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First Fundamental Theorem of Calculus

Without using...

Question

Without using the concept of inverse of a matrix, find the matrix $[\begin{matrix} x & y \\ z & u \end{matrix}]$ such that
$[\begin{matrix} 5 & - 7 \\ - 2 & 3 \end{matrix}] [\begin{matrix} x & y \\ z & u \end{matrix}] = [\begin{matrix} - 16 & - 6 \\ 7 & 2 \end{matrix}]$

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Solution

$Given : [\begin{matrix} 5 & - 7 \\ - 2 & 3 \end{matrix}] [\begin{matrix} x & y \\ z & u \end{matrix}] = [\begin{matrix} - 16 & - 6 \\ 7 & 2 \end{matrix}] \Rightarrow [\begin{matrix} 5 x - 7 z & 5 y - 7 u \\ - 2 x + 3 z & - 2 y + 3 u \end{matrix}] = [\begin{matrix} - 16 & - 6 \\ 7 & 2 \end{matrix}] The corresponding elements of two equal matrices are equal . ∴ 5 x - 7 z = - 16 . . . (1) 5 y - 7 u = - 6 . . . (2) - 2 y + 3 u = 2 \Rightarrow 3 u = 2 + 2 y \Rightarrow u = \frac{2 + 2 y}{3} . . . (3) - 2 x + 3 z = 7 \Rightarrow 3 z = 7 + 2 x \Rightarrow z = \frac{7 + 2 x}{3} . . . (4) Putting the value of z in eq . (1), we get 5 x - 7 (\frac{7 + 2 x}{3}) = - 16 \Rightarrow 5 x - \frac{49 + 14 x}{3} = - 16 \Rightarrow \frac{15 x - 49 - 14 x}{3} = - 16 \Rightarrow x - 49 = - 48 \Rightarrow x = - 48 + 49 ∴ x = 1 Putting the value of x in eq . (4), we get z = \frac{7 + 2 (1)}{3} \Rightarrow z = \frac{9}{3} = 3 Putting the value of u in eq . (2), we get 5 y - 7 (\frac{2 + 2 y}{3}) = - 6 \Rightarrow 5 y - \frac{14 + 14 y}{3} = - 6 \Rightarrow \frac{15 y - 14 - 14 y}{3} = - 6 \Rightarrow y - 14 = - 18 \Rightarrow y = - 18 + 14 ∴ y = - 4 Putting the value of y in eq . (3), we get u = \frac{2 + 2 (- 4)}{3} \Rightarrow u = - 2 ∴ [\begin{matrix} x & y \\ z & u \end{matrix}] = [\begin{matrix} 1 & - 4 \\ 3 & - 2 \end{matrix}]$

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be a matrix such that, inverse of

7 A

[\begin{matrix} - 1 & 2 4 & - 7 \end{matrix}]

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Using elementary transformations, find the inverse of the followng matrix.

$[\begin{matrix} 2 & 3 5 & 7 \end{matrix}]$

Q. Using elementary row transformations, find the inverse of the matrix

A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 3 2 & 5 & 7 - 2 & - 4 & - 5 \end{matrix} ⎤ ⎥ ⎦

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A = [a_{i j}]

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a_{i j} = {\begin{matrix} 0 i f i \neq j i i f i = j \end{matrix}

Statement - 1 : The inverse of A is matrix

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b_{i j} = {\begin{matrix} 0 i f i \neq j \frac{1}{i} i f i = j \end{matrix}

Statement - 2 : The inverse of a diagonal matrix is a scalar matrix.

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