Question Video: Solving a System of Three Equation Using Matrices | Nagwa Question Video: Solving a System of Three Equation Using Matrices | Nagwa

Question Video: Solving a System of Three Equation Using Matrices Mathematics • Third Year of Secondary School

Solve the system of linear equations −𝑥 + 𝑦 + 𝑧 = 8, −2𝑥 + 𝑦 − 𝑧 = −5, and 6𝑥 − 3𝑦 = −6 using the inverse of a matrix.

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Video Transcript

Solve the system of linear equations negative 𝑥 plus 𝑦 plus 𝑧 equals eight, negative two 𝑥 plus 𝑦 minus 𝑧 equals negative five, and six 𝑥 minus three 𝑦 equals negative six using the inverse of a matrix.

First, we need to represent our system of equations as a matrix equation. Our matrix equation will be of the form 𝐴𝑋 equals 𝐵, where 𝐴 is the coefficient matrix, 𝑋 is the variable matrix, and 𝐵 is the constant matrix.

Let’s suppose we had the system of linear equations 𝑎 one 𝑥 plus 𝑏 one 𝑦 plus 𝑐 one 𝑧 equals 𝑑 one, 𝑎 two 𝑥 plus 𝑏 two 𝑦 plus 𝑐 two 𝑧 equals 𝑑 two, and 𝑎 three 𝑥 plus 𝑏 three 𝑦 plus 𝑐 three 𝑧 equals 𝑑 three. We can represent this in matrix form by putting the coefficients of the 𝑥-, 𝑦-, and 𝑧-terms in the coefficient matrix. The variable matrix consists of our three variables 𝑥, 𝑦, and 𝑧. And the constant matrix consists of the constants which our three equations are equal to.

We may notice that our third equation doesn’t have a 𝑧-term. This just means that the coefficient of the 𝑧-term is zero, So we can write this in if we want to. Our equation in matrix form is therefore negative one, one, one, negative two, one, negative one, six, negative three, zero multiplied by 𝑥, 𝑦, 𝑧, is equal to eight, negative five, negative six.

Now we have our system in the form 𝐴𝑋 equals 𝐵. We need to use the inverse of a matrix to solve it. Since we’re trying to find 𝑋, we can multiply both sides of the equation by the inverse of 𝐴 on the left. Since 𝐴 inverse multiplied by 𝐴 is just the identity matrix, the left-hand side just becomes 𝑋. We now have that 𝑋 is equal to 𝐴 inverse multiplied by 𝐵.

Our next step is to find the inverse of the matrix 𝐴. Let’s clear some space. We need to find the inverse of our matrix 𝐴. And we can do this using the adjoint method, which is used to find the inverse of three-by-three matrices. There are five steps to finding the inverse of a matrix using the adjoint method. The first of these steps is finding the determinant of the matrix. This will also tell us whether the matrix is invertible. If the determinant is nonzero, then the matrix is invertible. Next, we find the matrix of minors, then the matrix of cofactors, then the adjoint matrix, and finally we multiply the adjoint matrix by the reciprocal of the determinant to find the inverse matrix.

For the first step, we need to find the determinant of 𝐴. We can do this by recalling the formula for finding the determinant of a three-by-three matrix. Now, we can apply this formula to find the determinant of 𝐴. Then, we simplify to find that the determinant of 𝐴 is equal to negative three. Since this is nonzero, we know that our matrix is invertible.

Next, we need to find the matrix of minors . We find each entry of the matrix of minors by crossing out the row and column that the entry is on and then finding the determinant of the remaining two-by-two matrix, as shown here. We need to apply this to our matrix 𝐴. Following this method, we find our matrix of minors. All we need to do now is evaluate the determinants in the matrix. After doing this, we find that the matrix of minors is negative three, six, zero, three, negative six, negative three, negative two, three, one.

Now that we have found the matrix of minors, we can move on to find the matrix of cofactors. When finding the matrix of cofactors, we simply take the matrix of minors and change some of the signs. We change the signs of the entries in the matrix of minors as shown to obtain the matrix of cofactors. We can apply this to the matrix of minors we found in the last step to get our matrix of cofactors.

For the penultimate step, we need to find the adjoint matrix. The adjoint matrix is simply the transpose of the matrix of cofactors. So in order to find it, we simply reflect the matrix of cofactors along its leading diagonal. Hence, the adjoint matrix is negative three, negative three, negative two, negative six, negative six, negative three, zero, three, one.

We are now ready to find the inverse matrix. In order to do this, we simply multiply the reciprocal of the determinant by the adjoint matrix.

We can now use this inverse matrix to solve the system of linear equations. We can recall that we have 𝑋 equals the inverse of 𝐴 multiplied by 𝐵. To find the values of 𝑥, 𝑦, and 𝑧, we need to carry out the multiplication. We multiply each row of the matrix by the column matrix and add together these products. Now, we simplify the column matrix which we are left with. Now, all we have to do is multiply our column matrix by the constant. We have that 𝑥, 𝑦, 𝑧 is equal to negative one, zero, seven.

Hence, our solution to the system of linear equations is 𝑥 equals negative one, 𝑦 equals zero, and 𝑧 equals seven.

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