Calculation of inverse of some special matrix with complex number

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Steps to calculate Inverse of matrix  using adjoint method

Step 1: calculating the  Minors of matrix
Step 2: then turn that into its  Cofactors,
Step 3: calculation of adjoint of matrix
Step 4: multiply that by 1/Determinant of the matrix.

Let A= \[\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}\]

Then \[A^{-1}=\frac{adjA}{|A|}\]

Note -For inverse to exist matrix must be non singular . That means determinant of A must not be 0.

Adjoint of a matrix is defined as the transpose of cofactor matrix.

Calculation of minor of matrix

Step 1-It is calculated by deleting the all the elements of row and column corresponding to the element for which minor is determined.

step 2 -then find out the determinant of determinant of remaining sub matrix .The value of determinant is minor of that element.

 Step 3-Step 1 and step 2 is repeated for all the element to get a minor of matrix of order same as that of order of given matrix

For example

Minor of a matrix A= \[\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}\] is given by \[ \begin{bmatrix}a_{22}&a_{21}\\a_{12}&a_{11}\end{bmatrix}\]

Cofactor is \[(-1)^{i+j}\]  where i represents the \[i^{th}\] row and \[j^{th}\] column.

Cofactor of A =\[ \begin{bmatrix}a_{22}& -a_{21}\\-a_{12}&a_{11}\end{bmatrix}\]

Transpose is defined as matrix obtained by changing corresponding rows with columns.If order of matrix A is \[ 2\times 3\] the the order of transpose of matrix will be \[ 3\times 2\]

Adjoint of a matrix is defined as the transpose of cofactor matrix.

transpose of cofactor matrix =\[ \begin{bmatrix}a_{22}& -a_{12}\\-a_{21}&a_{11}\end{bmatrix}\]

\[A^{-1}=\frac{adjA}{|A|}\]

Example -If A=\[\begin{bmatrix}\frac{1}{2}(e^{ix}+ e^{-ix}) &\frac{1}{2}(e^{ix}- e^{-ix}) \\ \frac{1}{2}(e^{ix}- e^{-ix}) & \frac{1}{2}(e^{ix}+ e^{-ix}) \\ \end{bmatrix}\] then \[ A^{-1}\] exists 

(a) for all real x

(b) for positive real x only.

(c) for negative real x only.

(d) none of these.

 Solution-

since \[ cos(x)= \frac{e^{ix}+e^{-ix}}{2}\] and \[ sin(x)= \frac{e^{ix}-e^{-ix}}{2i}\]

Therefore Matrix can be rewritten as \[ \begin{bmatrix} cos(x)& isin(x) \\ isin(x) &cos(x) \\ \end{bmatrix}\]

Determinant of A =\[ cos^2x-i^2sin^2x=cos^2x+sin^2x=1\]

For a given matrix A  if determinant of a matrix is not zero then inverse exists and its unique.

hence Inverse exits for all real x .  

For x=0 its inverse is \[ \begin{bmatrix} 1 & 0 \\0 & 1 \\ \end{bmatrix}\]