In order to find [tex]\text{adjoint}[/tex] of any [tex]2 \times 2[/tex] matrix, we can use a small trick of switching the elements of both the diagonals. The alternate diagonal elements will change their sign right when we are switching. It is appearing in attachments.
To find Determinant of a [tex]2 \times 2[/tex] matrix. Supposed there is a matrix [tex]x = \left[\begin{array}{cc}a&b\\c&d\end{array}\right][/tex] , [tex]|x|[/tex] sums equal to [tex](a)(d) - (b)(c)[/tex].
Whenever we multiply a matrix, matrix on the left side of [tex]\times[/tex] donates the [tex]i[/tex] (number of rows) and the matrix on right side will donate [tex]j[/tex] (number of columns) to their product. For example, resulting matrix of [tex]a_{ij} \times b_{xy}[/tex], will appear like [tex]ab_{iy}[/tex]. This is also why in this question [tex]C[/tex] was equal to [tex]A^{ -1}B[/tex] not [tex]B A^{-1}[/tex].
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Answers & Comments
Answer :
[tex]{\mathbf{C = \left[\begin{array}{cc}1&-8\\1&12\end{array}\right]}}[/tex]
Step-by-step explanation :
Alright mate, so here we start this question with. . .
What we are given :
[tex]A = \left[\begin{array}{cc}5&-3\\-3&2\end{array}\right][/tex] ,
[tex]B = \left[\begin{array}{cc}3&-4\\-1&0\end{array}\right][/tex]
& [tex]AC = B[/tex]
What are we supposed to find :
[tex]C = \: \:?[/tex]
Alright so here we start!
Solution :
[tex]\because AC = B[/tex]
[tex]\implies C = A^{-1} B[/tex]
In order to find matrix [tex]C[/tex], we'll first need to deal with [tex]A^{-1}[/tex]'s value.
To find [tex]A^{-1}[/tex] we may want to find [tex]adj A[/tex] and [tex]|A|[/tex].
[tex]adj A = \left[\begin{array}{cc}2&3\\3&5\end{array}\right][/tex]
now [tex]|A| = (5)(2) - (-3)(-3)[/tex]
[tex]\implies |A| = 1[/tex]
As we know,
[tex]A^{-1} = \dfrac{adjA}{|A|}[/tex]
[tex]A^{-1} = \dfrac{\left[\begin{array}{cc}2&3\\3&5\end{array}\right] }{1}[/tex]
[tex]\implies A^{-1} = \left[\begin{array}{cc}2&3\\3&5\end{array}\right][/tex]
Moving back to the equation :
[tex]C = A^{-1} B[/tex]
[tex]\implies C = \left[\begin{array}{cc}2&3\\3&5\end{array}\right] \times \left[\begin{array}{cc}2&-4\\-1&0\end{array}\right][/tex]
[tex]\implies C = \left[\begin{array}{cc}(2)(2)+(3)(-1)&(2)(-4)+(3)(0)\\(3)(2) + (5)(-1)&(3)(4)+(5)(0)\end{array}\right][/tex]
[tex]\implies C = \left[\begin{array}{cc}4-3&-8+0\\6-5&12+0\end{array}\right][/tex]
[tex]\implies {\mathbf{C = \left[\begin{array}{cc}1&-8\\1&12\end{array}\right]}}[/tex]
Wew, so this is how we will solve this question.
Key Points to remember!☝
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