Exercise 2.4 (Solutions)

The solutions of the Exercise 2.4 of book “Model Textbook of Mathematics for Class XI” published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan are given on this page. This exercise consists of the question related to the determinant and properties of the determinant.

Question. 1 Without expansion show that:

(i) $\left|\begin{array}{lll}9 & 27 & 36 \\ 18 & 54 & 24 \\ 27 & 81 & 28\end{array}\right|=0$

(ii) $\left|\begin{array}{lll}1 / a & b c & b+c \\ 1 / b & a c & a+c \\ 1 / c & a b & a+b\end{array}\right|=0$

(iii) $\left|\begin{array}{lll}0 & -a & -b \\ a & 0 & -c \\ b & c & 0\end{array}\right|=0$

(iv) $\left|\begin{array}{lll}\sin ^{2} \alpha & 1 &\cos ^{2} \alpha\\ \tan ^{2} \alpha & \sec ^{2} \alpha & 1\\ - \csc ^{2} \alpha & -\cot ^{2} \alpha & 1\end{array} \right|=0$
Solution:Question 1(i-iv)

Question. 1 Without expansion show that:
(v) $\left|\begin{array}{lll}(a-b)^{3} & a^{3}-b^{3} & a b(a-b) \\ (c-d)^{3} & c^{3}-d^{3} & c d(c-d) \\ (e-f)^{3} & e^{3}-f^{3} & e f(e-f)\end{array}\right|=0$

(vi) $\left|\begin{array}{lll}x & -z & 0 \\ 0 & y & -x \\ -y & 0 & z\end{array}\right|=0$

(vii) $\left|\begin{array}{lll}(a-b)^{2} & (a+b)^{2} & a b \\ (c-d)^{2} & (c+d)^{2} & c d \\ (e-f)^{2} & (e+f)^{2} & e f\end{array}\right|=0$
Solution:Question 1(v-vii)

Question. 2 Using the properties of the determinants prove the following.
(i) $\left|\begin{array}{lll} x & y & x+y \\ y & x+y & x \\ x+y & x & y \end{array}\right|=-2\left(x^{3}+y^{3}\right)$

(ii) $\left|\begin{array}{lll} a & b-c & b+c \\ a+c & b & c-a \\ a-b & a+b & c \end{array}\right|=(a+b+c)\left(a^{2}+b^{2}+c^{2}\right)$

(iii) $\left|\begin{array}{lll} n a_{1}+b_{1} & n a_{2}+b_{2} & n a_{3}+b_{3} \\ n b_{1}+c_{1} & n b_{2}+c_{2} & n b_{3}+c_{3} \\ n c_{1}+a_{1} & n c_{2}+a_{2} & n c_{3}+a_{3}\end{array}\right|=\left(n^{3}+1\right)\left|\begin{array}{lll} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} &c_{3}\end{array}\right|$

(iv) $\left|\begin{array}{lll} x & x^{2} & 1+\alpha x^{3} \\ y & y^{2} & 1+\alpha y^{3} \\ z & z^{2} & 1+\alpha z^{3} \end{array}\right|=(1+\alpha x y z)(x-y)(y-z)(z-x)$

(v) $\quad\left|\begin{array}{ccc}2 a b & 1+a^{2}-b^{2} & 2 b \\ 2 a & -2 b & 1-a^{2}-b^{2} \\ 1-a^{2}+b^{2} & 2 a b & -2 a\end{array}\right|=\left(1+a^{2}+b^{2}\right)^{3}$

(vi) $\left|\begin{array}{lll} 3 a & 1 & 2 a+1 \\ 2 a+1 & 1 & a+2 \\ 3 & 1 & 2 \end{array}\right|=(a-1)(a-2)$
Solution:Question 2(i-vi)

Question. 1 Without expansion show that:

(vii) $\left|\begin{array}{lll} b+c & a & a \\ b & c+a & b \\ c & c & a+b\end{array}\right|=4 a b c$

(viii) $\left|\begin{array}{lll}-b c & b^{2}+b c & c^{2}+b c \\ a^{2}+a c & -a c & c^{2}+a c \\ a^{2}+a b & b^{2}+a b & -a b\end{array}\right|=(a b+b c+c a)^{2}$

(ix) $\left|\begin{array}{lll}(b+c)^{2} & a b & c a \\ a b & (a+c)^{2} & b c \\ a c & b c & (a+b)^{2}\end{array}\right|=2 a b c(a+b+c)^{3}$

(x) $\left|\begin{array}{lll} b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{array}\right|=2\left|\begin{array}{lll} a & p & x \\ b & q & y \\ c & r & z \end{array}\right|$

(xi) If $|A B|=|A| \cdot|B|$ and $\left|A^{-1}\right|=1 /|A|$ then for a square matrix of order $3 \times 3$ prove that $|a d j A|=|A|^{2}$

(xii) If $\mathrm{A}$ is of order $3 \times 3$ such that $\mid$ adj $j \mid=64$ then find $\left|A^{-1}\right|$.

(xiii) If $a, b, c$ are real numbers and $\left|\begin{array}{lll}b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a\end{array}\right|=0$. Show that either $a+b+c=0$ or $a=b=c$
Solution:Question 2(vii-xiii)