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CBSE Class 12 Math 2018 Solved Paper

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Question : 25 of 29
Marks: +1, -0
If A = (2−3532−411−2)\begin{pmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{pmatrix}, find A−1A^{-1}. Use it to solve the system of equations
2x - 3y + 5z = 11
3x + 2y - 4z = - 5
x + y - 2z = - 3
OR
Using elementary row transformations, find the inverse of the matrix
A = (123257−2−4−5)\begin{pmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \end{pmatrix}
Solution:  
A = (2−3532−411−2)\begin{pmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{pmatrix}
AA−1AA^{-1} = I
(2−3532−411−2)\begin{pmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{pmatrix} A−1A^{-1} = (100010001)\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}
R1R_1 ↔ R3R_3
(11−232−42−35)\begin{pmatrix} 1 & 1 & -2 \\ 3 & 2 & -4 \\ 2 & -3 & 5 \end{pmatrix} A−1A^{-1} = (100010001)\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}
R2R_2 → R2−3R1R_2 - 3R_1 . R3R_3 → R3−2R1R_3 - 2R_1
(11−20−120−59)\begin{pmatrix} 1 & 1 & -2 \\ 0 & -1 & 2 \\ 0 & -5 & 9 \end{pmatrix} A−1A^{-1} = (00101−310−2)\begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & -3 \\ 1 & 0 & -2 \end{pmatrix}
R2R_2 → - R2R_2
(11−20−120−59)\begin{pmatrix} 1 & 1 & -2 \\ 0 & -1 & 2 \\ 0 & -5 & 9 \end{pmatrix} A−1A^{-1} = (0010−1310−2)\begin{pmatrix} 0 & 0 & 1 \\ 0 & -1 & 3 \\ 1 & 0 & -2 \end{pmatrix}
R1R_1 → R1−R2R_1 - R_2 , R3R_3 → R3+5R2R_3 + 5R_2
(10001−200−1)\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & -2 \\ 0 & 0 & -1 \end{pmatrix} A−1A^{-1} = (01−20−131−513)\begin{pmatrix} 0 & 1 & -2 \\ 0 & -1 & 3 \\ 1 & -5 & 13 \end{pmatrix}
R3R_3 → - R3R_3
(10001−200−1)\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & -2 \\ 0 & 0 & -1 \end{pmatrix} A−1A^{-1} = (01−20−13−15−13)\begin{pmatrix} 0 & 1 & -2 \\ 0 & -1 & 3 \\ -1 & 5 & -13 \end{pmatrix}
R2R_2 → R2+2R3R_2+2R_3
(100010001)\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} A−1A^{-1} = (01−2−29−23−15−13)\begin{pmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{pmatrix}
A−1A^{-1} = (01−2−29−23−15−13)\begin{pmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{pmatrix}
Consider,
AX = B where B = [11−53]\begin{bmatrix} 11 \\ -5 \\ 3 \end{bmatrix} and X = [xyz]\begin{bmatrix} x \\ y \\ z \end{bmatrix}
⇒ A−1A^{-1} AX = A−1A^{-1} B
⇒ X = A−1A^{-1} B
⇒ X = (01−2−29−23−15−13)\begin{pmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{pmatrix} [11−53]\begin{bmatrix} 11 \\ -5 \\ 3 \end{bmatrix}
⇒ [11−53]\begin{bmatrix} 11 \\ -5 \\ 3 \end{bmatrix} = [123]\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}
⇒ x = 1 , y = 2 , z = 3
OR
Given A = (123257−2−4−5)\begin{pmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \end{pmatrix}
Consider,
∣123257−2−4−5∣\begin{vmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \end{vmatrix} = 1 (- 25 + 28) - 2 (- 10 + 14) + 3 (- 8 + 10)
= 3 - 8 + 6
= 1 ≠ 0
A−1A^{-1} exist.
A A−1A^{-1} = I
(123257−2−4−5)\begin{pmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \end{pmatrix} A−1A^{-1} = (100010001)\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}
R2R_2 → R2−2R1,R3R_2-2R_1 , R_3 → R3+2R1R_3 + 2R_1
(123011001)\begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix} A−1A^{-1} = (100−210201)\begin{pmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 2 & 0 & 1 \end{pmatrix}
R1R_1 → R1−2R2R_1 - 2R_2
(100010001)\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} A−1A^{-1} = (5−20−210201)\begin{pmatrix} 5 & -2 & 0 \\ -2 & 1 & 0 \\ 2 & 0 & 1 \end{pmatrix}
R1R_1 → R1−R3,R2R_1 - R_3 , R_2 → R2−R3R_2 - R_3
(100010001)\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} A−1A^{-1} = (3−2−1−41−120−1)\begin{pmatrix} 3 & -2 & -1 \\ -4 & 1 & -1 \\ 2 & 0 & -1 \end{pmatrix}
A−1A^{-1} = (3−2−1−41−1201)\begin{pmatrix} 3 & -2 & -1 \\ -4 & 1 & -1 \\ 2 & 0 & 1 \end{pmatrix}
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