Question Video: Solving Equations Involving Scalar Multiplication | Nagwa Question Video: Solving Equations Involving Scalar Multiplication | Nagwa

# Question Video: Solving Equations Involving Scalar Multiplication Mathematics • First Year of Secondary School

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Consider the matrix equation [8, 1 and 9, β3] = π [3, 0 and 2, β1] + [β1, 1 and 3, 0]. Find the value of π which solves this equation.

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### Video Transcript

Consider the matrix equation eight, one, nine, negative three is equal to π times three, zero, two, negative one plus negative one, one, three, zero. Find the value of π which solves this equation.

In this question, weβre given an equation which involves matrices. And we need to determine the value of π which solves this equation. Remember, when we say that a value of π solves this equation, it means our equation must be true. Both sides of the equation need to be equal. And to make both sides of this equation equal, letβs start by evaluating the right-hand side of this equation.

The right-hand side of the equation is as shown. And it includes two things. First, weβre multiplying a matrix by the value π. This is scalar multiplication. Next, weβre adding these two values together. This will be the addition of two two-by-two matrices. So this is matrix addition. So to answer this question, we need to recall what both of these mean.

Letβs start with what it means to multiply matrix by a scalar. We need to recall when we multiply a matrix by a scalar, we need to multiply every single entry inside of our matrix by this scalar. In this case, weβre going to need to multiply every single entry inside of our matrix by π.

So we can start with the entry in row one column one. Thatβs three. We need to multiply this by π. This gives us a new entry of three π. We can do exactly the same in row two column one. The entry in this position is two. And we need to multiply this by π, giving us a new entry of two π. In row one column two, the entry in our matrix is zero. So when we multiply this by π, weβll get zero π, which is of course just equal to zero. And finally, in row two column one, weβre going to get negative one multiplied by π, which weβll just write as negative π.

So by evaluating the scalar multiplication of our first matrix, we get the two-by-two matrix three π, zero, two π, negative π. And remember, we still need to add our other matrix.

Now to simplify this expression even further, weβre going to need to recall how we add together two matrices. To add two matrices together, we first need to recall we can only do this if they are of the same order. And remember, saying the two matrices of the same order is just saying they have the same number of rows and columns. This is because to add two matrices together, all we do is add together the corresponding entries.

In our case, both of our matrices are two-by-two square matrices. So we can add these together. When we add two matrices together, we want to add the corresponding entries together. So in row one column one, weβll get the two entries in row one column one added together. Weβll get three π plus negative one. To find the entry in row one column two, we need to add together the two entries in row one and column two. Itβs going to be zero plus one. We can do the same in row two column one. We add together two π and three, giving us two π plus three. And we can do exactly the same in row two column two. We add together negative π and zero to get negative π plus zero. This gives us the following two-by-two matrix, which we can simplify.

Simplifying each entry inside of our matrix, we get the two-by-two matrix three π minus one, one, two π plus three, negative π.

Now remember, the question wants us to find the value of π which solves our equation. So both sides of the equation given to us in the question should be equal. And we found this two-by-two matrix by simplifying the right-hand side of this equation. So it should be equal to the matrix given in the left-hand side of this equation. Therefore, for π to solve this equation, these two matrices have to be equal. And to find the value of π which makes these matrices equal, we need to recall what it means for two matrices to be equal.

We recall that we say that two matrices are equal if they have exactly the same order and all of their corresponding entries are equal. So letβs start by checking that these two matrices have the same order. Both of them are square two-by-two matrices. So they do indeed have the same order.

Next, we need that all of their corresponding entries are equal. So for these two matrices to be equal, the entries in row one column one need to be equal. In other words, we must have that three π minus one is equal to eight. Similarly, the entries in row two column one must be equal. We must have that two π plus three is equal to nine. Next, we must also have that the entries in row one and column two are equal. However, both of these are just equal to one. So this just gives us that one is equal to one. And this is already true for any value of π, so it doesnβt help us find the value of π. Finally, the entries in row two column two must be equal. And this is only true if negative π is equal to negative three.

So for our matrices to be equal, all three of these equations must be equal. This is what we call a system of equations. In fact, because all of these are linear, we can also call this a system of linear equations. Thereβs a lot of different ways we could solve this. The easiest way is just to solve the equation negative π is equal to negative three. By dividing both sides of this equation through by negative one, we see this is only true when π is equal to three. And in fact, this is the correct answer. However, itβs also worth substituting this value into our other linear equations to check that these are also correct.

Substituting π is equal to three into our first linear equation, we get three multiplied by three minus one, which we can calculate is nine minus one, which is eight. Substituting π is equal to three into our second linear equation, we get two times three plus three, which is six plus three, which is of course just equal to nine. Therefore, because all of our linear equations are satisfied when π is equal to three, the two matrices must be equal. And therefore, the value of π is equal to three solves this matrix equation.

Thereβs one last thing worth pointing out. We could have also checked our answer by substituting the value of π is equal to three directly into our matrix equation. We could then evaluate the right-hand side of this equation and check itβs equal to the left-hand side of the equation.

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