Engineering Mathematics Assessment Questions

This set of Engineering Mathematics Assessment Questions and Answers 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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