Question

$find \: the \: inverse$​

Answer 1

Answer:

$\boxed{ \begin{array}{cc} \bf \: \to \: given: \\ \\ A=\left[ \begin{array} {ccc}1&3&2\\ 6& - 8&5 \\ 9&7& - 4\end{array}\right] \\ \\ \\ \pink{ \sf \: we \: have \: to \: find : } \\ \\ \: \sf \: inverse \: of \: the \: \: matrix \: = ( {A}^{ - 1}) \\ \\ \red{ \underline{ \bf \: Solution}} \\ \\ \sf \: we \: know \: that : \\ \\ \sf \: Inverse \: of \: a \: matrix \: exists \\ \sf \: if \: the \: matrix \: is \: non \: singular. \\ \sf \: that \: is \: mean \: its \: diterminant\:\:|A|\: \neq \: 0 \\ \\ \bf \: now \\ \\ |A|=\left|\begin{array} {ccc}1&3&2\\ 6& - 8&5 \\ 9& 7& - 4\end{array}\right| \\ \\ = \small{1 \{( - 8) \times ( - 4) - 7 \times 5 \} - 3 \{6 \times ( - 4) - 9 \times 5 \} + 2 \{ 6 \times 7 - 9 \times ( - 8)\}}\\ \\ =1(32-35)-3( - 24-45)+2(42 + 72) \\ \\ =1 \times( - 3) + 3 \times 69 + \times 2 \times 114 \\ \\ = - 3 + 207 + 228\\ \\ =432\\ \\ \therefore \: |A| \neq \: 0 \\ \\ \bf \: so \: inverse \: of \: the \: matrix \: \: is \: exist \: \\ \\ \\ \bf \: we \: know \: that \\ \\ \pink{ \boxed{{A}^{ -1 } = \frac{1}{ |A| } \times Adj(A) }}\\ \\ \end{array}}$

Now we have to find Adj(A)

→ To find Adjoint of matrix A, we have to find co - factors.

$\boxed{ \begin{array}{cc}\bf \: We \: know \: that, \: \\ \\ \sf \: \red{\boxed{ \orange{ \bf \: { c_{ij} \: = \: {( - 1)}^{i \: + \: j} \: m_{ij} \: }}}} \\ \\ \sf \: Now, \\ \\ \rm \:c_{11} = {( - 1)}^{1 + 1}\begin{array}{|cc|}\sf - 8 &\sf 5\\ \sf 7&\sf - 4\\\end{array} = ( 32 - 35) = - 3\\ \\ \rm \:c_{12} = {( - 1)}^{1 + 2}\begin{array}{|cc|}\sf 6 &\sf 5 \\ \sf 9&\sf - 4 \\\end{array} = - ( - 2 4 - 45) = 69\rm \\ \\ \:c_{13} = {( - 1)}^{1 + 3}\begin{array}{|cc|}\sf 6 &\sf - 8 \\ \sf 9 &\sf 7\\\end{array} = (42 + 72) =114 \\ \end{array}}$

$\boxed{ \begin{array}{cc}\rm\:c_{21} = {( - 1)}^{2 + 1}\begin{array}{|cc|}\sf 3 &\sf 2\\ \sf 7&\sf - 4\\\end{array} = - ( - 12 - 14) = 26 \\ \\ \rm \: c_{22} = {( - 1)}^{2 + 2}\begin{array}{|cc|}\sf 1&\sf 2 \\ \sf 9&\sf - 4 \\\end{array} = ( - 4 - 18) = - 22 \\ \\ \rm \:c_{23} = {( - 1)}^{2 + 3}\begin{array}{|cc|}\sf 1&\sf 3 \\ \sf 9 &\sf 7\\\end{array} = - (7 - 27) = 20 \\ \end{array}}$

$\boxed{ \begin{array}{cc} \rm\:c_{31} = {( - 1)}^{3 + 1}\begin{array}{|cc|}\sf 3&\sf 2 \\ \sf - 8 &\sf 5 \\\end{array} =(15 + 16 ) = 31 \\ \\ \rm \:c_{32} = {( - 1)}^{3 + 2}\begin{array}{|cc|}\sf 1 &\sf 2\\ \sf 6 &\sf 5 \\\end{array} = - (5 - 12) = 7 \\ \\ \rm \:c_{33} = {( - 1)}^{3 + 3}\begin{array}{|cc|}\sf 1&\sf 3 \\ \sf 6 &\sf - 8\\\end{array} =( - 8 - 18) = - 26 \\ \end{array}}$

$\boxed{ \begin{array}{cc} \bf \: So, \rm \:\:adj(A) = \begin{gathered}\sf\left[\begin{array}{ccc} c_{11} & c_{12}&c_{13}\\ c_{21}& c_{22}& c_{23}\\ c_{31}&c_{32}& c_{33}\end{array}\right]\end{gathered} ' \\ \\ \rm \: \implies\:adj(A) = \begin{gathered}\sf\left[\begin{array}{ccc} - 3& 69&114\\ 26& - 22 &20 \\ 31& 7& - 26\end{array}\right]\end{gathered} ' \\ \\ \rm \: \implies\:adj(A )= \begin{gathered}\sf\left[\begin{array}{ccc} - 3& 26& 31 \\ 69& - 22& 7\\114& 20& -2 6\end{array}\right]\end{gathered} \end{array}}$

${\boxed{\boxed{\begin{array}{cc}\bf \: now \\ \\ {A}^{ - 1} = \frac{1}{ |A| } \times Adj(A) \\ \\ = \frac{1}{432}\begin{gathered}\sf\left[\begin{array}{ccc} - 3& 26& 31 \\ 69& - 22& 7\\114& 20& -2 6\end{array}\right]\end{gathered} \end{array}}}}$

Answer 2

Explanation:

Anaylatic solution

Writing the transpose of the matrix of cofactors, known as an adjugate matrix, can also be an efficient way to calculate the inverse of small matrices, but this recursive method is inefficient for large matrices. To determine the inverse, we calculate a matrix of cofactors:

${\displaystyle \mathbf {A} ^{-1}={1 \over {\begin{vmatrix}\mathbf {A} \end{vmatrix}}}\mathbf {C} ^{\mathrm {T} }={1 \over {\begin{vmatrix}\mathbf {A} \end{vmatrix}}}{\begin{pmatrix}\mathbf {C} _{11}&\mathbf {C} _{21}&\cdots &\mathbf {C} _{n1}\\\mathbf {C} _{12}&\mathbf {C} _{22}&\cdots &\mathbf {C} _{n2}\\\vdots &\vdots &\ddots &\vdots \\\mathbf {C} _{1n}&\mathbf {C} _{2n}&\cdots &\mathbf {C} _{nn}\\\end{pmatrix}}}{\displaystyle \mathbf {A} ^{-1}={1 \over {\begin{vmatrix}\mathbf {A} \end{vmatrix}}}\mathbf {C} ^{\mathrm {T} }={1 \over {\begin{vmatrix}\mathbf {A} \end{vmatrix}}}{\begin{pmatrix}\mathbf {C} _{11}&\mathbf {C} _{21}&\cdots &\mathbf {C} _{n1}\\\mathbf {C} _{12}&\mathbf {C} _{22}&\cdots &\mathbf {C} _{n2}\\\vdots &\vdots &\ddots &\vdots \\\mathbf {C} _{1n}&\mathbf {C} _{2n}&\cdots &\mathbf {C} _{nn}\\\end{pmatrix}}}$

so that

${\displaystyle \left(\mathbf {A} ^{-1}\right)_{ij}={1 \over {\begin{vmatrix}\mathbf {A} \end{vmatrix}}}\left(\mathbf {C} ^{\mathrm {T} }\right)_{ij}={1 \over {\begin{vmatrix}\mathbf {A} \end{vmatrix}}}\left(\mathbf {C} _{ji}\right)}{\displaystyle \left(\mathbf {A} ^{-1}\right)_{ij}={1 \over {\begin{vmatrix}\mathbf {A} \end{vmatrix}}}\left(\mathbf {C} ^{\mathrm {T} }\right)_{ij}={1 \over {\begin{vmatrix}\mathbf {A} \end{vmatrix}}}\left(\mathbf {C} _{ji}\right)}$

where |A| is the determinant of A, C is the matrix of cofactors, and CT represents the matrix transpose

Now solution

${\displaystyle A={\begin{bmatrix}2&-5\\-1&3\end{bmatrix}}}</p><p>{\displaystyle A^{-1}={\frac {1}{|A|}}\mathrm {adj} {\begin{pmatrix}{\begin{bmatrix}2&-5\\-1&3\end{bmatrix}}\end{pmatrix}}={\frac {1}{|A|}}\mathrm {cof} {\begin{pmatrix}{\begin{bmatrix}2&-5\\-1&3\end{bmatrix}}\end{pmatrix}}^{\mathrm {T} }}</p><p>{\displaystyle ={\frac {1}{6-5}}{\begin{bmatrix}3&1\\5&2\end{bmatrix}}^{\mathrm {T} }={\begin{bmatrix}3&5\\1&2\end{bmatrix}}}$

This is indeed the inverse of A, as

${\displaystyle {\begin{bmatrix}2&-5\\-1&3\end{bmatrix}}{\begin{bmatrix}3&5\\1&2\end{bmatrix}}={\begin{bmatrix}1&0\\0&1\end{bmatrix}}}$