Question

Find the inverse of the matrix ( if exists ) and state the reason if it doesn't exists.

$\left[ \begin{array} {ccc}2&3&4 \\ 4&3&1 \\ 1&2&1\end{array}\right]$
​

Answer 1

Step-by-step explanation:

Given:$\left[ \begin{array} {ccc}2&3&4 \\ 4&3&1 \\ 1&2&1\end{array}\right]$

To find: Inverse of matrix (if exists)

Solution:Inverse of a matrix exists if given matrix is non-singular.

Step 1: Find the determinant of given matrix

$|A|=\left|\begin{array} {ccc}2&3&4 \\ 4&3&1 \\ 1&2&1\end{array}\right|$

|A|=2(3-2)-3(4-1)+4(8-3)

|A|=2×1-3×3+4×5

|A|=2-9+20

|A|=13

|A|≠0

Thus,

Inverse of matrix exist.

Step 2: Write matrix of Minors and co-factor matrix

$M=\left[ \begin{array} {ccc}1&3&5 \\ -5&-2&1\\ -9&-14&-6\end{array}\right]$

$C=\left[ \begin{array} {ccc}1&-3&5 \\ 5&-2&-1 \\ -9&14&-6\end{array}\right]$

Step 3:Write adjoint matrix by writing transpose of C.

$A_{adj}=\left[ \begin{array} {ccc}1&5&-9 \\ -3&-2&14 \\ 5&-1&-6\end{array}\right]$

Step 4: Write inverse of A

$A^{-1}=\frac{1}{|A|}A_{adj}$

$A^{-1}=\frac{1}{13}\left[ \begin{array} {ccc}1&5&-9 \\ -3&-2&14 \\ 5&-1&-6\end{array}\right]$

Final answer:

Inverse of matrix is

$\frac{1}{13}\left[ \begin{array} {ccc}1&5&-9 \\ -3&-2&14 \\ 5&-1&-6\end{array}\right]$

Hope it helps you.

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Answer 2

[ Note: Kindly see this answer from brainly app. Cause, some latex may be not visible in website. ]

$\boxed{ \begin{array}{cc} \bf \: \to \: let: \\ \\ A=\left[ \begin{array} {ccc}2&3&4 \\ 4&3&1 \\ 1&2&1\end{array}\right] \\ \\ \sf \: we \: have \: to \: find : \: \sf \: inverse \: of \: matrix \: ( {A}^{ - 1}) \\ \\ \red{ \underline{ \bf \: Solution}} \\ \\ \sf \: we \: know \: that : \\ \\ \sf \: Inverse \: of \: a \: matrix \: exists \\ \sf \: if \: the \: matrix \: is \: non \: singular. \\ \sf \: that \: is \: mean \: its \: diterminant\:\:|A|\: \neq \: 0 \\ \\ \bf \: now \\ \\ |A|=\left|\begin{array} {ccc}2&3&4 \\ 4&3&1 \\ 1&2&1\end{array}\right| \\ \\ =2(3-2)-3(4-1)+4(8-3) \\ \\ =2 \times 1-3 \times 3+4 \times 5 \\ \\ =2-9+20 \\ \\ =13\\ \\ \therefore \: |A| \neq \: 0 \\ \\ \bf \: so \: inverse \: of \: the \: matrix \: \: is \: exist \: \\ \\ \\ \bf \: we \: know \: that \\ \\ \pink{ \boxed{{A}^{ -1 } = \frac{1}{ |A| } \times Adj(A) }}\\ \\ \end{array}}$

Now we have to find Adj(A)

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

$\boxed{ \begin{array}{cc}\bf \: We \: know, \: \\ \\ \sf \: \boxed{ \bf \: { c_{ij} \: = \: {( - 1)}^{i \: + \: j} \: m_{ij} \: }} \\ \\ \sf \: Now, \\ \\ \rm \:c_{11} = {( - 1)}^{1 + 1}\begin{array}{|cc|}\sf 3 &\sf 1 \\ \sf 2&\sf 1\\\end{array} = ( 3 - 2) = 1 \\ \\ \rm \:c_{12} = {( - 1)}^{1 + 2}\begin{array}{|cc|}\sf 4 &\sf 1 \\ \sf 1 &\sf 1 \\\end{array} = - ( 4 - 1) = - 3\rm \\ \\ \:c_{13} = {( - 1)}^{1 + 3}\begin{array}{|cc|}\sf 4 &\sf 3 \\ \sf 1 &\sf 2\\\end{array} = (8 - 3) =5 \\ \\ \rm\:c_{21} = {( - 1)}^{2 + 1}\begin{array}{|cc|}\sf 3 &\sf 4\\ \sf 2 &\sf 1\\\end{array} = - ( 3 - 8) = 5 \\ \\ \rm \: c_{22} = {( - 1)}^{2 + 2}\begin{array}{|cc|}\sf 2&\sf 4 \\ \sf 1&\sf 1 \\\end{array} = ( 2 - 4) = - 2 \\ \\ \rm \:c_{23} = {( - 1)}^{2 + 3}\begin{array}{|cc|}\sf 2&\sf 3 \\ \sf 1 &\sf 2\\\end{array} = - (4 - 3) = - 1 \\ \\ \rm\:c_{31} = {( - 1)}^{3 + 1}\begin{array}{|cc|}\sf 3&\sf 4 \\ \sf3 &\sf 1 \\\end{array} =(3 - 12) = - 9 \\ \\ \rm \:c_{32} = {( - 1)}^{3 + 2}\begin{array}{|cc|}\sf 2 &\sf 4\\ \sf 4 &\sf 1 \\\end{array} = - (2 - 16) = 14 \\ \\ \rm \:c_{33} = {( - 1)}^{3 + 3}\begin{array}{|cc|}\sf 2&\sf 3 \\ \sf4 &\sf 3\\\end{array} =(6 - 12) = - 6 \\ \\ \bf \: So, \\ \\ \rm \: \longmapsto\:adj(A) = \begin{gathered}\sf\left[\begin{array}{ccc} - 1& - 3&5\\ 5& - 2& - 1\\ - 9& 14& - 6\end{array}\right]\end{gathered} ' \\ \\ \rm \: \longmapsto\:adj(A )= \begin{gathered}\sf\left[\begin{array}{ccc} 1& 5& - 9\\ - 3& - 2& 14\\5& - 1& - 6\end{array}\right]\end{gathered} \end{array}}$

${\boxed{\boxed{\begin{array}{cc}\bf \: now \\ \\ {A}^{ - 1} = \frac{1}{ |A| } \times Adj(A) \\ \\ = \frac{1}{13}\sf\left[\begin{array}{ccc} 1& 5& - 9\\ - 3& - 2& 14\\5& - 1& - 6\end{array}\right] \end{array}}}}$