====== 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.

=====Question 13(i)=====
Solve for $x,$ $\left| \begin{matrix}x & 2 & 3  \\0 & -1 & 1  \\0 & 4 & 5 \end{matrix} \right|=9$
====Solution====
Given
$$\left| \begin{matrix}
   x & 2 & 3  \\
   0 & -1 & 1  \\
   0 & 4 & 5  \\
\end{matrix} \right|=9$$
$$x(-5-4)-2(0)+3(0)=9$$
$$-9x=9$$
$$x=-1$$ 


=====Question 13(ii)=====
Solve for $x,$  $\left| \begin{matrix}-1 & 0 & 1  \\x^2 & 1 & x  \\2 & 3 & 4 \end{matrix} \right|=-6$
====Solution====
Given
$$\left| \begin{matrix}
   -1 & 0 & 1  \\
   x^2 & 1 & x  \\
   2 & 3 & 4  \\
\end{matrix} \right|=-6$$
$$-1(4-3x)-0+1(3x^2-2)=-6$$
$$3x-4+3x^2-2=-6$$
$$3x-4+3x^2-2+6=0$$
$$3x+3x^2=0$$
$$3x(1+x)=0$$
$$x=0,3\ne 0$$
$$(1+x)=0$$ 
$$x=-1$$ 
$$x=0,-1$$ 

=====Question 13(iii)=====
Solve for $x,$  $\left| \begin{matrix}x+2 & 3 & 4  \\2 & x+3 & 4  \\2 & 3 & x+4\end{matrix} \right|=0$
====Solution====
Given
$$\left| \begin{matrix}
   x+2 & 3 & 4  \\
   2 & x+3 & 4  \\
   2 & 3 & x+4  \\
\end{matrix} \right|=0$$
$$(x+2)(x^2+7x+12-12)-3(2x+8-8)+4(6-2x-6)=0$$
$$(x+2)(x^2+7x)-6x-8x=0$$
$$x^3+2x^2+7x^2+14x-14x=0$$
$$x^3+9x^2=0$$
$$x^2(x+9)=0$$
$$x=0,-9$$


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