Question 3, Exercise 2.1
Solutions of Question 3 of Exercise 2.1 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 3(i)
If A=[xyz], B=[ahghbfgfc] and C=[xyz]. Verify that (AB)C=A(BC).
Solution
Given: A=[xyz], B=[ahghbfgfc]
and C=[xyz].
We have to prove that
(AB)C=A(BC)
First, we take
AB=[xyz][ahghbfgfc]=[ax+hy+gzhx+by+fzgx+fy+cz]=[ax+hy+gzhx+by+fzgx+fy+cz][xyz]=[x(ax+hy+gz)+y(hx+by+fz)+z(gx+fy+cz)]=[ax2+hxy+gxz+hxy+by2+fyz+gxz+fyz+cz2]=[ax2+2hxy+2gxz+by2+2fyz+cz2]…(1)
Now we take
BC=[ahghbfgfc][xyz]=[ax+hy+gzhx+by+fzgx+fy+cz]
R.H.S=A(BC)=[xyz][ax+hy+gzhx+by+fzgx+fy+cz]=[x(ax+hy+gz)+y(hx+by+fz)+z(gx+fy+cz)]=[ax2+hxy+gxz+hxy+by2+fyz+gxz+fyz+cz2]=[ax2+2hxy+2gxz+by2+2fyz+cz2]…(2)
From (1) and (2), we have
(AB)C=A(BC).
Question 3(ii)(a)
If A=[13−14], B=[21−3042] and C=[3−15210]. Verify that A(B+C)=AB+AC.
Solution
Given: A=[13−14], B=[21−3042] and C=[3−15210].
Now
B+C=[21−3042]+[3−15210]=[2+31−1−3+50+24+12+0]=[502252]
This gives
A(B+C)=[13−14][502252]=[5+60+152+6−5+80+20−2+8]⟹A(B+C)=[111583206]...(1)
Now, we take
AB=[13−14][21−3042]=[2−01+12−3+6−2+0−1+163+8]=[2133−21511].
and
AC=[13−14][3−15210]=[3+6−1+35+0−3+81+4−5+0]=[92555−5]
Now we take
AB+AC=[2133−21511]+[92555−5]=[2+913+23+5−2+515+511−5]⟹AB+AC=[111583206]...(2)
From (1) and (2), we have
A(B+C)=AB+AC.
Question 3(ii)(b)
If A=[13−14], B=[21−3042] and C=[3−15210]. Verify that A(B−C)=AB−AC.
Solution
Given: A=[13−14], B=[21−3042] and
C=[3−15210].
Now
B−C=[21−3042]−[3−15210]=[2−31+1−3−50−24−12−0]⟹B−C=[−12−8−232]
Now
A(B−C)=[13−14][−12−8−232]=[−1−62+9−8+61−8−2+128+8]⟹A(B−C)=[−711−2−71016]...(1)
Now we take
AB=[13−14][21−3042]=[2+01+12−3+6−2+0−1+163+8]=[2133−21511]
Now
AC=[13−14][3−15210]=[3+6−1+35+03+81+4−5+0]=[92555−5]
AB−AC=[2133−21511]−[92555−5]=[2−913−23−5−2−515−511+5]⟹AB−AC=[−711−2−71016]...(2)
From (1) and (2), we have
A(B−C)=AB−BC.
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