Reduce the following matrices to row echelon form and row reduced echelon forms:(i)begin{bmatrix}1 & p & -1 2 & 1 & 7 -3 & 3 & 2 end{bmatrix}(ii) begin{bmatrix}p & -1 & 7&2 2 & 1 & -5 & 3 1 & 3 & 2 & 0 end{bmatrix}Find also the ra of these matrices.Notice:p=4

Kaycee Roche 2021-02-22 Answered

Reduce the following matrices to row echelon form and row reduced echelon forms:
$(i) [\begin{matrix} 1 & p & - 1 \\ 2 & 1 & 7 \\ - 3 & 3 & 2 \end{matrix}] (i i) [\begin{matrix} p & - 1 & 7 & 2 \\ 2 & 1 & - 5 & 3 \\ 1 & 3 & 2 & 0 \end{matrix}]$
*Find also the ra
of these matrices.
*Notice: $p = 4$

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Expert Answer

mhalmantus

Answered 2021-02-23 Author has 106 answers

Step 1
i) Let the matrix be,

A = [\begin{matrix} 1 & 4 & - 1 \\ 2 & 1 & 7 \\ - 3 & 3 & 2 \end{matrix}]

Apply

R_{2} \to R_{2} - 2 \cdot R_{1}

\Rightarrow [\begin{matrix} 1 & 4 & - 1 \\ 0 & - 7 & 9 \\ - 3 & 3 & 2 \end{matrix}]

Apply

R_{3} \to R_{3} + 3 \cdot R_{1}

\Rightarrow [\begin{matrix} 1 & 4 & - 1 \\ 0 & - 7 & 9 \\ 0 & 15 & - 1 \end{matrix}]

Step 2
Apply

R_{3} \to R_{3} + \frac{15}{7} R_{2}

\Rightarrow [\begin{matrix} 1 & 4 & - 1 \\ 0 & - 7 & 9 \\ 0 & 0 & \frac{128}{7} \end{matrix}]

Apply

R_{2} \to - \frac{1}{7} \cdot R_{2}

\Rightarrow [\begin{matrix} 1 & 4 & - 1 \\ 0 & 1 & - \frac{7}{9} \\ 0 & 0 & \frac{128}{7} \end{matrix}]

Step 3
Apply

R_{3} \to \frac{7}{128} \cdot R_{3}

\Rightarrow [\begin{matrix} 1 & 4 & - 1 \\ 0 & 1 & - \frac{7}{9} \\ 0 & 0 & 1 \end{matrix}]

This is the Row Echelon Form of the matrix.
Also, the rank of this matrix A is 3.
Apply

R_{1} \to R_{1} - 4 R_{2}

\Rightarrow [\begin{matrix} 1 & 0 & \frac{19}{9} \\ 0 & 1 & - \frac{7}{9} \\ 0 & 0 & 1 \end{matrix}]

Step 4
Apply

R_{1} \to R_{1} - \frac{19}{9} R_{3}

\Rightarrow [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & - \frac{7}{9} \\ 0 & 0 & 1 \end{matrix}]

Apply

R_{2} \to R_{2} + \frac{7}{9} R_{3}

\Rightarrow [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]

This is the Reduced Row Echelon Form of the given matrix.
Step 5
ii) Let the matrix be represented as,

B = [\begin{matrix} 4 & - 1 & 7 & 2 \\ 2 & 1 & - 5 & 3 \\ 1 & 3 & 2 & 0 \end{matrix}]

Apply

R_{1} \leftrightarrow R_{3}

\Rightarrow [\begin{matrix} 1 & 3 & 2 & 0 \\ 2 & 1 & - 5 & 3 \\ 4 & - 1 & 7 & 2 \end{matrix}]

Apply

R_{2} \to R_{2} - 2 R_{1}

\Rightarrow [\begin{matrix} 1 & 3 & 2 & 0 \\ 0 & - 5 & - 9 & 3 \\ 4 & - 1 & 7 & 2 \end{matrix}]

Step 6
Apply

R_{3} \to R_{3} - 4 R_{1}

\Rightarrow [\begin{matrix} 1 & 3 & 2 & 0 \\ 0 & - 5 & - 9 & 3 \\ 0 & - 13 & - 1 & 2 \end{matrix}]

Apply

R_{2} \to \frac{- 1}{5} R_{2}

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Jeffrey Jordon

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