Question 4 and 5, Review Exercise

Solutions of Question 4 and 5 of Review Exercise of Unit 02: Matrices and Determinants. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Without expanding show that $\left|\begin{array}{ccc}a+1 & l & l \\ l & a+1 & l \\ l & l & a+1\end{array}\right|=(a+1+2 l)(a+1-l)^{2}$.

Solution. $$\begin{align*} L.H.S &= \left|\begin{array}{ccc}a+1 & l & l \\ l & a+1 & l \\ l & l & a+1\end{array}\right|\\ &=\left|\begin{array}{ccc}a+1+2l & a+1+2l & a+1+2l \\ l & a+1 & l \\ l & l & a+1\end{array}\right|\quad R_1+R_2+R_3\\ &=(a+1+2l)\left|\begin{array}{ccc}1 & 1 & 1 \\ l & a+1 & l \\ l & l & a+1\end{array}\right|\\ &=(a+1+2l)[1((a+1)^2-l^2)-1(al+l-l^2)+1(l^2-al-l)]\\ &=(a+1+2l)[a^2+2a+1-l^2-al-l+l^2+l^2-al-l]\\ &=(a+1+2l)[a^2+2a+1-2al-2l+l^2]\\ &=(a+1+2l)(a+1-l)^2\\ &=R.H.S. \end{align*}$$

Find the value of $\lambda$ so that the following system has infinite many solutions.
$2 x-3 y+z=1 ; x-2 y+\lambda z=2 ; 3 y+z=-1$

Solution.

< Question 2 & 3