We have a system of two equations in two variables. Let [latex]x=[/latex] the amount invested at 10.5% interest, and [latex]y=[/latex] the amount invested at 12% interest.

[latex]\begin{gathered}x+y=12,000 \\ 0.105x+0.12y=1,335 \end{gathered}[/latex]

As a matrix, we have

[latex]\left[\left.\begin{array}{rr}\hfill 1& \hfill 1\\ \hfill 0.105& \hfill 0.12\end{array}\right\rvert\begin{array}{r}\hfill 12,000\\ \hfill 1,335\end{array}\right][/latex]

Multiply row 1 by [latex]-0.105[/latex] and add the result to row 2.

[latex]\left[\left.\begin{array}{rr}\hfill 1& \hfill 1\\ \hfill 0& \hfill 0.015\end{array}\right\rvert\begin{array}{r}\hfill 12,000\\ \hfill 75\end{array}\right][/latex]

Then,

[latex]\begin{gathered}0.015y=75 \\ y=5,000 \end{gathered}[/latex]

So [latex]12,000 - 5,000=7,000[/latex].

Thus, $5,000 was invested at 12% interest and $7,000 at 10.5% interest.

We have a system of three equations in three variables. Let [latex]x[/latex] be the amount invested at 5% interest, let [latex]y[/latex] be the amount invested at 8% interest, and let [latex]z[/latex] be the amount invested at 9% interest. Thus,

[latex]\begin{gathered}x+y+z=10,000 \\ 0.05x+0.08y+0.09z=770 \\ 2x-z=0 \end{gathered}[/latex]

As a matrix, we have

[latex]\left[\left.\begin{array}{rrr}\hfill 1& \hfill 1& \hfill 1\\ \hfill 0.05& \hfill 0.08& \hfill 0.09\\ \hfill 2& \hfill 0& \hfill -1\end{array}\right\rvert\begin{array}{r}\hfill 10,000\\ \hfill 770\\ \hfill 0\end{array}\right][/latex]

Now, we perform Gaussian elimination to achieve row-echelon form.

[latex]-0.05{R}_{1}+{R}_{2}={R}_{2}\to \left[\left.\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill 1& \hfill & \hfill 1& \hfill \\ \hfill 0& \hfill & \hfill 0.03& \hfill & \hfill 0.04& \hfill \\ \hfill 2& \hfill & \hfill 0& \hfill & \hfill -1& \hfill \end{array}\right\rvert\begin{array}{rr}\hfill & \hfill 10,000\\ \hfill & \hfill 270\\ \hfill & \hfill 0\end{array}\right][/latex]

[latex]-2{R}_{1}+{R}_{3}={R}_{3}\to \left[\left.\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill 1& \hfill & \hfill 1& \hfill \\ \hfill 0& \hfill & \hfill 0.03& \hfill & \hfill 0.04& \hfill \\ \hfill 0& \hfill & \hfill -2& \hfill & \hfill -3& \hfill \end{array}\right\rvert\begin{array}{rr}\hfill & \hfill 10,000\\ \hfill & \hfill 270\\ \hfill & \hfill -20,000\end{array}\right][/latex]

[latex]\frac{1}{0.03}{R}_{2}={R}_{2}\to \left[\left.\begin{array}{rrrrrr}\hfill 0& \hfill & \hfill 1& \hfill & \hfill 1& \hfill \\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill \frac{4}{3}& \hfill \\ \hfill 0& \hfill & \hfill -2& \hfill & \hfill -3& \hfill \end{array}\right\rvert\begin{array}{rr}\hfill & \hfill 10,000\\ \hfill & \hfill 9,000\\ \hfill & \hfill -20,000\end{array}\right][/latex]

[latex]2{R}_{2}+{R}_{3}={R}_{3}\to \left[\left.\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill 1& \hfill & \hfill 1& \hfill \\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill \frac{4}{3}& \hfill \\ \hfill 0& \hfill & \hfill 0& \hfill & \hfill -\frac{1}{3}& \hfill \end{array}\right\rvert\begin{array}{rr}\hfill & \hfill 10,000\\ \hfill & \hfill 9,000\\ \hfill & \hfill -2,000\end{array}\right][/latex]

The third row tells us [latex]-\frac{1}{3}z=-2,000[/latex]; thus [latex]z=6,000[/latex].

The second row tells us [latex]y+\frac{4}{3}z=9,000[/latex]. Substituting [latex]z=6,000[/latex], we get

[latex]\begin{array}{r}\hfill y+\frac{4}{3}\left(6,000\right)=9,000\\ \hfill y+8,000=9,000\\ \hfill y=1,000\end{array}[/latex]

The first row tells us [latex]x+y+z=10,000[/latex]. Substituting [latex]y=1,000[/latex] and [latex]z=6,000[/latex], we get

[latex]\begin{gathered}x+1,000+6,000=10,000 \\ x=3,000 \end{gathered}[/latex]

The answer is $3,000 invested at 5% interest, $1,000 invested at 8%, and $6,000 invested at 9% interest.