Determining the value of the product of a matrix A with another known matrix without knowing what is the matrix A (math GRE subject test 9768 Q.43)

Question

I have given the entries of the matrix the names $a$,$b$,$c$,$\dots$,$i$, and I multiplied the matrix by the given $2$ matrices. But the problem is that I have for every $3$ entries $2$ equation in $3$ unknowns. Could anyone show me the intuition behind solving this question please?

Answer 1

As a general rule, GRE Math Subject Exam questions do not require much nitty-gritty computation. There is almost always a "trick," or some clever way of getting around computation. In this case, it is to use the linearity of matrix multiplication:

Using the fact that matrix multiplication is linear (i.e. $A(a\vec{u} + b\vec{v}) = aA\vec{u} + bA\vec{v}$ for any scalars $a$, $b$ and any vectors $\vec{u}$, $\vec{v}$), we can rewrite the last vector given as a linear combination of the other two, and obtain \begin{align*} &\begin{pmatrix}6\\ 7\\ 8\end{pmatrix} = -\begin{pmatrix}0\\1\\2\end{pmatrix} + 2\begin{pmatrix}3\\4\\5\end{pmatrix} \\ &\quad\implies A\begin{pmatrix}6\\ 7\\ 8\end{pmatrix} = A\left(-\begin{pmatrix}0\\1\\2\end{pmatrix} + 2\begin{pmatrix}3\\4\\5\end{pmatrix}\right) = -A\begin{pmatrix}0\\1\\2\end{pmatrix} + 2A\begin{pmatrix}3\\4\\5\end{pmatrix} = -\begin{pmatrix}1\\0\\0\end{pmatrix} + 2\begin{pmatrix}0\\1\\0\end{pmatrix} = \begin{pmatrix}-1\\2\\0\end{pmatrix}, \end{align*} which is option (B).

Answer 2

If we know that $A \vec v_1 = b_1$ and $A \vec v_2 = b_2$, then for any other vector which is a linear combination of them $\vec v = \alpha \vec v_1 + \beta \vec v_2$, we know that

$$A\vec{v} = \alpha A \vec v_1 + \beta A \vec v_2.$$

In our specific case, we can search for that linear combination by finding values of $\alpha$ and $\beta$ such that

$$\alpha \begin{bmatrix}0\\1\\2\end{bmatrix} + \beta \begin{bmatrix}3\\4\\5\end{bmatrix} = \begin{bmatrix}6\\7\\8\end{bmatrix}.$$

To start, we know from the top entry that $\beta = 2$. Hence the middle entry requires that $\alpha = -1$. We check the final entry just to be sure. Our values of α and β are consistent with the final entry —if they weren't, the problem wouldn't have enough information.

We have that

$$ \begin{align*} A\left(\begin{bmatrix}6\\7\\8\end{bmatrix}\right) &= A\left(-1 \begin{bmatrix}0\\1\\2\end{bmatrix} + 2 \begin{bmatrix}3\\4\\5\end{bmatrix}\right)\\ &= -1\cdot A\left(\begin{bmatrix}0\\1\\2\end{bmatrix}\right) + 2\cdot A\left( \begin{bmatrix}3\\4\\5\end{bmatrix}\right)&\text{linearity of }A\\ &= -1 \cdot \begin{bmatrix}1\\0\\0\end{bmatrix} + 2 \cdot \begin{bmatrix}0\\1\\0\end{bmatrix}&\text{given values of }A\\ &= \begin{bmatrix}-1\\2\\0\end{bmatrix}. \end{align*} $$

So the answer is (B).