Test Index

CBSE Class 12 Math 2008 Solved Paper

© examsnet.com
Question : 13 of 29
Marks: +1, -0
Let A = ∣325413067∣\begin{vmatrix} 3 & 2 & 5 \\ 4 & 1 & 3 \\ 0 & 6 & 7 \end{vmatrix}. Express A as sum of two matrices such that one is symmetric and the other is skew symmetric.
OR
If A = ∣122212221∣\begin{vmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{vmatrix}, verify that A2A^2 - 4A - 5I = 0
Solution:  
A = [325413067]\begin{bmatrix} 3 & 2 & 5 \\ 4 & 1 & 3 \\ 0 & 6 & 7 \end{bmatrix}
A' = [340216537]\begin{bmatrix} 3 & 4 & 0 \\ 2 & 1 & 6 \\ 5 & 3 & 7 \end{bmatrix}
Now, A can be written as:
A = 12\frac{1}{2} A + A' + 12\frac{1}{2} A - A'
A + A' = [325413067]+[340216537]\begin{bmatrix} 3 & 2 & 5 \\ 4 & 1 & 3 \\ 0 & 6 & 7 \end{bmatrix} + \begin{bmatrix} 3 & 4 & 0 \\ 2 & 1 & 6 \\ 5 & 3 & 7 \end{bmatrix}
= [3+32+45+04+21+13+60+56+37+7]\begin{bmatrix} 3+3 & 2+4 & 5+0 \\ 4+2 & 1+1 & 3+6 \\ 0+5 & 6+3 & 7+7 \end{bmatrix}
= [6656295914]\begin{bmatrix} 6 & 6 & 5 \\ 6 & 2 & 9 \\ 5 & 9 & 14 \end{bmatrix}
12\frac{1}{2} A + A' = 12[6656295914]\frac{1}{2} \begin{bmatrix} 6 & 6 & 5 \\ 6 & 2 & 9 \\ 5 & 9 & 14 \end{bmatrix} = [3352319252927]\begin{bmatrix} 3 & 3 & \frac{5}{2} \\ 3 & 1 & \frac{9}{2} \\ \frac{5}{2} & \frac{9}{2} & 7 \end{bmatrix} = P, say
Now, P' = [3352319252927]\begin{bmatrix} 3 & 3 & \frac{5}{2} \\ 3 & 1 & \frac{9}{2} \\ \frac{5}{2} & \frac{9}{2} & 7 \end{bmatrix}
Thus, P = 12\frac{1}{2} A + A' is a symmetric matrix.
Now, A - A' = [3−32−45−04−21−13−60−56−37−7]\begin{bmatrix} 3-3 & 2-4 & 5-0 \\ 4-2 & 1-1 & 3-6 \\ 0-5 & 6-3 & 7-7 \end{bmatrix} = [0−2520−3−530]\begin{bmatrix} 0 & -2 & 5 \\ 2 & 0 & -3 \\ -5 & 3 & 0 \end{bmatrix}
12\frac{1}{2} A - A' = [0−15210−32−52320]\begin{bmatrix} 0 & -1 & \frac{5}{2} \\ 1 & 0 & -\frac{3}{2} \\ -\frac{5}{2} & \frac{3}{2} & 0 \end{bmatrix} = Q, say
Now, Q' = [01−52−103252−320]\begin{bmatrix} 0 & 1 & -\frac{5}{2} \\ -1 & 0 & \frac{3}{2} \\ \frac{5}{2} & -\frac{3}{2} & 0 \end{bmatrix} = - [0−15210−32−52320]\begin{bmatrix} 0 & -1 & \frac{5}{2} \\ 1 & 0 & -\frac{3}{2} \\ -\frac{5}{2} & \frac{3}{2} & 0 \end{bmatrix} = - Q
Thus, Q = 12\frac{1}{2} A - A' is a skew symmetric matrix.
∴ A = [325413067]\begin{bmatrix} 3 & 2 & 5 \\ 4 & 1 & 3 \\ 0 & 6 & 7 \end{bmatrix} = [3352319252927]\begin{bmatrix} 3 & 3 & \frac{5}{2} \\ 3 & 1 & \frac{9}{2} \\ \frac{5}{2} & \frac{9}{2} & 7 \end{bmatrix} + [0−15210−32−52320]\begin{bmatrix} 0 & -1 & \frac{5}{2} \\ 1 & 0 & -\frac{3}{2} \\ -\frac{5}{2} & \frac{3}{2} & 0 \end{bmatrix}
OR
A2A^2 = [122212221]\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} [122212221]\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}
=
[1×1+2×2+2×22×1+2×1+2×21×2+2×2+2×12×1+1×2+2×22×2+1×1+2×22×2+1×2+2×12×1+2×2+1×22×2+2×1+1×22×2+2×2+1×1]\begin{bmatrix} 1 \times 1 + 2 \times 2 + 2 \times 2 & 2 \times 1 + 2 \times 1 + 2 \times 2 & 1 \times 2 + 2 \times 2 + 2 \times 1 \\ 2 \times 1 + 1 \times 2 + 2 \times 2 & 2 \times 2 + 1 \times 1 + 2 \times 2 & 2 \times 2 + 1 \times 2 + 2 \times 1 \\ 2 \times 1 + 2 \times 2 + 1 \times 2 & 2 \times 2 + 2 \times 1 + 1 \times 2 & 2 \times 2 + 2 \times 2 + 1 \times 1 \end{bmatrix}
=
[1+4+42+2+42+4+22+2+44+1+44+2+22+4+24+2+24+4+1]\begin{bmatrix} 1+4+4 & 2+2+4 & 2+4+2 \\ 2+2+4 & 4+1+4 & 4+2+2 \\ 2+4+2 & 4+2+2 & 4+4+1 \end{bmatrix}
= [988898889]\begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix}
4A = 4[122212221]4 \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} =
[1×42×42×42×41×42×42×42×41×4]\begin{bmatrix} 1 \times 4 & 2 \times 4 & 2 \times 4 \\ 2 \times 4 & 1 \times 4 & 2 \times 4 \\ 2 \times 4 & 2 \times 4 & 1 \times 4 \end{bmatrix}
= [488848884]\begin{bmatrix} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{bmatrix}
5I = 5 [100010001]\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = [500050005]\begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix}
A2A^2 - 4A - 5I = [988898889]\begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix} - [488848884]\begin{bmatrix} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{bmatrix} - [500050005]\begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix}
=
[9−4−58−8−08−8−08−8−09−4−58−8−08−8−08−8−09−4−5]\begin{bmatrix} 9-4-5 & 8-8-0 & 8-8-0 \\ 8-8-0 & 9-4-5 & 8-8-0 \\ 8-8-0 & 8-8-0 & 9-4-5 \end{bmatrix}
= [000000000]\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = 0 = R.H.S.
© examsnet.com
Go to Question:
Prev Question
Next Question