Question Video: Expressing a Given Vector as a Linear Combination in Terms of Another Two Given Vectors | Nagwa Question Video: Expressing a Given Vector as a Linear Combination in Terms of Another Two Given Vectors | Nagwa

# Question Video: Expressing a Given Vector as a Linear Combination in Terms of Another Two Given Vectors Mathematics • First Year of Secondary School

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If 𝐀 = <1, 2, 1>, 𝐁 = <−1, −1, 0>, and 𝐂 = <−2, −1, 1>, express 𝐂 in terms of 𝐀 and 𝐁.

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### Video Transcript

If vector 𝐀 is equal to one, two, one; vector 𝐁 is equal to negative one, negative one, zero; and vector 𝐂 is equal to negative two, negative one, one, express vector 𝐂 in terms of vector 𝐀 and vector 𝐁.

In order to express vector 𝐂 in terms of vector 𝐀 and 𝐁, we will firstly rewrite vector 𝐂 equal to the scalar 𝑚 multiplied by vector 𝐀 plus scalar 𝑛 multiplied by vector 𝐁. This means that negative two, negative one, one is equal to 𝑚 multiplied by one, two, one plus 𝑛 multiplied by negative one, negative one, zero.

We know that when multiplying a vector by a scalar, we simply multiply each of the components of the vector by the scalar. Multiplying the vector 𝐀 by the scalar 𝑚 gives us the vector 𝑚, two 𝑚, 𝑚. And multiplying vector 𝐁 by the scalar 𝑛 gives us negative 𝑛, negative 𝑛, zero. We can now compare the corresponding components to create three equations.

Firstly, we have negative two is equal to 𝑚 plus negative 𝑛. This can be rewritten as negative two is equal to 𝑚 minus 𝑛. We will call this equation one. Our second equation gives us negative one is equal to two 𝑚 plus negative 𝑛, which can be rewritten as negative one is equal to two 𝑚 minus 𝑛. Finally, we have the equation one is equal to 𝑚 plus zero.

From equation three, we see that the scalar or constant 𝑚 is equal to one. We can substitute this value into equation one, giving us negative two is equal to one minus 𝑛. Adding two and then adding 𝑛 to both sides of this equation gives us 𝑛 is equal to three. This would also be true had we substituted 𝑚 is to equal one into equation two.

Substituting in the scalars to our initial equation gives us vector 𝐂 is equal to one multiplied by vector 𝐀 plus three multiplied by vector 𝐁. This simplifies to a final answer of 𝐂 is equal to 𝐀 plus three 𝐁.

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