5x + 3y + 7z = 4
3x + 26y + 2z = 9
7x + 2y + 10z = 5
The matrix equation corresponding to the given systematic
\({\begin{bmatrix}5&3&7\\3&26&2\\7&2&10\end{bmatrix}} \begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}4\\9\\5\end{bmatrix} \)
[A] [X] = B
Obviously, the last equivalent matrix is in the echelon form.
It has two non-zero rows.
p([A, B]) = 2, p(A) = 2
p(A) = p([A, B]) = 2 < Number of unknowns
The given system is consistent and has infinitely many solutions.
The given system is equivalent to the matrix equation
\({\begin{bmatrix}1&5&1\\0&-11&1\\0&0&0\end{bmatrix}} \begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}2\\-3\\0\end{bmatrix} \)
x + 5y + z = 2 .......(1)
– 11y + z = – 3 .......(2)
Let z = k
Equation (1),
\(x + 5\left[ \frac 1{11} (3 + k)\right] + k = 2\)
\(x = 2 - k - \frac 5{11} (3 + k)\)
\(x = \frac{22 - 11k - 15 - 5 k}{11}\)
\(x = \frac 1{11} (7 - 16 k)\)
Equation (2),
– 11y + k = – 3
3 + k = 11y
\(y = \frac 1{11} (3 + k)\)
(x, y, z) = \(\left(\frac 1{11} (7 - 16 k), \frac 1{11} (3+ k), k\right)\)
Where K ∈ R.
