If A =\( \begin{bmatrix} 1 & 0 & 0 \\[0.3em] 2 & 1 & 0 \\[0.3em] 3 &3

1.	If A =\( \begin{bmatrix} 1 & 0 & 0 \\[0.3em] 2 & 1 & 0 \\[0.3em] 3 &3 &1 \end{bmatrix}\)then reduce it to I3 by using column transformations.[1,0,0,2,1,0,3,3,1]
Answer» Given, [1,0,0,2,1,0,3,3,1] \|A\| =\( \begin{vmatrix}1 & 0 & 0 \\[0.3em]2 & 1 & 0 \\[0.3em]3 &3 &1\end{vmatrix}\) = 1(1 – 0) – 0 + 0 = 1 ≠ 0 ∴ A is a non-singular matrix. Hence, The required transformation is possible. Now, A = \( \begin{bmatrix}1 & 0 & 0 \\[0.3em]2 & 1 & 0 \\[0.3em]3 &3 &1\end{bmatrix}\) By C₁ – 2C₂, we get, A ~\( \begin{bmatrix}1 & 0 & 0 \\[0.3em]0 & 1 & 0 \\[0.3em]-3 &3 &1\end{bmatrix}\) By C₁ + 3C₃ and C₂ – 3C₃, we get, A ~\( \begin{bmatrix}1 & 0 & 0 \\[0.3em]0 & 1 & 0 \\[0.3em]0 &0 &1\end{bmatrix}\) = I₃.

If A =\( \begin{bmatrix} 1 & 0 & 0 \\[0.3em] 2 & 1 & 0 \\[0.3em] 3 &3 &1 \end{bmatrix}\)then reduce it to I3 by using column transformations.[1,0,0,2,1,0,3,3,1]

Answer»

Given,

[1,0,0,2,1,0,3,3,1]

|A| =\( \begin{vmatrix}1 & 0 & 0 \\[0.3em]2 & 1 & 0 \\[0.3em]3 &3 &1\end{vmatrix}\)

= 1(1 – 0) – 0 + 0 = 1 ≠ 0

∴ A is a non-singular matrix.

Hence,

The required transformation is possible.

Now,

A = \( \begin{bmatrix}1 & 0 & 0 \\[0.3em]2 & 1 & 0 \\[0.3em]3 &3 &1\end{bmatrix}\)

By C₁ – 2C₂, we get,

A ~\( \begin{bmatrix}1 & 0 & 0 \\[0.3em]0 & 1 & 0 \\[0.3em]-3 &3 &1\end{bmatrix}\)

By C₁ + 3C₃ and C₂ – 3C₃, we get,

A ~\( \begin{bmatrix}1 & 0 & 0 \\[0.3em]0 & 1 & 0 \\[0.3em]0 &0 &1\end{bmatrix}\) = I₃.

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If A =\( \begin{bmatrix} 1 &amp; 0 &amp; 0 \\[0.3em] 2 &amp; 1 &amp; 0 \\[0.3em] 3 &amp;3 &amp;1 \end{bmatrix}\)then reduce it to I3 by using column transformations.[1,0,0,2,1,0,3,3,1]

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If A =\( \begin{bmatrix} 1 & 0 & 0 \\[0.3em] 2 & 1 & 0 \\[0.3em] 3 &3 &1 \end{bmatrix}\)then reduce it to I3 by using column transformations.[1,0,0,2,1,0,3,3,1]