Linear Algebra Questions and Answers - Rank of Matrix in PAQ and Normal Form

This set of Engineering Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Rank of Matrix in PAQ and Normal Form”.

1. Reduce the given matrix to normal form, and hence find its rank.
\(\begin{bmatrix}1& 2& -2& 3\\ 2& 5& -4& 6\\ -1& -3& 2& -2\\ 2& 4& -1& 6\end{bmatrix}\)
a) 1
b) 4
c) 3
d) 2
View Answer

Answer: b
Explanation: In this Question we have,
A=\(\begin{bmatrix}1& 2& -2& 3\\ 2& 5& -4& 6\\ -1& -3& 2& -2\\ 2& 4& -1& 6\end{bmatrix}\)
By R₂-2R₁ and R₃+R₁ and R₄-2R₁
A=\(\begin{bmatrix}1& 2& -2& 3\\ 0& 1& 0 & 0\\ 0& -1& 0 & 1\\ 0& 0 & 3& 0\end{bmatrix}\)
By C₂-2C₁, C₃+2C₁, C₄-3C₁
A=\(\begin{bmatrix}1& 0& 0 & 0\\ 0& 1 & 0 & 0\\ 0& -1 & 0 &1 \\ 0& 0 & 3 &0 \end{bmatrix}\)
By R₂+R₃
A=\(\begin{bmatrix}1& 0 & 0 &0 \\0 & 1 & 0 & 0\\0 & 0 & 0& 1\\ 0& 0 & 3 & 0\end{bmatrix}\)
By C₃₄
A=\(\begin{bmatrix}1& 0 & 0 &0 \\0 & 1 & 0 &0 \\ 0& 0 & 1 & 0\\ 0& 0 & 0 & 3\end{bmatrix}\)
By \(\frac{1}{3}\) C₄
A=\(\begin{bmatrix}1& 0 & 0 &0 \\0 & 1 & 0 &0 \\ 0& 0 & 1 & 0\\ 0& 0 & 0 & 1\end{bmatrix}\)
This is in Normal Form.
Hence the rank of the given Matrix is 4.

2. Find the value of p for which, the rank of the given matrix is 1.
\(\begin{bmatrix}3&p&p\\p&3&p\\p&p&3\end{bmatrix}\)
a) 4
b) 2
c) 3
d) 1
View Answer

Answer: c
Explanation: In this Question we have,
A=\(\begin{bmatrix}3&p&p\\p&3&p\\p&p&3\end{bmatrix}\)
If we put the value of p=3,
The matrix becomes:
A=\(\begin{bmatrix}3&3&3\\3&3&3\\3&3&3\end{bmatrix}\)
By R₂-R₁ and R₃-R₁
A=\(\begin{bmatrix}3&3&3\\0&0&0\\0&0&0\end{bmatrix}\)
Thus, the rank of the given matrix is 1.
Therefore, the value of p must be 3.

3. Find the value of non singular matrices P and Q, such that PAQ is in the normal form, where A is
\(\begin{bmatrix}1&1&2\\1&2&3\\0&-1&-1\end{bmatrix}\)

a) \(\begin{bmatrix}1&0&0\\-1&1&0\\-1&1&1\end{bmatrix}\), \(\begin{bmatrix}1&-1&-1\\0&1&-1\\0&0&1\end{bmatrix}\)
b)\(\begin{bmatrix}1&1&0\\-1&1&0\\-1&1&1\end{bmatrix}\), \(\begin{bmatrix}1&-1&-1\\0&1&-1\\0&0&1\end{bmatrix}\)
c)\(\begin{bmatrix}1&0&0\\-1&1&0\\-1&1&1\end{bmatrix}\), \(\begin{bmatrix}1&-1&1\\0&1&-1\\0&0&1\end{bmatrix}\)
d)\(\begin{bmatrix}1&0&0\\-1&1&0\\-1&0&1\end{bmatrix}\), \(\begin{bmatrix}1&-1&-1\\0&1&-1\\0&0&-1\end{bmatrix}\)
View Answer

Answer: a
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\(\begin{bmatrix}1&-1&-2\\0&1&0\\0&0&1\end{bmatrix}\)

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

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