Question Video: Finding an Unknown Matrix Using Operations on Matrices | Nagwa Question Video: Finding an Unknown Matrix Using Operations on Matrices | Nagwa

# Question Video: Finding an Unknown Matrix Using Operations on Matrices Mathematics • First Year of Secondary School

## Join Nagwa Classes

Attend live Mathematics sessions on Nagwa Classes to learn more about this topic from an expert teacher!

Consider the matrices π΄ = [0, β4 and 2, 1], π΅ = [β3, β7, 6 and 6, β4, 3], and πΆ = [7, β7 and β7, 2 and β7, 7]. Determine the matrix π that satisfies βπ^(π) = π΄Β² + (π΅πΆ)^(π).

06:28

### Video Transcript

Consider the matrices π΄ is the two-by-two matrix zero, negative four, two, one; π΅ is the three-by-two matrix negative three, negative seven, six, six, negative four, three; and πΆ is the two-by-three matrix seven, seven, negative seven, two, negative seven, seven. Determine the matrix π that satisfies negative π transpose is equal to π΄ squared plus π΅πΆ all transposed.

In this question, weβre given a matrix equation involving three known matrices π΄, π΅, and πΆ. We need to use this to determine the matrix π. To do this, letβs start by evaluating the right-hand side of this equation, and we can do this term by term. First, we need to evaluate π΄ squared. And we recall squaring a matrix means we multiply it by itself. π΄ squared is π΄ times π΄. Therefore, π΄ squared is given by the following expression.

And we recall to multiply two matrices together, we need to find the sum of the products of the corresponding entries in the rows of the first matrix and the columns of the second matrix. For example, to find the entry in row one, column one of matrix π΄ squared, we need to find the product of zero and zero and add this to the product of negative four and two. And we can evaluate this expression. Itβs equal to negative eight. This is the entry in row one, column one of matrix π΄ squared.

We can follow the same process to find the entry in row one, column two of matrix π΄ squared. Itβs equal to zero multiplied by negative four plus negative four multiplied by one, which we can calculate is equal to negative four. And we can follow the same procedure to find the final two elements of π΄ squared. We see that matrix π΄ squared is equal to the two-by-two matrix negative eight, negative four, two, negative seven. And now that we found an expression for matrix π΄ squared, letβs find an expression for the second term of the right-hand side of our equation. Thatβs the transpose of π΅πΆ. And to find the transpose of matrix π΅πΆ, letβs start by multiplying matrix π΅ by matrix πΆ. And although itβs not necessary, letβs check that we can multiply matrix π΅ by matrix πΆ.

Remember, to multiply two matrices together, we need the number of columns of the first matrix to be equal to the number of rows of the second matrix. We can see that these are both equal to three. Then the order of the products of these two matrices will be the number of rows of the first matrix by the number of columns of the second matrix. The product of these two matrices is a two-by-two matrix. And we find the products of these two matrices in the same way we did for the matrix π΄ squared.

We need to find the sum of the products of the corresponding elements of the rows of matrix π΅ with the columns of matrix πΆ. For example, the entry in row one, column one of matrix π΅πΆ will be negative three times seven plus negative seven times negative seven plus six multiplied by negative seven, which if we evaluate we see is equal to negative 14. So the entry in row one, column one of π΅πΆ is negative 14.

We can follow the same process to find the entry in row one, column two. Itβs negative three times seven plus negative seven times two plus six multiplied by seven, which if we calculate is equal to seven. We can do the same to find the remaining two entries of this matrix. π΅ times πΆ is the two-by-two matrix negative 14, seven, 49, 55. Now weβre ready to find an expression for the second term of the right-hand side of our equation. We need to take the transpose of matrix π΅πΆ. And to do this, letβs recall how we find the transpose of a matrix. This means we write the rows of the matrix as the columns of the new matrix. So the first column of the transpose of this matrix is found by using the first row of this matrix.

The first column of the transpose is negative 14, seven. And we can do the same to find the second column of this matrix. We use the second row of π΅πΆ. The second column is 49, 55. Therefore, weβve shown the transpose of π΅πΆ is the two-by-two matrix negative 14, 49, seven, 55. And now that we found π΄ squared and the transpose of π΅πΆ, we can evaluate the right-hand side of the equation. We just need to add π΄ squared to the transpose of matrix π΅πΆ.

Remember, to add two matrices of the same order together, we just need to add the corresponding entries together. For example, the entry in row one, column one of the sum of these two matrices will be negative eight plus negative 14, which we can calculate is negative 22. Similarly, the element in row one, column two of our matrix will be negative four plus 49, which we can calculate is 45. And we can continue this process to find the final two elements of the matrix. We have two plus seven is nine and negative seven plus 55 is 48. So our matrix is the two-by-two matrix negative 22, 45, nine, 48.

Now, we can substitute the matrix we found for the entire right-hand side of the equation weβre given. This gives us that negative π transpose is the two-by-two matrix negative 22, 45, nine, 48. And remember, we want to solve this equation for the matrix π, and thereβs a few different ways of doing this. For example, we could notice multiplying by the scalar and taking the transpose of a matrix does not change its order. Therefore, we can conclude that matrix π is also a two-by-two matrix. So we could write π as a two-by-two matrix and find all of the elements of matrix π. And this would work. However, we can solve this matrix equation directly.

First, weβll multiply both sides of our equation through by negative one. When we do this, on the left-hand side of our equation, we get negative one times negative one which is one multiplied by the transpose of π. And on the right-hand side of our equation, we get negative one multiplied by our matrix. And remember, to multiply a matrix by a scalar, we just multiply all of the entries of the matrix by our scalar. And since our scalar is negative one, this just means we switch the sign of all of the entries of the matrix. This gives us the two-by-two matrix 22, negative 45, negative nine, negative 48.

Now, we can solve the matrix π by taking the transpose of both sides of the equation. When we do this, on the left-hand side of the equation, we get the transpose of the transpose of π. And on the right-hand side of our equation, we get the transpose of the given two-by-two matrix. And we can then simplify this. First, for any matrix π, the transpose of the transpose of π is just equal to π. So the left-hand side of our equation is equal to π. π is just the transpose of the given two-by-two matrix. And therefore, we can find matrix π by taking the transpose of this matrix. π is the two-by-two matrix 22, negative nine, negative 45, negative 48.

## Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

• Interactive Sessions
• Chat & Messaging
• Realistic Exam Questions