \Rarr \begin{vmatrix} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1\end{vmatrix}= \pm \text{ 70} \\ \Rarr 2(4×1-4×1)+6(5×1-k×1)+1 (5×4-k×4)\\=\pm\text{ 70} \\ \Rightarrow \text{0 + 6(5 – k) + (20 – 4k)=} \pm 70\\ \Rarr \text{50 – 10k = }\space \pm 70 \\ \text{On taking positive sign,}\\ \text{50 – 10k =} \text{ 70}\\ \Rarr \text{k = – 2}\\ \text{On taking positive sign,}\\ \text{50 – 10k = -70}\\ \Rarr \text{10k = 120}\\ \Rarr \text{k = 12.}\\ \text{Thus, k = -2, 12.}$$

$$ \textbf{Q. Examine the consistency of the system of equations x + 2y = 2 and 2x + 3y = 3. } $$

$$\textbf{Ans. } \text{The given system of equations can be represented in the matrix form as} \\ AX = B \\ \text{where} \space \space A = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}, X = \begin{bmatrix} x \\ y \end{bmatrix} \text{and}\space B = \begin{bmatrix} 2 \\ 3 \end{bmatrix} \\ \text{Now,} \space \space |A| = \begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix} \\ = 1×3 – 2×2 \\ = -1 \not = 0 \\ \therefore \text{A is a non-singular matrix.} \\ \text{As a result, its inverse can be calculated.} \\ \text{So, given system of equations is consistent.}$$