Mathematics 9 (Complete Solutions)

Q4
Verify that

If
A =
 
1 2
4 6
 
,
B =
 
3 -1
2 -2
 
, then verify that.

(i) A(Adj A) = (Adj A) A = (∣ A ∣)I
(ii) BB-1 = I = B-1B

Solution:

(i) A(Adj A) = (Adj A) A = (∣ A ∣)I
First we will find A(Adj A)
Adj A =
 
6 -2
-4 1
 
So,
A(Adj A) =
 
1 2
4 6
 
 
6 -2
-4 1
 
A(Adj A) =
 
1(6) + 2(-4) 1(-2) + 2(1)
4(6) + 6(-4) 4(-2) + 6(1)
 
A(Adj A) =
 
6 - 8 -2 + 2
24 - 24 -8 + 6
 
A(Adj A) =
 
-2 0
0 -2
 
    ---------- ( i )
Now we will find (Adj A)A
(Adj A)A =
 
6 -2
-4 1
 
 
1 2
4 6
 
A(Adj A) =
 
6(1) + (-2)(4) 6(2) + (-2)(6)
-4(1) + 1(4) -4(2) + 1(6)
 
A(Adj A) =
 
6 - 8 12 - 12
-4 + 4 -8 + 6
 
A(Adj A) =
 
-2 0
0 -2
 
    ---------- ( ii )
Now we will find (∣ A ∣)I
∣ A ∣ =
 
1 2
4 6
 
= 1(6) - 2(4)
∣ A ∣ = 6 - 8
∣ A ∣ = -2
So,
(∣ A ∣)I = -2
 
1 0
0 1
 
(∣ A ∣)I =
 
-2(1) -2(0)
(-2)(0) (-2)(1)
 
(∣ A ∣)I =
 
-2 0
0 -2
 
    ---------- ( iii )
Hence from equations ( i ), ( ii ) and ( iii ) it is verified that A(Adj A) = (Adj A) A = (∣ A ∣)I

(ii) BB-1 = I = B-1B
First we will find BB-1
B-1 =
Adj B
∣ B ∣
Now we will find ∣ B ∣
∣ B ∣ =
 
3 -1
2 -2
 
= (3)(-2) - (-1)(2)
∣ B ∣ = -6 + 2
∣ B ∣ = -4
Now we will find Adj B
Adj B =
 
-2 1
-2 3
 
So,
B-1 =
 
-2 1
-2 3
 
-4
B-1 =
-
1
4
 
-2 1
-2 3
 
B-1 =
 
-2 (-
1
4
)
1 (-
1
4
)
-2 (-
1
4
)
3 (-
1
4
)
 
B-1 =
 
2
4
-
1
4
2
4
-
3
4
 
B-1 =
 
1
2
-
1
4
1
2
-
3
4
 
So,
BB-1 =
 
3 -1
2 -2
 
 
1
2
-
1
4
1
2
-
3
4
 
BB-1 =
 
3
(
1
2
)
+ (-1)
(
1
2
)
3
( -
1
4
)
+ (-1)
( -
3
4
)
2
(
1
2
)
+ (-2)
(
1
2
)
2
( -
1
4
)
+ (-2)
( -
3
4
)
 
BB-1 =
 
3
2
-
1
2
-
3
4
+
3
4
2
2
-
2
2
-
2
4
+
6
4
 
BB-1 =
 
3
2
-
1
2
-
3
4
+
3
4
1 - 1
-
1
2
+
3
2
 
BB-1 =
 
3 - 1
2
-3 + 3
4
0
3 - 1
2
 
BB-1 =
 
2
2
0
4
0
2
2
 
BB-1 =
 
1 0
0 1
 
You know that I =
 
1 0
0 1
 
Now we will find B-1B
B-1B =
 
1
2
-
1
4
1
2
-
3
4
 
 
3 -1
2 -2
 
BB-1 =
 
1
2
(3) +
( -
1
4
)
(2)
1
2
(-1) +
( -
1
4
)
(-2)
1
2
(3) +
( -
3
4
)
(2)
1
2
(-1) +
( -
3
4
)
(-2)
 
BB-1 =
 
3
2
-
2
4
-
1
2
+
2
4
3
2
-
6
4
-
1
2
+
6
4
 
BB-1 =
 
3
2
-
1
2
-
1
2
+
1
2
3
2
-
3
2
-
1
2
+
3
2
 
BB-1 =
 
3 - 1
2
-1 + 1
2
3 - 3
2
-1 + 3
2
 
BB-1 =
 
2
2
0
2
0
2
2
2
 
BB-1 =
 
1 0
0 1
 
Hence it is verified that BB-1 = I = B-1B

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