Animated Solution for Mathematics - Matrices and Determinants: Let M be a $3\times 3$ matrix satisfying $M\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]=\left[\begin{array}{c}-1\\ 2\\ 3\end{array}\right],M\left[\begin{array}{c}1\\ -1\\ 0\end{array}\right]=\left[\begin{array}{c}1\\ 1\\ -1\end{array}\right]$, and $M\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 12\end{array}\right]$. Then the sum of the diagonal entries of M is

Welcome, fellow traveler in the realm of linear algebra. Today, we are not just solving a matrix problem; we are decoding a transformation.

Imagine matrix $M$ as a machine. You feed it a vector, and it spits out a new one. We have been given three specific inputs and their corresponding outputs. Our mission is to find the trace of $M$, which is the sum of its diagonal entries: $a_1+b_2+c_3$.

The Filter

Let us look at our first condition:

In the language of linear algebra, multiplying a matrix by the vector $\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]$ is like using a filter. It selects the second column of the matrix and ignores the rest.

If we define $M$ as having columns $C_1,C_2,C_3$, then this operation tells us exactly what $C_2$ is. We immediately see that the second column is $\left[\begin{array}{c}-1\\ 2\\ 3\end{array}\right]$.

This gives us $b_1=-1$, $b_2=2$, and $b_3=3$. We have already secured one-third of our diagonal sum: $b_2=2$.

The Difference Game

Next, we tackle the second condition:

This operation is a bit more complex. It tells us that $1\cdot C_1-1\cdot C_2+0\cdot C_3=\left[\begin{array}{c}1\\ 1\\ -1\end{array}\right]$.

In other words, the first column minus the second column equals our result vector. We can write this as $C_1-C_2=\left[\begin{array}{c}1\\ 1\\ -1\end{array}\right]$.

Since we already know $C_2$, we can isolate $C_1$ by adding $C_2$ to both sides:

Now we have $a_1=0$, $a_2=3$, and $a_3=2$. We have found our second diagonal piece: $a_1=0$.

The Final Summation

Finally, we face the third condition:

This operation is a simple summation: $C_1+C_2+C_3=\left[\begin{array}{c}0\\ 0\\ 12\end{array}\right]$. We only need $c_3$ to complete our trace.

Let us look at the third row of this equation: $a_3+b_3+c_3=12$. We already know $a_3=2$ and $b_3=3$.

Substituting these values, we get $2+3+c_3=12$, which simplifies to $5+c_3=12$. Thus, $c_3=7$.

We have arrived at the finish line. We have our diagonal entries: $a_1=0$, $b_2=2$, and $c_3=7$.

The trace of $M$ is simply the sum of these values: $0+2+7=9$.

See how the complexity melted away? By understanding the geometric meaning of matrix multiplication, we didn't need to solve a massive system of nine equations. We simply listened to what the matrix was telling us. Keep this perspective, and no matrix will ever intimidate you again. The final answer is 9.