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If A = 1 -1 2 …

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If A = 1 -1 2 and B = 2 2 -4

2 3 4 -4 2 -4

0 1 2 2 -1 5

are two square matrix, find AB and hence solve the system of linear equation:

x - y = 3; 2x + 3y + 4z = 17; y + 2z = 7

Posted by Prabhnoor Singh 7 years, 3 months ago

CBSE > Class 12 > Mathematics

2 answers

Naveen Sharma 7 years, 3 months ago

From 1

We can say that

$X = \frac{1}{6} B C$

=> $X = \frac{1}{6} (\begin{matrix} 2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5 \end{matrix}) (\begin{matrix} 3 \\ 17 \\ 7 \end{matrix})$

=> $X = \frac{1}{6} (\begin{matrix} 6 + 34 - 28 \\ - 12 + 34 - 28 \\ 6 - 17 + 35 \end{matrix})$

=> $X = \frac{1}{6} (\begin{matrix} 12 \\ - 6 \\ 24 \end{matrix})$

=> $X = (\begin{matrix} 2 \\ - 1 \\ 4 \end{matrix})$

X = 2

y = -1

z = 4

Naveen Sharma 7 years, 3 months ago

Ans.

$A = (\begin{matrix} 1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{matrix}), B = (\begin{matrix} 2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5 \end{matrix})$

=> $A B = (\begin{matrix} (2 + 4 + 0) & (2 - 2 - 0) & (- 4 + 4 + 0) \\ (4 - 12 + 8) & (4 + 6 - 4) & (- 8 - 12 + 20) \\ (0 - 4 + 4) & (0 + 2 - 2) & (0 - 4 + 10) \end{matrix})$

=> $A B = (\begin{matrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{matrix})$

=> $A B = 6 I W h e r e I i s i d e n t i t y m a t r i x$

=> $A^{- 1} = \frac{1}{6} B . . . . . . . . . . . (1)$

x - y = 3

2x + 3y + 4z = 17

y + 2z = 7

We can represent these eqution In Matrix

$A = (\begin{matrix} 1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{matrix})$ , $X = (\begin{matrix} x \\ y \\ z \end{matrix})$ , $C = (\begin{matrix} 3 \\ 17 \\ 7 \end{matrix})$

=> $A X = C = > X = A^{- 1} C$

From 1

=>

ANSWER

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