Engineering Oasis

Category: Solution of linear systems

How to use the different techniques to solve linear systems? A discussion of Lower upper factorization is presented with full detailed solved problems for 2×2 matrices and 3×3 matrices.

  • 5A- Second solved problem-LU decomposition-3×3 matrix.

    5A- Second solved problem-LU decomposition-3×3 matrix.

    Table Of Contents
    1. The second solved problem- LU Decomposition-3×3 matrix.

    The second solved problem- LU Decomposition-3×3 matrix.

    Step-1- Develop U1 and L1 matrices for the second solved problem-LU decomposition

    This is the second solved problem for Lu decomposition is required to get (x1,x2, x3) values of three linear simultaneous equations. matrix A is 3×3 and matrix b is the (3×1) matrix.

    The first check is conducted for all sub-matrices values A1 A2, and A3, that their values are nonzeros as we can see that A1=1 and A2=+2 and A3=+6.

    The matrix A is invertible and has L and U decomposition.

    To get the Lower Matrix, we write Matrix A as shown in the slide. We use a11 as a pivot; we know that for the lower Matrix, L11=L22=L33=1. We need to find out the value of L21 and L31.
    For L21 it is equal to 3/1, while l31 is equal to 2/1; please check the arrows.

    The elements of L1 are completed. Still, we need to estimate the value of L32, but this depends on the pivot u22 of the matrix U1.

    We proceed to find the matrix u1. The first three elements in the first Row of the U1 Matrix are the same as the first Row of matrix A ( 1 2 4).

    We change the sign of L21 by multiplying by -1* the first Row and adding the value to the second Row. We change the sign of L31 by multiplying by -1* the first Row and adding the value to the third Row. We can see the elements of the matrix U1 as ( 1 2 4, 0 2 2, 0 2 5).

    Second solved problem-LU decomposition

    Step-2-Develop U and L matrices for the second solved problem-LU decomposition

    The next step is finding L32, the last unknown for the lower matrix L. The elements of L1 matrix are (1  0  0, 3 1 0,2 L32 1).

    From the previous slide, we have U1, consider u22 as a pivot divide U32/U22 to get the value of l32, and check the arrow which goes from 2 to 2, the division of 2/2 is equal to 1, which is the value of L32.

    Now we have completed the elements of the lower Matrix. These elements are (1 0 0, 3 1 0, 2 1 1).

    Step-3-Develop a new matrix C.

    We have L and U matrices written as shown. We know that LU=A, for the expression A*x=b, we can rewrite it as (L)*(U)*x=b, and we can write U*x=c.

    The final expression will be LC=b, for l and b are known values; hence we can get the value of matrix C, C=L-1*b, but we need to estimate the inverse of matrix L.

    The matrix C expression.

    Step-4- Get the inverse matrix L-1.

    For the second solved problem-LU decomposition, the determinant value of L=+1.We estimate the cofactor elements and consider the inverse equals adjugate/ determinant.

    The adjugate of the lower Matrix is the cofactor transpose. The cofactor C11 is equal to +1; the cofactor C12 is equal to -3. The cofactor C13 is equal to +1; cofactor C21 is equal to 0.

    The expression for inverse matrix L-1.

    The cofactor C22 is equal to +1; cofactor C23 is equal to -1. The cofactor C31 is equal to 0; cofactor C32 is equal to 0. The cofactor C33 is equal to 1.

    The expression for inverse matrix L-1.

    Step 4-Get the value of the C matrix.

    The inverse of matrix L is a 3×3 matrix with the first Row as ( 1  0 0). The Second Row is (-3 1 0), and the last is (1 -1 1).
    We can estimate the C matrix by multiplying L-1 by b.

    The product of the inverse of L by b will be (3×1) Matrix with the elements (3 4 -6). But U*X=c. We need to estimate the inverse of matrix U to get the column vector X.

    Check the value of matrix C.

    Step 5-Get the (x1, x2, x3) values for the 3 x 3 matrix.

    The inverse of Matrix U is a 3×3 matrix with the first Row as ( 1  -1 -2/3). The Second Row is (0 ½- 1/3), and the last Row is (0 0 1/3). The product of the inverse of U by c will give the value of the column vector X. X1=3, x2=4, x3=-2.

    It is essential to check that the value of the three unknowns is correct; we double-check by multiplying A*x and check that the product is equal to vector b, which can be found valid.

    Check if AX is =B or not.

    We can get the value of the X column vector by multiplying U-1L-1b to get the values of x1,x2, and x3 as shown in the next slide image.

    Alternative method for X value.

    This is the pdf file used for the illustration of this post.

    The next post: Solved problem for x-y-z values by Lu decomposition.

    HELM-Helping Engineers Learn Mathematics.

    This is the Omni calculator for estimating various items of linear algebra -LU Decomposition Calculator.

    This is a link to another –Calculator for matrices.

  • 5- First solved problem-LU decomposition for the 3×3 matrix.

    5- First solved problem-LU decomposition for the 3×3 matrix.

    Table Of Contents
    1. First solved problem-LU decomposition for the 3×3 matrix.

    First solved problem-LU decomposition for the 3×3 matrix.

    For the First solved problem-LU decomposition, check that the given matrix is invertible

    First I will refer to one of the math sites which is, HELM, which is the abbreviation for helping engineers learn mathematics, there is a valuable source HELM, please refer to Item# 30. Introduction to Numerical Methods, article 30.3, from which I have used the information to create this post.

    Start with the first question, do all matrices have an LU decomposition? The answer is sometimes it is not possible to write the matrix in the form of a Lower triangular multiplied by the Upper triangular. Why not?

    An invertible matrix has an LU decomposition provided that all the leading submatrices have non-zero determinants.
    We will move to the second slide, to apply this statement.

    The first solved problem-Lu decomposition.

    We check the submatrices determinants to be nonzeros, as follows: the first submatrix is A1 which is the corner element of the matrix simply a11, it has a value of 3 which is not a zero value. The second submatrix is A2, which is a matrix of 2×2, elements are( 3 1 -6 0) the determinant value is not zero it equals +6.

    The value of sub-matrices should be non-zero to check that the matrix will have Lu decomposition.

    Finally, the entire matrix which is called A3 is a determinant is -6 not zero. Since all the submatrices have fulfilled the requirements, then the given matrix can have LU decomposition.

    The first solved problem-Lu decomposition.

    Step-1-Develop an expression for U1 and L1 matrices.

    We will proceed to estimate L and U matrices for the given 3×3 matrix. For the lower matrix L by definition, we have L11=L22=L33=1, then it is required to find the values of L21 and L31.

    L21=a21/a11 while L31=a31/a11, consider a11 as a pivot. We can see the two arrows going from -6 to 3 will give -3 for L21 and the second arrow from 0  to 3, then  l31= 0/3.

    The next step is to apply Gauss elimination to get the upper matrix U1. Multiply the value of L21*(-1)R1 and add it to the second Row.

    Multiply the value of L31*(-1)R1 and add it to the third Row. We can see the matrix u1 as a 3×3 matrix that has the first row of (3 1 16 ) which is identical to the first row of the original matrix. The second row is ( 0 2 -4). The

    The third row is (0 8 -17).
    The next step is to find L32 which is the last unknown for the lower matrix L.

    How to find U1 and L1 matrix?

    Step-2-Derive the expression for the final U and L matrices.

    From the previous slide, we have U1, consider a22 as a pivot, and divide a32/a22 to get the value of l32, check the arrow which goes from 8 to 2, the division of 8/2 is equal to 4 which is the value of L32.

    Now we have completed the elements of the lower matrix. These elements are ( 1 0 0,  -2 1 0, 0 4 1).

    We proceed to find the final upper matrix, multiply -l32*R2 and add it to R3, then we multiply -4 by the second row of the U1 matrix and add the value to the third row. The final Upper matrix is ( 3 1 6, 0 2 4, 0 0 -1).

    The final value of L and U matrices.

    Step-3-Check LU=A for the 3×3 matrix.

    Now, we would like to check the expression of L*U=A. We will perform row-column operations of the multiplication of L by U matrices.

    We multiply row 1 by column 1 (1 0 0 ) by (3  0 0) will give the value of 3 this is a11.


    We multiply row 1 by column 2 (1 0 0 ) by (1 2 0) will give the value of 1 which is a12. We multiply row 1 by column 3 (1 0 0 ) by (6 -4 -1) will give the value of 6 which is a13. We multiply row 2 by column 1 (-2 1 0 ) by (3  0 0) will give the value of -6 this is a21. We multiply row 2 by column 2 (-2  1  0 ) by (1 2 0) will give the value of 0 which is a22.

    We multiply row 2 by column 3 (-2  1 0 ) by (6 -4 -1) will give the value of -16 which is a23.

    We multiply row 3 by column 1 ( 0 4 1) by (3  0 0) will give the value of 0 this is a31. We multiply row 3 by column 2 ( 0 4 1) by (1 2 0) will give the value of 8 which is a32.

    Finally, we multiply row 3 by column 3 ( 0 4 1) by (6 -4 -1) will give the value of -17 which is a33. The final Upper matrix is ( 3 1 6, 0 2 4, 0 0 -1).

    The matrix A can be set to be the product (L*U). A routine check of the product of L*U will produce the same elements of matrix A to be performed.

    This is the next post: The Second solved problem– LU decomposition for the 3×3 matrix.

    HELM-Helping Engineers Learn Mathematics.

    This is the Omni calculator for estimating various items of linear algebra -LU Decomposition Calculator.

    This is a link to another –Calculator for matrices.

  • 4-Derive the expression for L-U values for-3×3 matrix.

    4-Derive the expression for L-U values for-3×3 matrix.

    Table Of Contents
    1. How to derive the expression for L-U values for-3×3 matrix?

    How to derive the expression for L-U values for-3×3 matrix?

    This is an explanation for the terms for both the upper and lower L-u values for the 3 x 3 matrix.

    The matrix A of 3×3 is represented by(aij) where i is the row number and j is the column number.
    The matrix a 3×3 can be represented by the multiplication of the Lower triangular matrix L( with 3 rows and 3 columns), which is to be multiplied by the upper triangular matrix (3×3) U.

    For the terms of the L matrix, we use the notation aij as well, for that L matrix, the diagonal is =1, and we have three diagonals a11=a22=a33=1.

    While for a12, the element n the first row at the second column =a13=0. Below the diagonal, all remaining elements are nonzeros.

    On the contrary, the upper triangular matrix has U11, U12, and U13, and all the diagonals are non-zero, while all elements below the diagonal represented by U11, U22, and U33 are all zeros.

    Step-1 Equating the product of L-U values for-3×3 matrix by the value of matrix A.

    Since matrix A is the product of LU matrices, we can equate the value of the elements of A with the product of LU for the corresponding items.

    The product of L*u is represented by a matrix 3×3 as we can see in the next slide image. The first row contains (U11, U12, and U13) which are equal to the first row of matrix A, hence we can proceed to the next step.

    How to derive the expression for L-U values?

    Step-2 Get the expression for L-U elements for the upper and lower matrix for 3 x 3 matrix.

    1-Equate the first element of (L*U) decomposition to the first element a11 of the matrix A as follows: U11=a11, similarly we can equate the second and the third element to the corresponding elements of matrix A.
    So U12=a12 and U13=a13.
    2-We will move to the second row of the L*U matrix – to get the expression for L-U values.

    3- We equate L21*U11 to a21, also, L21*U12+U22 to a22, and L21*U13+U23 to a23, we will use the derived values for U11 as a11, U12 as a12 from which we have estimated from step 1.

    How to get the values of the elements of L*U?

    4- We can get values of L21, and L31 concerning values and also the value of U22 as shown in the next slide image.

    How to get the values of the elements of L*U for 3x3 matrix?

    5- We can get values of U23, and L32 for values as shown in the next slide image.

    The value of U23 and L32 for both the upper and lower matrix.

    The product of LU is equated to matrix A. The product of the LU matrix of 3×3 is written in symbols. From the equating of both Lu and A, we can find out the values of U11=a11, U22= a22, and U33=a33.

    To create an upper matrix we need to let L21*U11=0 this product is equal to a21, to get the L21*U11 to be zero, we will subtract this value from a21. The details of the subtraction are shown in the slide image.

    To create an upper matrix we need to let L31*U11=0 this product is equal to a31, to get the L21*U11 to be zero, we will subtract this value from a31. The details of the subtraction are shown in the slide image.

    The details of the subtraction are shown in the slide image.

    L31 value with respect to a31&a11.

    I have written the A matrix in the form aij, the process of subtraction for both a21 and a31 is shown in the next slide image. We can get the value of L21=a21/a11. The value of L31=a31/a11.

    To get the expression for L-U elements with regards to the value of U22, we will subtract (L21*U12+U22) from a22, but we will substitute the value of L21 as a21/a11, while U12=a12,.

    To get the values of U23, we will subtract (L21*U13+U23) from a23, but we will substitute the value of L21 as a21/a11, while U13=a13.

    The values of L21,L31.

    How to find the elements of the lower matrix L1 is shown in the next slide image where we divide a11/a12 and set it as equal to L21 and divide (a11) by a13 and set the division =L31. The equations used for the estimations are shown in the next slide image.

    L & U matrices side by side.

    The next slide image shows how to get the value of L32, It is equal to ((a32-(a31/a11)*a12/a22-a21*(a12/a11). We take U21 as a pivot.

    How to get the values of l32?

    Thus we have come to the end of the method to derive the expression for L-U values for A 3×3 matrix.

    In the next post, we will use a solved problem to apply this method. A discussion on whether all matrices will have Lu decomposition or not.

    This is the pdf file used for the illustration of this post.

    This is the next post which is the First solved problem-LU decomposition for the 3×3 matrix.

    HELM-Helping Engineers Learn Mathematics.

    This is the Omni calculator for estimating various items of linear algebra -LU Decomposition Calculator.

    This is a link to another –Calculator for matrices.


  • 3-How to solve for x-y by L-U Decomposition?

    3-How to solve for x-y by L-U Decomposition?

    Table Of Contents
    1. How to solve for x-y by L-U Decomposition?

    How to solve for x-y by L-U Decomposition?

    Illustration by using the Gauss elimination method for the U matrix formation.

    We have a matrix as A with two rows and two columns as (2 3 3 4), how to develop the upper matrix U for matrix A?

    We use Gauss elimination through elementary row operations to convert the matrix A into an upper matrix and later, use the back substitution to solve for the values of unknowns.

    For matrix A 2×2 which is given as (2 3 3 4) To convert this matrix into an upper matrix, we want this diagonal to be non-zeros and let  a21 be zero.

    Using a11 as a  pivot, divide a21/a11 multiply by the minus sign by R1, and add the result to row R2. The second row  R2 will be changed.

    We will have a new second row as (0- ½). As we have noticed, there was no operation in the first row of matrix A, which is the same in the upper matrix. Element U22, second row /2nd column is equal to (a22-(a12/a11).

    What about the Lower Matrix? We have by definition L11=L22=1 and L12=0. While for L21 it is equal to (a21/a11), considering that a11 is a pivot. Place the result as a new second row, we can get a zero value for U21, the first element in the second row and first column.

    How to find the value of U matrix by Gauss elimination?

    If we want to check whether  L*U  multiplication will lead us to ( 2 3 3 4), we can use row-column multiplication. (1*2+0*0)=2, and (1*3+(0*-1/2)=3. (3/2*2+1*0) =3+0=3 which we will lead us to 3 and (3/2 * 3+1*(-1/2)=4.50-0.50=4. We have a valid L*U  through a valid multiplication. We will use the same matrix A to solve for the values of two unknowns x&y.

    Now the A matrix as the product of L*U is shown by using the matrix symbols.

    Express the matrix A as a product of L*U.

    Write an expression of (L*C)=b as the first step to developing an expression to solve for x-y for two equations-L-U.

    Consider the two simultaneous linear equations, The two equations can be expressed by using matrices as A*X=b, where A is a matrix 2×2 for our example, Matrix X, is a ( 2×1) matrix that contains (X Y), the unknown variables, while C is a matrix (2×1) of value (13 18). we use matric C as an intermediate stage to solve for x-y by L-U Decomposition.


    Since we have expressed that matrix A is a product of(L*U), we can further write a new expression as (L*U)*X=b.
    The values of L& U and B are known, the only unknown is the column vector matrix X.

    A new expression as (U*x)=C, C is a new item that is unknown. Finally, A*x=b can be expressed as(L)*C=b. To solve for a value of C, an inverse matrix of L-1 is to be presented and to be multiplied by the two sides of the given equation. That process will enable us to get a value for matrix C. The modified expression is shown in the next slide image.

    How to solve for x-y by L-U Decomposition?

    Find the inverse of the L matrix as the second step to solve for x-y by L-U decomposition.

    In the lower matrix  L, the element is (1  0, +3/2 1), then its inverse of the lower matrix is L-1, we can estimate the determinant value of l, which is =1-0=1.

    We divide the adjoint matrix by the determinant. We will swap and make a change of sign of (3/2) to (-3/2), and( 0) Will be the same. The final value of L-1 can be expressed as a matrix of 2×2 and is = ( 1 0 -3/2 1).

    Get the value of C as the third step to develop an expression to solve for x-y by L-U decomposition.

    Get the value of C as the third step to develop an expression to solve for x-y by L-U decomposition. The matrix c is the product of (L-1)*(B), since both two values are known we can then get the value of C, multiplying two matrices(2×2) by(2×1) will yield a new matrix with (2×1) with a value of (13 -1.5). The details of the estimation is shown in the next slide image.

    Evaluation of the value of matrix C.

    Get the value of X as the fourth step to solve for x-y by L-U decomposition.

    Get the value of X as the fourth step to solve for x-y f by L-U decomposition. Remember that we have the expression of U*x=C, get an inverse matrix of C which is a new matrix of 2×2, the X=(U-1)*X, Then the final value of X is( 2 3), the matrix x is a (2×1) matrix. The value of x=2 and y=3.

    How to get the final value of x and y?

    Substitute by the values of x and y to check whether x,y values are right or wrong.

    After checking the two given equations we find that the estimated value of x and Y satisfy the two equations. We have x = 2 &  y = 3. In the last step, we would like to check if our solution is correct. 2x+3*y=13 and + 3x + 4 y=18. We introduce x value as=2 and y value is 3, then (2*2+3*3)=13, which will give us the right-hand side as 13.

    While 3*2+4 *3=6+12=18, which is already equal to the right-hand side of the second equation. Both are OK. In the end, we have used L&U the inverse Of L, and the inverse of U to estimate the values for x& y for simultaneous equations.

    Check the final values of x and y.

    This is a link to download the PDF file used for the illustration of this post.

    The next post: step-by-step on how to derive the expression for L-U values for-3×3 matrix.

    HELM-Helping Engineers Learn Mathematics.

    This is the Omni calculator for estimating various items of linear algebra -LU Decomposition Calculator.

    This is a link to another –Calculator for matrices.


  • 2- Easy illustration for LU decomposition for 2×2 matrix.

    2- Easy illustration for LU decomposition for 2×2 matrix.

    Table Of Contents
    1. How to use the LU decomposition method for the 2×2 matrix?

    How to use the LU decomposition method for the 2×2 matrix?

    LU decomposition method for the 2×2 matrix?

    Now we will start with the lower /upper decomposition of LU’s triangulation method.

    The lower matrix property is a matrix. That has a diagonal, which is not zero and the lower corner has a value that is not zero while for the upper corner, each element will be=0.

    On the contrary, the upper matrix is a matrix that has a diagonal of zero and the upper corner is also non-zero, while the lower corner is =zero. Just why this is called a lower matrix it is enclosed by a diagonal and lower bottom While the upper matrix is enclosed by the diagonal and the upper right-hand side.

    Lu decomposition method for 2x2 matrix.

    For the symbols. thus the matrix can be decomposed by using the Lu decomposition method.

    LU decomposition method by using an example of 2×2 matrix.

    For the symbols, we will write A matrix (2 3 3 4, we are saying a11, a12, a21, and a22. We have six unknowns, which are L11,L21,l22,U11, U12 and U22.

    Due the multiplication of the lower by the upper Matrix, we have only four equations.

    According to the LU  triangulation method, the two elements  L11 and L22 will be=1. Hence we can solve the four equations to get the values of the remaining unknowns.

    Since A = the lower multiplication by the upper matrices, we will perform the multiplication of lower and upper metrics and equate the product to the elements of matrix A. For example, if we multiply the first row by the First Column, we will get the value of 2, which we express as  (1*U11+0)=2.

    The multiplication of the  First row by the second column will give us 3. While a22, which is equal to 4, will be equivalent to the multiplication of the second row by the second column,( L21 *U12)+(1*U22)=4. From equation I, we have U11=2.

    From equation II, we have U12=3. Substitute the value of U11 in equation three and get the value of L21 is equal to 3/2.

    Finally, we can get the value of u22 from equation four, after substituting the values of L21 and U12. U22 will be equal to  -1/2. Remaining for us the minus(-) (1/2) is denoted by U22, which is = 4 – 3, which is L21, 2 * 3 = -1/2. or U22=a22-L21*U12.

    How to find lower and upper composition values?

    If we put the Matrix on the left-hand side and the corresponding values after solving the 4 equations, we will find an exciting remark that a11 will go to U11. And a12 will go to U12. As if we are taking these values. From here to there.

    Consider a11 as a pivot divide a21/a11 which is equal to 3/2 will come to constitute a value of L21. This is from the equations which we have solved.

    While multiply a21/a11 by minus 1 will give the value of U21.

    An important remark for L and U

    Does all matrix 2×2 have LU decomposition?


    Not all 2×2 matrix has Lu decomposition, for the case of determinant value=0, LU decomposition can not be considered because it will lead to a U matrix with zero value at the diagonal.

    Lu decomposition for a given matrix

    To estimate the values of elements of the upper Matrix, we have U11=a11=1, and U12 is equal to a22=5, by definition. U12=0. Finally, U22=a12-L21*u12=10*(2*5)=0. U22 cannot equal zero since U is an upper matrix with the diagonal elements being nonzeros. This is the upper matrix for matrix B.

    The upper matrix elements.

    The expression of B=LU is not acceptable.

    The matrix 2x2 will not have Lu decomposition

    This is the pdf file used for the illustration of this post.

    The next post, post 3, Best illustration how to solve for x-y for two equations-L/U Decomposition?

    HELM-Helping Engineers Learn Mathematics.

    This is the Omni calculator for estimating various items of linear algebra -LU Decomposition Calculator.

    This is a link to another –Calculator for matrices.


  • 1- Easy introduction to the Inverse of a Matrix-2×2 matrix.

    1- Easy introduction to the Inverse of a Matrix-2×2 matrix.

    Table Of Contents
    1. The inverse of a Matrix-inverse of a 2×2 matrix.
    2. The possible operations to a matrix.

The inverse of a Matrix-inverse of a 2×2 matrix.

A brief description of the content of the lecture.

The objective of this lecture is to determine the inverse of a matrix by elementary row operation and use it for solving the simultaneous equation by lower/upper factorization or Doolittle’s factorization for the lower /upper factorization.

The content of the lecture. post 1

How to find the inverse of a matrix?

First, let us discuss how to get the inverse of a matrix by using elementary operations, we are using the Augmented matrix on the left-hand side will be written a matrix denoted by A, separated by a line and this is the identity matrix I.

For example, it is 3 by 3 or 2 by 2 or 4 by 4 and it’s written in the capital letter L.

Through elementary operations, we are converting our original matrix A to the identity matrix I, and through the elementary operations the I which was in the augmented matrix on the right-hand side of the Augmented matrix, will be converted into A-1, where A-1 is the inverse of A. Our original matrix, which is the one we are looking for, what are the elementary row operations.

How to create an identity matrix on the left side?

The possible operations to a matrix.

There are three operations one is the swap of rows changing from R1 to R2 changes location for example and multiplying or dividing each element in the row by a constant. The 3rd of operation is replacing rows by adding or subtracting multiple other rows. We can say R1 +2*R2 or R1-R3 where R1 is the first-row donation and R3 is the third row, etc.

The possible operations that can be done to a matrix.

A solved example of how to get the inverse of a matrix of a 2×2 matrix?

For example, if we have a matrix that is 2×2, it’s required to find the inverse of a 2×2 matrix where is A (-3 -2, 7 4), respectively.

The solution that we are going to use is the augmented matrix, we write on the left-hand side our matrix A.
We divide the matrix by a vertical line and we are putting this identity matrix in our case it will be 2 by 2 the diagonal as we can see is (1 and 1) and the other elements are zeros.

We want to convert the left-hand side into (1 1 0 0), how this can be done?

First, to make -3 converted to +1, we multiply R1*1/3, and the result we are going to put it in the place of R1, so this (-3*-1/3) to 1,(-2*-1/3), it will give us (+2/3) and the right-hand-side will be (-1/3,0).

While the second row will be left unchanged because there is no operation, it will move the same elements which are a(7 4 0 1) here.

The next step will be making this 7 = 0 and this can be done by multiplying the first row by(-7) and adding to the second row R2, so (-7*1+7) will become 0, here 2/ 3  -7 +4=-2/3, -7*-1/3 will be 7/3  and the application of 0 will be 0 + R2. It will become here, now we want to make this one equal to one. The next step is to multiply the second row by (-3/2) and place it in our to zero, which is the same this is reciprocal would become 1 and 7/3 RI.(-3/2) it will become (-7/3) -3/2 or the last item which is equal 1.

The Second part of the solved example to get the inverse of a matrix.

The next target is to make the 2/3 =0, this can be done by multiplying the second row which is 1 by -2/3, and adding to 1.
We get (1 0 2 +1 ), and since the second row is unchanged so it will be maintained at ( 0 1 – /2 (-3/2), on the left-hand side we find that the diagonal is=1 and the other diagonal is 0.

This is the inverse we are looking for it is divided and lies in the right-hand portion of this line.

Another way to get the inverse of a matrix is by the determinant.

There is a rule for the matrix A if it is (2×2), we replace this we swap these -3  will go down, while we are going to change the sign of 7, it will become -7, and the(-2) will become the +2.

But again we are going to divide on the determinant which is(-3*4-1*-2), which at the end will become +2 so the inverse of the matrix is called adjoint divided by the determinant, the adjoint is (4 -2, and  (-7 -3/2). The details of the operations can be seen in the next slide image.

The Second part of the solved example to get the inverse of a matrix.

The result, we have obtained by using the determinant is identical to the result obtained by the augmented matrix.

This is the pdf file used for the illustration of this post.

The next post is  Easy Illustration for LU decomposition method for 2×2 matrix.

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This is the Omni calculator for estimating various items of linear algebra -LU Decomposition Calculator.

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