How do I find the unique solution to this linear system?

Question

The question is based on an exercise in linear algebra:

THOUGHTS:

By looking at the determinant, I know that a unique solution occurs when $c \neq 2$ or $-3$ since a square matrix is singular if and only if its determinant is $0$ (see this article). How can I really find what this solution is concretely?

Answer 1

Hint: The system has a unique solution if you can invert the matrix $$ \begin{bmatrix} 1&1&-1\\ 1&c&3\\ 2&3&c \end{bmatrix} $$ And the solution is $$ \begin{bmatrix} X_1\\X_2\\X_3 \end{bmatrix} = \begin{bmatrix} 1&1&-1\\ 1&c&3\\ 2&3&c \end{bmatrix}^{-1} \begin{bmatrix} 1\\2\\3 \end{bmatrix} $$

Answer 2

If, in your system, you subtract the first line from the second one and twice the first line from the third line, you get$$\left\{\begin{array}{l}X_1+X_2-X_3=1\\(c-1)X_2+4X_3=1\\X_2+(c+2)X_3=1.\end{array}\right.$$Now, if you subtract the third line times $c-1$ from the second one, you get$$\left\{\begin{array}{l}X_1+X_2-X_3=1\\(-c^2-c+6)X_3=-c+2\\X_2+(c+2)X_3=1.\end{array}\right.$$Can you take it from here?

Answer 3

We can proceed by elimination for the augmented matrix

$$\begin{bmatrix} 1&1&-1&1\\ 1&c&3&2\\ 2&3&c&3 \end{bmatrix} \to \begin{bmatrix} 1&1&-1&1\\ 0&c-1&4&1\\ 0&1&c+2&1 \end{bmatrix}\to \begin{bmatrix} 1&1&-1&1\\ 0&c-1&4&1\\ 0&0&(c+2)(c-1)-4&c-2 \end{bmatrix}$$

$$\to \begin{bmatrix} 1&1&-1&1\\ 0&c-1&4&1\\ 0&1&c+2&1 \end{bmatrix}\to \begin{bmatrix} 1&1&-1&1\\ 0&c-1&4&1\\ 0&0&(c-2)(c+3)&c-2 \end{bmatrix}$$

and obtain all from that.

Notably form the third row we obtain that the system is consistent for $x\neq 2$ and $x\neq -3$ therefore we obtain

$$\to \begin{bmatrix} 1&1&-1&1\\ 0&c-1&4&1\\ 0&0&c+3&1 \end{bmatrix}$$

and we obtain

• from the third row $X_3=\frac 1{c+3}$
• from the second row $X_2=\frac{1-4X_3}{c-1}=\frac{c+3-4}{(c-1)(c-3)}=\frac 1{c+3}$
• from the first row $X_1=1-X_2+X_3=1$

Answer 4

To obtain in practice the solutions when they exist, you can start from the augmented matrix and use row operations until the submatrix corresponding to the l.h.s. is the unit matrix. The vector of solutions is then the last column.

Here is how it goes, if my computations are correct: \begin{align} \left[\begin{array}{ccc|c} 1&1&-1&1\\1&c&3&2\\2&3&c&3 \end{array}\right]&\rightsquigarrow \left[\begin{array}{ccc|c} 1&1&-1&1\\0&c-1&4&1\\0&1&c+2&1 \end{array}\right]\rightsquigarrow \left[\begin{array}{ccc|c} 1&1&-1&1\\0&1&c+2&1\\0&c-1&4&1 \end{array}\right]\rightsquigarrow \\[1ex] \left[\begin{array}{ccc|c} 1&1&-1&1\\0&1&c+2&1\\0&c-1&4&1 \end{array}\right]&\rightsquigarrow \left[\begin{array}{ccc|c} 1&1&-1&1\\0&1&c+2&1\\0&0&\substack{6-c-c^2\\=-(c-2)(c+3)}&2-c \end{array}\right]\rightsquigarrow \left[\begin{array}{ccc|c} 1&1&-1&1\\0&1&c+2&1\\0&0&1&\frac1{c+3} \end{array}\right]\rightsquigarrow \\[1ex] \text{(we suppose here }c\ne2,-3)\\[1ex]\left[\begin{array}{ccc|c} 1&1&0&\frac{c+4}{c+3}\\0&1&0&\frac1{c+3}\\0&0&1&\frac1{c+3} \end{array}\right]&\rightsquigarrow\left[\begin{array}{ccc|c} 1&0&0&1\\0&1&0&\frac{1}{c+3}\\0&0&1&\frac1{c+3} \end{array}\right]. \end{align}