Question

Rewrite the following system of linear equations in matrix equation form and in vector equation form. Solve the system.

Answer 1

Answer:

The set of solutions is $\{\left[\begin{array}{c}a\\b\\x\\y\\z\end{array}\right] = \left[\begin{array}{c}-26+503y+543z\\-37+655y+724z\\-4+80y+90z\\y\\z\end{array}\right] : \text{y, z are real numbers}\}$

Step-by-step explanation:

The matrix associated to the problem is $A=\left[\begin{array}{ccccc}1&-1&2&-8&1\\2&-1&-4&1&-2\\-4&1&4&-3&-1\end{array}\right]$ and the vector of independent terms is (3,1,-1)^t. Then the matrix equation form of the system is Ax=b.

The vector equation form is $a\left[\begin{array}{c}1\\2\\-4\end{array}\right]+b\left[\begin{array}{c}-1\\-1\\1\end{array}\right] + x\left[\begin{array}{c}2\\-4\\4\end{array}\right]+y\left[\begin{array}{c}-8\\1\\-3\end{array}\right] + z\left[\begin{array}{c}1\\-2\\-1\end{array}\right]=\left[\begin{array}{c}3\\1\\-1\end{array}\right]$.

Now we solve the system.

The aumented matrix of the system is $\left[\begin{array}{cccccc}1&-1&2&-8&1&3\\2&-1&-4&1&-2&1\\-4&1&4&-3&-1&-1\end{array}\right]$.

Applying rows operations we obtain a echelon form of the matrix, that is $\left[\begin{array}{cccccc}1&-1&2&-8&1&3\\0&1&-8&-15&-4&-5\\0&0&1&-80&-9&-4\end{array}\right]$

Now we solve for the unknown variables:

• x-80y-90z=-4 then x=-4+80y+90z
• b-8x-15y-4z=-5, b-8(-4+80y+90z)-15y-4z=-5 then b=-37+655y+724z.
• a-b+2x-8y+z=3, a-(-37+655y+724z)+2(-4+80y+90z)-8y+z=3, then a=-26+503y+543z

Since the system has two free variables then has infinite solutions.

The set of solutions is $\{\left[\begin{array}{c}a\\b\\x\\y\\z\end{array}\right] = \left[\begin{array}{c}-26+503y+543z\\-37+655y+724z\\-4+80y+90z\\y\\z\end{array}\right] : \text{y, z are real numbers}\}$