Linear Algebra Questions and Answers – System of Equations and their Consistencies

5x + 3y + 7z = 4
3x + 26y + 2z = 9
7x + 2y + 10z = 5

Answer: d
Explanation: In this Question we have,
\(\begin{bmatrix}5&3&7\\3&26&2\\7&2&10\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}4\\9\\5\end{bmatrix}\)
By 7R₁ and 5R₃
\(\begin{bmatrix}35&21&49\\3&26&2\\35&10&50\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}28\\9\\25\end{bmatrix}\)
By R₃-R₁ and 5R₂
\(\begin{bmatrix}35&21&49\\15&130&10\\0&-11&1\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}28\\45\\-3\end{bmatrix}\)
By \(R_2 – \frac{3}{7} R_1\)
\(\begin{bmatrix}35&21&49\\0&121&-11\\0&-11&1\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}28\\33\\-3\end{bmatrix}\)
By \(R_3 + \frac{1}{11}R_2\)
\(\begin{bmatrix}35&21&49\\0&121&-11\\0&0&0\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}28\\33\\0\end{bmatrix}\)
By \(\frac{1}{7} R_1 and \frac{1}{11} R_2\)
\(\begin{bmatrix}5&3&7\\0&11&-1\\0&0&0\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}4\\3\\0\end{bmatrix}\)
The rank of both the coefficient matrix and the augmented matrix is equal (i.e. 2)
Thus the equations are consistent.
However the rank of the matrix 2 is less than the total number of unknowns 3.
Hence the given set of equations has infinite number of solutions.

2x - 3y+ 7z = 5
3x + y - 3z = 13
2x + 19y - 47z = 32

Answer: c
Explanation: In this Question we have,
\(\begin{bmatrix}2&-3&7\\3&1&-3\\2&19&-47\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}5\\13\\32\end{bmatrix}\)
By R₁-R₂
\(\begin{bmatrix}-1&-4&10\\3&1&-3\\2&19&-47\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}-8\\13\\32\end{bmatrix}\)
By -R₁
\(\begin{bmatrix}1&4&-10\\3&1&-3\\2&19&-47\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}8\\13\\32\end{bmatrix}\)
By R₂-3R₁ and R₃-2R₁
\(\begin{bmatrix}1&4&-10\\0&-11&27\\0&11&-27\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}8\\-11\\16\end{bmatrix}\)
By R₃+R₂
\(\begin{bmatrix}1&4&-10\\0&-11&27\\0&0&0\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}8\\-11\\5\end{bmatrix}\)
Here we see that the rank of the co-efficient matrix is 2, while the rank of the augmented matrix is 3.
Since the two ranks are not equal, the given system of equations is inconsistent.