1.Algebra Booster

7.80 Algebra Booster
55.
(iii) Let
Êxˆ
Ê1ˆ
X = Áy˜
and B=
Á0˜
Á ˜ Á ˜
Ëz¯
Ë0¯
Thus, A 1
X = B has infinite number of solutions
A 6
X = B has no solution
A 8
X = B has no solution
A 10
X = B has no solution
A 12
X = B has infinite number of solutions
We know that,
det(AdjA) = (detA) n–1
and B is a skew-symmetric matrix, so
det (B) = 0
det (adj B) = det B) 3–1 = (det B) 2 = 0
now,
2k - 1 2 k 2 k
| A| = 2 k 1 -2k
-2 k 2k
-1
2k - 1 2 k 2 k
= 0 1+ 2 k -(1+ 2 k) ( R2Æ R2 + R3)
-2 k 2k
-1
2k - 1 2 k 4 k
= 0 1+ 2k 0 ( C3Æ C3+
C2)
-2 k 2k 2k
-1
= (2k – 1)(4k 2 – 1) + 8k(1 + 2k)
= (2k + 1)[(2k – 1) 2 + 8k]
= (2k + 1)[4k 2 + 4k + 1]
= (2k + 1)(2k + 1) 2
= (2k + 1) 3
It is given that,
det (adj A) + det (adj B) = 10 6
fi (2k + 1) 6 + 0 = 10 6
fi (2k + 1) 6 = 10 6
fi (2k + 1) = 10
fi k = 4.5
Thus, the value of [k] = [4.5] = 4
56. The given system of equations can be written in matrix
form as
AX = B
It has either
(i) a unique solution
or (ii) infinite solutions
or(iii) no solution.
Thus, there can not exist any matrix A such that
Èx˘
È1˘
A
Í
y
˙ Í
0
˙
Í ˙
=
Í ˙
has two distinct solutions.
ÍÎz
˙˚
ÍÎ0
˙˚
58. Given
MN = NM
Now, M 2 N 2 (M T N) –1 (MN –1 ) T
= M 2 N 2 N –1 (M T N) –1 (MN –1 ) T
= M 2 N ◊ (M T ) –1 (N –1 ) T M T
= –M 2 N ◊ (M) –1 (N T ) –1 M T
= +M 2 N ◊ (M) –1 N –1 M T
= –MNMM –1 N –1 M
= –MNN –1 M
= –M 2
Êa b cˆ
59. Let M = Ád Á
e f˜
˜
Ëg h i¯
Êa b cˆÊ0ˆ Ê-1ˆ
Now,
Ád e f˜Á1˜ = Á 2˜
Á ˜Á ˜ Á ˜
Ëg h i¯Ë0¯ Ë 3¯
fi b = –1, e = 2, h = 3
Êa b cˆÊ 1ˆ Ê 1ˆ
Also, Ád e f˜Á- 1˜ = Á 1˜
Á ˜Á ˜ Á ˜
Ëg h i¯Ë 0¯ Ë-1¯
fi a = 0, d = 3, g = 2
Êa b cˆÊ1ˆ Ê 0ˆ
and, Ád e f˜Á1˜ = Á 0˜
Á ˜Á ˜ Á ˜
Ëg h i¯Ë1¯ Ë12¯
fi g + h + i = 12
fi i = 7
Thus, sum of the main diagonal
= a + e + i = 0 + 2 + 7 = 9
60. Given
P T = 2P + I
fi (P T ) T = (2P + I) T
fi P = (2P T + I)
fi P = 2(2P + I) + I
fi 3(P + I) = O
fi (P + I) = O
fi PX + X = O
fi PX = –X
Ê1 4 4ˆ
61. Given adj ( P) = Á2 Á
1 7˜
˜
Ë1 1 3¯
fi
1 4 4
|adj ( P )| = 2 1 7 = 4
1 1 3
As we know that,
|adj(P)| = |P| 3–1 = |P| 2
Thus,
|P| 2 = 4
fi |P| = 2, –2