# Video: Solving Systems of Linear Equations by Applying Operations on Matrices

Given that 𝑀 = [5, 6 and −5, −4], find the values of 𝑥 and 𝑦 that satisfy 𝑀² + 𝑥𝑀 + 𝑦𝐼 = 𝑂, where 𝑂 is the zero matrix of order 2 × 2 and 𝐼 is the unit matrix of order 2 × 2.

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### Video Transcript

Given that the matrix 𝑀 is equal to five, six, negative five, negative four, find the values of 𝑥 and 𝑦 that satisfy 𝑀 squared plus 𝑥𝑀 plus 𝑦𝐼 is equal to 𝑂, where 𝑂 is the zero matrix of order two by two and 𝐼 is the unit matrix of order two by two.

We recall that any zero matrix has all elements equal to zero. Therefore, the matrix 𝑂 is equal to zero, zero, zero, zero. The unit or identity matrix has ones on its main or leading diagonal and zeros elsewhere. Therefore, 𝐼 is equal to one, zero, zero, one. When multiplying any matrix by a constant, in this question 𝑥 and 𝑦, we simply multiply each of the elements in the matrix by the constant. This means that the matrix 𝑦𝐼 is equal to 𝑦, zero, zero, 𝑦. As 𝑀 is equal to five, six, negative five, negative four, then 𝑥𝑀 is equal to five 𝑥, six 𝑥, negative five 𝑥, negative four 𝑥. Finally, we need to calculate the matrix 𝑀 squared. This is equal to matrix 𝑀 multiplied by itself. We need to multiply five, six, negative five, negative four by five, six, negative five, negative four.

When multiplying matrices, we multiply each of the rows in the first matrix by each of the columns in the second matrix. Five multiplied by five plus six multiplied by negative five is equal to negative five. Multiplying the elements of the first row in the first matrix by the second column in the second matrix gives us six. Repeating this for the second row of the first matrix, we get the elements negative five and negative 14. Substituting the four matrices into our equation, we have negative five, six, negative five, negative 14 plus five 𝑥, six 𝑥, negative five 𝑥, negative four 𝑥 plus 𝑦, zero, zero, 𝑦 is equal to zero, zero, zero, zero. We can now set up four equations comparing the corresponding components or elements.

Comparing the top-left elements, we have negative five plus five 𝑥 plus 𝑦 is equal to zero. We will call this equation one. The top-right elements give us six plus six 𝑥 plus zero is equal to zero. As there was only one unknown in this equation, we can solve it. Subtracting six from both sides gives us six 𝑥 is equal to negative six. Dividing both sides of this equation by six gives us 𝑥 is equal to negative one. We can then substitute this value back into equation one to calculate the value of 𝑦. Simplifying the left-hand side, we have negative 10 plus 𝑦 is equal to zero. This means that 𝑦 is equal to 10. The values of 𝑥 and 𝑦 are negative one and 10, respectively.

We do need to check, however, that these values hold for the bottom row of our matrices. Here, we have the two equations: negative five plus negative five 𝑥 plus zero is equal to zero and negative 14 plus negative four 𝑥 plus 𝑦 is equal to zero. Adding five 𝑥 to both sides of the first equation gives us five 𝑥 is equal to negative five. Dividing both sides by five, we see that 𝑥 once again is equal to negative one. In the second equation, substituting 𝑥 is negative one gives us negative 14 plus four plus 𝑦 is equal to zero. Once again, this confirms that 𝑦 is equal to 10. If the matrix 𝑀 equals five, six, negative five, negative four, then the values of 𝑥 and 𝑦 that satisfy the equation 𝑀 squared plus 𝑥𝑀 plus 𝑦𝐼 equals 𝑂 are 𝑥 is equal to negative one and 𝑦 is equal to 10.