Question Video: Solving a System of Linear Equations Using an Inverse Matrix | Nagwa Question Video: Solving a System of Linear Equations Using an Inverse Matrix | Nagwa

Question Video: Solving a System of Linear Equations Using an Inverse Matrix Mathematics • First Year of Secondary School

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By expressing the following simultaneous equations as a matrix equation and finding the inverse of the coefficient matrix, determine the values of π‘₯ and 𝑦 that solve the equations π‘₯ + 3𝑦 = βˆ’4, 2π‘₯ + 5𝑦 = 10.

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Video Transcript

By expressing the following simultaneous equations as a matrix equation and finding the inverse of the coefficient matrix, determine the values of π‘₯ and 𝑦 that solve the equations π‘₯ plus three 𝑦 equals negative four and two π‘₯ plus five 𝑦 equals 10.

In this question, we are given a pair of simultaneous equations in two variables and asked to determine the solutions of the system by finding the inverse of the coefficient matrix. To do this, we first recall that we can write a pair of linear simultaneous equations as a matrix equation 𝐴𝑋 equals 𝐡, where 𝐴 is the matrix of coefficients of the variables in the simultaneous equations. In this case, it’s the two-by-two matrix one, three, two, five.

𝑋 is the matrix of variables. It is a column matrix containing the variables of the simultaneous equation. We have that 𝑋 is the two-by-one matrix π‘₯, 𝑦. Finally, matrix 𝐡 is a column matrix containing the constants on the right-hand side of the simultaneous equations. This is usually called the constant matrix. We have that 𝐡 is the two-by-one matrix negative four, 10.

If we were to evaluate the matrix multiplication, we would see that the matrix equation only holds true when the simultaneous equations are satisfied and vice versa. If 𝐴 is an invertible matrix, then we can multiply on the left by the inverse of 𝐴 to obtain 𝑋 equals 𝐴 inverse times 𝐡. In other words, we can solve the system by calculating the inverse of 𝐴 and then evaluating this expression. To find the inverse of the coefficient matrix, we can recall that the inverse of a two-by-two matrix π‘Ž, 𝑏, 𝑐, 𝑑 is given by one over π‘Žπ‘‘ minus 𝑏𝑐 times the two-by-two matrix 𝑑, negative 𝑏, negative 𝑐, π‘Ž, provided the determinant is nonzero.

To find the inverse of matrix 𝐴, we first need to calculate its determinant, which we can do by finding the difference in the product of its diagonals. We have the determinant of 𝐴 is one times five minus three times two, which evaluates to give us negative one. We can now find the inverse of 𝐴 by using the formula for the inverse of a two-by-two matrix. We have that the inverse of 𝐴 is equal to one over negative one times the two-by-two matrix five, negative three, negative two, one. Evaluating this expression by multiplying all of the elements of the matrix by negative one yields the two-by-two matrix negative five, three, two, negative one.

We can now use the fact that 𝑋 is equal to 𝐴 inverse times 𝐡 to solve the system of equations. Substituting our matrices into this equation gives us that 𝑋 is equal to the two-by-two matrix negative five, three, two, negative one times the column matrix negative four, 10.

We can now evaluate these products by multiplying the elements in the rows of the first matrix with the corresponding elements in the column of the second matrix and finding their sum. We get the following column matrix for 𝑋. We can evaluate each entry separately. This gives us the two-by-one matrix 50, negative 18. The first entry of this matrix is the value of π‘₯, and the second entry is the value of 𝑦 that solves the simultaneous equations. We can verify that these are the solutions by substituting them back into the system. However, this is not necessary to answer the question.

Therefore, we were able to show that the solution to the given system of equations is π‘₯ equals 50 and 𝑦 equals negative 18.

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