Tabansi

2020-11-14

To calculate : The reduced form of the provided matrix, $\left[\begin{array}{cccc}1& 0& 4& 0\\ 0& 1& -3& -1\\ 0& 0& -2& 2\end{array}\right]$ with the use of row operations.

yunitsiL

Skilled2020-11-15Added 108 answers

Calculation:

Consider the matrix,

$\left[\begin{array}{cccc}1& 0& 4& 0\\ 0& 1& -3& -1\\ 0& 0& -2& 2\end{array}\right]$

Now, for a matrix to be in the reduced form, it must have the following properies.

1. The first non-zero element in each row must be 1, which is also known as the leading entry.

2. Each leading entry is in a column to the right of the entry in the previous row of the matrix.

3. If there are any rows with all zero elements, then they must be below the rows having a non-zero element.

Make the zeros in column 1 except the leading entry in row 1 and column 1. But here, it is observed that all the elements are already zero.

$\left[\begin{array}{cccc}1& 0& 4& 0\\ 0& 1& -3& -1\\ 0& 0& -2& 2\end{array}\right]$

Similarly, make the zeros in column 2 except the leading entry in row 2 and column 2.

But here, it is observed that all the elements are already zero.

$\left[\begin{array}{cccc}1& 0& 4& 0\\ 0& 1& -3& -1\\ 0& 0& -2& 2\end{array}\right]$

Then make the zeros in row 1 column 3 and row 2 column 3, except the leading entry

in row 3 and column 3 with the of the operation,${R}_{1}\to {R}_{1}+2{R}_{3}.$

$\left[\begin{array}{cccc}1& 0& 4& 0\\ 0& 1& -3& -1\\ 0& 0& -2& 2\end{array}\right]{R}_{1}\to {R}_{1}+2{R}_{3}\left[\begin{array}{cccc}1& 0& 0& 4\\ 0& 1& -3& -1\\ 0& 0& -2& 2\end{array}\right]$

Similarly, use the operations${R}_{3}\to \frac{{R}_{3}}{-2}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}{R}_{2}\to {R}_{2}+3\left({R}_{3}\right)$ to get the desired result as,

$\left[\begin{array}{cccc}1& 0& 0& 4\\ 0& 1& -3& -1\\ 0& 0& -2& 2\end{array}\right]{R}_{3}\to \frac{{R}_{3}}{-2}\left[\begin{array}{cccc}1& 0& 0& 4\\ 0& 1& -3& -1\\ 0& 0& 1& -1\end{array}\right]$

$\left[\begin{array}{cccc}1& 0& 0& 4\\ 0& 1& -3& -1\\ 0& 0& 1& -1\end{array}\right]{R}_{2}\to {R}_{2}+3{R}_{3}\left[\begin{array}{cccc}1& 0& 0& 4\\ 0& 1& 0& -4\\ 0& 0& 1& -1\end{array}\right]$

The matrix obtained above by performing certain row transformation satisfies all the properties of the reduced form of a matrix.

Hence, the reduced form of the matrix$\left[\begin{array}{cccc}1& 0& 4& 0\\ 0& 1& -3& -1\\ 0& 0& -2& 2\end{array}\right]$ is

$\left[\begin{array}{cccc}1& 0& 0& 4\\ 0& 1& 0& -4\\ 0& 0& 1& -1\end{array}\right].$

Consider the matrix,

Now, for a matrix to be in the reduced form, it must have the following properies.

1. The first non-zero element in each row must be 1, which is also known as the leading entry.

2. Each leading entry is in a column to the right of the entry in the previous row of the matrix.

3. If there are any rows with all zero elements, then they must be below the rows having a non-zero element.

Make the zeros in column 1 except the leading entry in row 1 and column 1. But here, it is observed that all the elements are already zero.

Similarly, make the zeros in column 2 except the leading entry in row 2 and column 2.

But here, it is observed that all the elements are already zero.

Then make the zeros in row 1 column 3 and row 2 column 3, except the leading entry

in row 3 and column 3 with the of the operation,

Similarly, use the operations

The matrix obtained above by performing certain row transformation satisfies all the properties of the reduced form of a matrix.

Hence, the reduced form of the matrix

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