Question

ANswer Fast.....(12MathS) ..​

Answer 1

Step-by-step explanation:

$\sf \star \: Question \:10 \: \star \\ \\ \sf A = \left [ \begin {array} {cc} -1 & 5 \\ -3 & 2 \end{array} \right ] \\ \\ \sf |A| = - 2 + 15 = 13 \\ \\ \sf {C}_{11} = 2 \\ \sf {C}_{12} = 3 \\ \sf {C}_{21} = - 5 \\ \sf {C}_{22} = 1 \\ \\ \sf C = \left [ \begin {array} {cc} 2&3 \\ - 5 & 1 \end{array} \right] \\ \\ \sf {C}^{T} = \left [ \begin {array}{cc} 2&-5 \\ 3 & 1 \end{array} \right] \\ \\ \sf {A}^{-1} = \frac{1}{|A|} × {C}^{T} \\ \\ \sf {A}^{-1} = \frac{1}{13} \left [ \begin {array} {cc} 2&-5 \\ 3 & 1 \end{array} \right]$

Answer 2

$a = ( \frac{ - 1}{ - 3} \frac{5}{2} \: )$

$|A| = - 2 + 15 = 13$

$c11 = 2$

$c12 = 3$

$c21 = - 5$

$c22 = 1$

$c = ( \frac{ 2}{-5} \frac{3}{1} )$

$ct \: = (\frac{2}{3} \frac{ - 5}{1} )$

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