# Question 13, Exercise 2.2

Solutions of Question 13 of Exercise 2.2 of Unit 02: Matrices and Determinants. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Solve for $x,$ $\left| \begin{matrix}x & 2 & 3 \\0 & -1 & 1 \\0 & 4 & 5 \end{matrix} \right|=9$

Given $$\left| \begin{matrix} x & 2 & 3 \\ 0 & -1 & 1 \\ 0 & 4 & 5 \\ \end{matrix} \right|=9$$ $$x(-5-4)-2(0)+3(0)=9$$ $$-9x=9$$ $$x=-1$$

Solve for $x,$ $\left| \begin{matrix}-1 & 0 & 1 \\x^2 & 1 & x \\2 & 3 & 4 \end{matrix} \right|=-6$

Given $$\left| \begin{matrix} -1 & 0 & 1 \\ x^2 & 1 & x \\ 2 & 3 & 4 \\ \end{matrix} \right|=-6$$ $$-1(4-3x)-0+1(3x^2-2)=-6$$ $$3x-4+3x^2-2=-6$$ $$3x-4+3x^2-2+6=0$$ $$3x+3x^2=0$$ $$3x(1+x)=0$$ $$x=0,3\ne 0$$ $$(1+x)=0$$ $$x=-1$$ $$x=0,-1$$

Solve for $x,$ $\left| \begin{matrix}x+2 & 3 & 4 \\2 & x+3 & 4 \\2 & 3 & x+4\end{matrix} \right|=0$

Given $$\left| \begin{matrix} x+2 & 3 & 4 \\ 2 & x+3 & 4 \\ 2 & 3 & x+4 \\ \end{matrix} \right|=0$$ $$(x+2)(x^2+7x+12-12)-3(2x+8-8)+4(6-2x-6)=0$$ $$(x+2)(x^2+7x)-6x-8x=0$$ $$x^3+2x^2+7x^2+14x-14x=0$$ $$x^3+9x^2=0$$ $$x^2(x+9)=0$$ $$x=0,-9$$