Question

Reduce matrix in echelon form and find its rank​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Given:      

Let the given matrix be A = $\left[\begin{array}{cccc}1^{2} &2^{2} &3^{2}&4^{2} \\2^{2} &3^{2} &4^{2}&5^{2} \\3^{2} &4^{2} &5^{2}&6^{2}\\4^{2}&5^{2}&6^{2}&7^{2} \end{array}\right]$

To find:

We have to find the echelon form of the given matrix and the rank of the given matrix

Solution:

In echelon form only row operations are performed

A = $\left[\begin{array}{cccc}1&4&9&16\\4&9&16&25\\9&16&25&36\\16&25&36&49\end{array}\right]$

Now R₂→R₂-4R₁ , R₃→R₃-9R₁ and R₄→R₄-16R₁

A = $\left[\begin{array}{cccc}1&4&9&16\\0&-7&-20&-39\\0&-20&-56&-108\\0&-39&-108&-177\end{array}\right]$

      R₃→7R₃-20R₂ ,R₄→7R₄-39R₂

A = $\left[\begin{array}{cccc}1&4&9&16\\0&-7&-20&-39\\0&0&8&-14\\0&0&14&282\end{array}\right]$

Now  R₄→4R₄-7R₃

A =  $\left[\begin{array}{cccc}1&4&9&16\\0&-7&-20&-39\\0&0&8&-14\\0&0&0&1226\end{array}\right]$

Now R₂→R₂/-7    R₃→R₃/8    R₄→R₄/1226

A =  $\left[\begin{array}{cccc}1&4&9&16\\0&1&\frac{20}{7} &\frac{39}{7} \\0&0&1&\frac{-7}{4} \\0&0&0&1\end{array}\right]$

Hence this is the ECHELON form of given matrix  $\left[\begin{array}{cccc}1&4&9&16\\0&1&\frac{20}{7} &\frac{39}{7} \\0&0&1&\frac{-7}{4} \\0&0&0&1\end{array}\right]$

In the above form number of non zero rows = 4

∴The rank of the given matrix = 4