Question

linear help anyone cant find the third number

Answer 1

The first two answers you got are coefficients for the first two matrices in your basis

Answer 2

In other words,


\[\Large
a*\begin{bmatrix}
-1 & 1 \\
0 & 0 \\
\end{bmatrix}
+
b*\begin{bmatrix}
0 & 1 \\
0 & 2 \\
\end{bmatrix}
+
c*\begin{bmatrix}
0 & 0 \\
0 & -2 \\
\end{bmatrix}
=
M
\]

Answer 3

You've already found a & b, they are a = -7 and b = -14, so

\[\Large
-7*\begin{bmatrix}
-1 & 1 \\
0 & 0 \\
\end{bmatrix}
+
(-14)*\begin{bmatrix}
0 & 1 \\
0 & 2 \\
\end{bmatrix}
+
c*\begin{bmatrix}
0 & 0 \\
0 & -2 \\
\end{bmatrix}
=
M
\]

Answer 4

this is what i did but cant get the third value right

Answer 5

So we'll have


\[\Large
-7*\begin{bmatrix}
-1 & -1 \\
0 & 0 \\
\end{bmatrix}
+
(-14)*\begin{bmatrix}
0 & 1 \\
0 & 2 \\
\end{bmatrix}
+
c*\begin{bmatrix}
0 & 0 \\
0 & -2 \\
\end{bmatrix}
=
M
\]


\[\Large
\begin{bmatrix}
7 & 7 \\
0 & 0 \\
\end{bmatrix}
+
\begin{bmatrix}
0 & -14 \\
0 & -28 \\
\end{bmatrix}
+
\begin{bmatrix}
0 & 0 \\
0 & -2c \\
\end{bmatrix}
=
\begin{bmatrix}
7 & -7 \\
0 & 9 \\
\end{bmatrix}
\]


\[\Large
\begin{bmatrix}
7+0+0 & 7+(-14)+0 \\
0+0+0 & 0+(-28)+(-2c) \\
\end{bmatrix}
=
\begin{bmatrix}
7 & -7 \\
0 & 9 \\
\end{bmatrix}
\]


\[\Large
\begin{bmatrix}
7 & -7 \\
0 & -28-2c \\
\end{bmatrix}
=
\begin{bmatrix}
7 & -7 \\
0 & 9 \\
\end{bmatrix}
\]

Answer 6

I made a typo and used -1 1 in the first row of the first matrix, but I fixed that typo

Answer 7

Anyways, from there you equate the components in row2, column2 to get

-28 - 2c = 9

and solve for c