Question

The matrix below represents a system of equations.

Answer 1

Answer:

The first choice: $a = 1$, $b = -1$, and $c = 2$.

Step-by-step explanation:

$\left[ \begin{array}{ccc|c} 2 & -1 & 1 & 5 \cr 1 & 1 & -1 & -2 \cr 1 & 0 & -3 & -5\end{array} \right]$.

Add the first row to the second row to obtain:

$\left[ \begin{array}{ccc|c} 2 & -1 & 1 & 5 \cr 2 + 1 & (-1) + 1 & 1 + (-1) & 5 + (-2) \cr 1 & 0 & - 3& -5\end{array} \right]$.

Simplify that matrix to obtain:

$\left[ \begin{array}{ccc|c} 2 & -1 & 1 & 5 \cr 3 & 0 & 0 & 3\cr 1 & 0 & -3 & -5\end{array} \right]$.

Divide the second row by $3$:

$\left[ \begin{array}{ccc|c} 2 & -1 & 1 & 5 \cr 1 & 0 & 0 & 1\cr 1 & 0 & -3& -5\end{array} \right]$.

Hence, $a = 1$.

Subtract the current row two from row three:

$\left[ \begin{array}{ccc|c} 2 & -1 & 1 & 5 \cr 1 & 0 & 0 & 1\cr 1 - 1 & 0 & -3 & (-5)- 1 \end{array} \right]$.

Simplify that matrix to obtain:

$\left[ \begin{array}{ccc|c} 2 & -1 & 1 & 5 \cr 1 & 0 & 0 & 1\cr 0 & 0 & -3 & -6 \end{array} \right]$.

Invert the signs in the third row and divide it by $3$ to obtain:

$\left[ \begin{array}{ccc|c} 2 & -1 & 1 & 5 \cr 1 & 0 & 0 & 1\cr 0 & 0 & 1 & 2 \end{array} \right]$.

Hence, $c = 2$.

Subtract two times the current second row from the first row to obtain:

$\left[ \begin{array}{ccc|c} 2 - (2 \times 1) & -1 & 1 & 5 - 2\times 1\cr 1 & 0 & 0 & 1\cr 0 & 0&1 & 2 \end{array} \right]$.

That simplifies to

$\left[ \begin{array}{ccc|c} 0 & -1 & 1 & 3\cr 1 & 0 & 0 & 1\cr 0 & 0 & 1& 2\end{array} \right]$.

Subtract the current third row from the first row to obtain:

$\left[ \begin{array}{ccc|c} 0 & -1 & 1-1 & 3-2\cr 1 & 0 & 0 & 1\cr 0 & 0 & 1& 2\end{array} \right]$.

That simplifies to

$\left[ \begin{array}{ccc|c} 0 & -1 & 0& 1\cr 1 & 0 & 0 & 1\cr 0 & 0 & 1& 2\end{array} \right]$.

Invert the signs in the first row to obtain:

$\left[ \begin{array}{ccc|c} 0 & 1 & 0& -1\cr 1& 0 & 0 & 1\cr 0 & 0 & 1& 2\end{array} \right]$.

Hence, $b = -1$.

$\left[ \begin{array}{ccc|c}1& 0 & 0 & 1 \cr0 & 1 & 0& -1 \cr 0 & 0 & 1& 2\end{array} \right]$.