If A=
1
3
2
1
Then show that A2-2A-5I = 0 where 0 is the 2×2 null matrix and I be the identity matrix of order 2.
If A=
1
3
2
1
1
3
2
1
Then show that A2-2A-5I = 0 where 0 is the 2×2 null matrix and I be the identity matrix of order 2.
Solution
A × A =
1
3
2
1
×
1
3
2
1
A² =
1×1+2×3
3×1+1×3
1×2+2×1
3×2+1×1
A² =
1+6
3+3
2+2
6+1
∴ A² =
7
6
4
7
1
3
2
1
×
1
3
2
1
A² =
1×1+2×3
3×1+1×3
1×2+2×1
3×2+1×1
A² =
1+6
3+3
2+2
6+1
∴ A² =
7
6
4
7
2A = 2
1
3
2
1
2A =
2 × 1
2 × 3
2 × 2
2 × 1
2A =
2
6
4
2
1
3
2
1
2A =
2 × 1
2 × 3
2 × 2
2 × 1
2A =
2
6
4
2
5I = 5
1
0
0
1
5I =
5 × 1
5 × 0
5 × 0
5 × 1
5I =
5
0
0
5
1
0
0
1
5I =
5 × 1
5 × 0
5 × 0
5 × 1
5I =
5
0
0
5
A² -2A-5I =
7
6
4
7
2
6
4
2
5
0
0
5
A² -2A-5I =
7-2-5
6-6-0
4-4-0
7-2-5
A² -2A-5I =
0
0
0
0
7
6
4
7
2
6
4
2
5
0
0
5
A² -2A-5I =
7-2-5
6-6-0
4-4-0
7-2-5
A² -2A-5I =
0
0
0
0
A2 – 2A – 5I = 0 proved.
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