Question

a) \(\begin{bmatrix}1&0&0\\-1&1&0\\-1&1&1\end{bmatrix}\),  \(\begin{bmatrix}1&-1&-1\\0&1&-1\\0&0&1\end{bmatrix}\)

(a) \(\begin{bmatrix}1&1&0\\-1&1&0\\-1&1&1\end{bmatrix}\),  \(\begin{bmatrix}1&-1&-1\\0&1&-1\\0&0&1\end{bmatrix}\)

(b) \(\begin{bmatrix}1&0&0\\-1&1&0\\-1&1&1\end{bmatrix}\),  \(\begin{bmatrix}1&-1&1\\0&1&-1\\0&0&1\end{bmatrix}\)

(c) \(\begin{bmatrix}1&0&0\\-1&1&0\\-1&0&1\end{bmatrix}\),  \(\begin{bmatrix}1&-1&-1\\0&1&-1\\0&0&-1\end{bmatrix}\)

I got this question by my college professor while I was bunking the class.

This interesting question is from Rank of Matrix in PAQ and Normal Form topic in section Matrices of Engineering Mathematics

Answer 1

The correct answer is (a) \(\begin{bmatrix}1&1&0\\-1&1&0\\-1&1&1\end{bmatrix}\),  \(\begin{bmatrix}1&-1&-1\\0&1&-1\\0&0&1\end{bmatrix}\)

For explanation: In this Question we have,

A=\(\begin{bmatrix}1&1&2\\1&2&3\\0&-1&-1\end{bmatrix}\)

Let A=I3AI3

 

\(\begin{bmatrix}1&1&2\\1&2&3\\0&-1&-1\end{bmatrix}\)=\(\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\)A\(\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\)

By C2-C1 and C3-2C1

        

\(\begin{bmatrix}1&0&0\\1&1&1\\0&-1&-1\end{bmatrix}\)=\(\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\)A\(\begin{bmatrix}1&-1&-2\\0&1&0\\0&0&1\end{bmatrix}\)

By R2-R1

\(\begin{bmatrix}1&0&0\\0&1&1\\0&-1&-1\end{bmatrix}\)=\(\begin{bmatrix}1&0&0\\-1&1&0\\0&0&1\end{bmatrix}\)A\(\begin{bmatrix}1&-1&-2\\0&1&0\\0&0&1\end{bmatrix}\)

By C3-C2

\(\begin{bmatrix}1&0&0\\0&1&0\\0&-1&0\end{bmatrix}\)=\(\begin{bmatrix}1&0&0\\-1&1&0\\0&0&1\end{bmatrix}\)A\(\begin{bmatrix}1&-1&-1\\0&1&-1\\0&0&1\end{bmatrix}\)

By R3+R2

\(\begin{bmatrix}1&0&0\\0&1&0\\0&0&0\end{bmatrix}\)=\(\begin{bmatrix}1&0&0\\-1&1&0\\-1&1&1\end{bmatrix}\)A\(\begin{bmatrix}1&-1&-1\\0&1&-1\\0&0&1\end{bmatrix}\)

This is in Normal Form.

\(\begin{bmatrix}I2&0\\0&0\end{bmatrix}\)=PAQ

P=\(\begin{bmatrix}1&0&0\\-1&1&0\\-1&1&1\end{bmatrix}\) and

Q=\(\begin{bmatrix}1&-1&-1\\0&1&-1\\0&0&1\end{bmatrix}\).