1.Algebra Booster

7.46 Algebra Booster
90. We know that,
|[adj{adj ( A)}]|
= | A| n-
fi
2
( n -1) 16
| A| = | A|
fi (n – 1) 2 = 16
fi (n – 1) = 4
fi n = 5
Ê1 4 4ˆ
91. Given adj ( P) = Á2 Á
1 7˜
˜
Ë1 1 3¯
fi
2
( 1)
1 4 4
|adj ( P )| = 2 1 7 = 4
1 1 3
As we know that,
-
|adj ( P)| = | P| = | P|
3 1 2
Thus, |P| 2 = 4
fi |P| = 2, –2
Êa11 a12 a13ˆ
92. Let P= Áa 21 a22 a ˜
23 , where |P| = 2
Á ˜
Ëa a a ¯
31 32 33
Ê
2 3 4
2 a11 2 a12 2 a ˆ
13
Á
3 4 5
˜
Now, Q= Á2 a21 2 a22 2 a23˜
Á 4 5 6 ˜
Ë2 a 2 a 2 a ¯
fi
31 32 33
2 3 4
a11 a12 a13
3 4 5
21 22 23
4 5 6
a31 a32 a33
2 2 2
| Q| = 2 a 2 a 2 a
2 2 2
a a a
2 3 4
= 2 ◊2 ◊2 2a 2a 2a
2 2 2
2 3 4 2
= 2 ◊2 ◊2 ◊2◊2
= 2
12
a a a
a a a
a a a
11 12 13
21 22 23
2 2 2
a31 a32 a33
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
11 12 13
21 22 23
31 32 33
= 2 12 ◊ 2 = 2 13
93. We know that,
|adj (A)| = |A| n–1
fi |adj (A)| = |A| 2
= 16
fi 1(12 – 12) – a(4 – 6) + 3(4 – 6) = 16
fi 2a – 6 = 16
fi 2a = 16 + 6 = 22
fi a = 1
94. Since the determinant of a skew-symmetric matrix is
zero, so the inverse does not exist
95. We know that,
BB –1 = 1
fi |BB –1 | = |1|
fi |B||B –1 | = |I|
fi
–1 1
| B | =
| B|
Now,
|B –1 AB| = |B –1 ||A||B|
= 1 | AB || | = | A|
| B|
96. We have,
R = (P cos q + Q sin q)
Êcos q 0 ˆ Ê 0 sin qˆ
= Á +
Ë 0 cosq
˜
¯
Á
Ë-sinq
0
˜
¯
Ê cos q sin qˆ
= Á
Ë–sin
q cos q ˜
¯
adj ( R)
Êcos q –sin qˆ
Now, R –1 = = .
| R|
Á
Ësin
q cos q ˜
¯
97. Given B –1 AB = A 2
Now,
B –3 AB 3 = B –2 (B –1 AB)B 2
= B –2 (A 2 )B 2
= B –1 (B –1 (A 2 )B)B
= B –1 (A 4 )B
= (A 4 ) 2 = A 8
98. Given,
I + A + A 2 + A 3 + … + A k = O
fi A –1 (I + A + A 2 + A 3 + … + A k ) = O
fi (A –1 + I + A + A 2 + … + A k–1 ) = O
fi A –1 + (–A k ) = O
fi A –1 = A k
99. Given,
A –2 –A + I = O
fi A –1 (A 2 – A + I) = A –1 O = O
fi (A –1 A 2 – A –1 A + A –1 ) = O
fi (A – I + A –1 ) = O
fi A –1 = I – A
100. We know that,
|adj (A)| = |A| n–1
= |A| 3–1 = |A| 2
fi |adj (A –1 )| = |A –1 | 2
–1 Ê 1 ˆ 1
fi |adj ( A )| = Á =
2
Ë| A| ˜
¯ | A|
1 1
= =
2
| A|
25
101. We have,
|A –1 adj (A)| = |A –1 ||adj (A)|
2