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Chapter:

Laplace-Transform

1. Which of following matrix is Skew Symmetric?

\(\begin{bmatrix}0&1\\-1&0\end{bmatrix}\) \(\begin{bmatrix}0&1\\1&0\end{bmatrix}\) \(\begin{bmatrix}0&3\\-1&9\end{bmatrix}\) \(\begin{bmatrix}8&-2\\-1&3\end{bmatrix}\)

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

We know for a Skew Symmetric Matrix,
A=-A^T
So for option \(\begin{bmatrix}0&1\\-1&0\end{bmatrix}\),
A^T=\(\begin{bmatrix}0&-1\\1&0\end{bmatrix}\)=(-1)\(\begin{bmatrix}0&1\\-1&0\end{bmatrix}\)=A.

Hint Answer

2. Which of following matrix are singular?

\(\begin{bmatrix}31&12\\26&8\end{bmatrix}\) \(\begin{bmatrix}1&11\\2&8\end{bmatrix}\) \(\begin{bmatrix}13&12\\22&8\end{bmatrix}\) \(\begin{bmatrix}3&12\\2&8\end{bmatrix}\)

\(\begin{bmatrix}3&12\\2&8\end{bmatrix}\)

We know that,
A singular matrix is one whose determinant = 0
\(\begin{bmatrix}3&12\\2&8\end{bmatrix}\)
Determinant=24-24=0
Thus it is a singular matrix.

Hint Answer

3. If \(A = \begin{bmatrix}
a+ib & c+id\\
-c+id & a-ib\\
\end{bmatrix} and \, a^2+b^2+c^2+d^2=1\) n find A^-1.

\(\begin{bmatrix}
a+ib & -c+id\\
-a+id & a-ib\\
\end{bmatrix} \) \(\begin{bmatrix}
a-ib & c-id\\
-c-id & a+ib\\
\end{bmatrix} \) \(\begin{bmatrix}
a-ib & -c-id\\
c-id & a+ib\\
\end{bmatrix} \) \(\begin{bmatrix}

\(\begin{bmatrix}
a-ib & -c-id\\
c-id & a+ib\\
\end{bmatrix} \)

We have
\(|A| = (a+ib) (a-ib) – (-c+id) (c+id) \)
\( = a^2+b^2+c^2+d^2 \)
\(s=1 \)
Also \(adj (A) = \begin{bmatrix}
a-ib & -c-id\\
c-id & a+ib\\
\end{bmatrix} \)
? A^-1 = \(\frac{1}{|A|} .adj (A) = \frac{1}{1} \begin{bmatrix}
a-ib & -c-id\\
c-id & a+ib\\
\end{bmatrix} \)
\(\begin{bmatrix}
a-ib & -c-id\\
c-id & a+ib\\
\end{bmatrix} \).

Hint Answer

4. The inverse of a symmetric matrix (if it exists) is?

A symmetric matrix A skew symmetric matrix A diagonal matrix A triangular matrix

A symmetric matrix

Let A be an invertible matrix.
We have AA^-1 = A^-1A = I_N
(AA^-1)? = (A^-1A)? = (I_N)?
(A^-1)? A? = A? (A^-1)? = I_N
(A^-1)? A = A (A^-1)? = I_N
(A^-1)? = A^-1 (inverse of a matrix is unique).

Hint Answer

5. Find rank of matrix \(\begin{bmatrix}
4 & 2 & -1 & 2\\
1 & -1 & 2 & 1\\
2 & 2 & -2 & 0\\
\end{bmatrix} \)?

0 1 2 3

3

Say,
A = \(\begin{bmatrix}
4 & 2 & -1 & 2\\
1 & -1 & 2 & 1\\
2 & 2 & -2 & 0\\
\end{bmatrix} \)
\( ~ \begin{bmatrix}
1 & -1 & 2 & 1\\
4 & 2 & -1 & 2\\
2 & 2 & -2 & 0\\
\end{bmatrix} \)R₂↔R₁
\( ~ \begin{bmatrix}
1 & -1 & 2 & 1\\
0 & 6 & -9 & -2\\
0 & 4 & -6 & -2\\
\end{bmatrix}\) R₂-4R₁, R₃-2R₁
\( ~ \begin{bmatrix}
1 & -1 & 2 & 1\\
0 & 6 & -9 & -2\\
0 & 0 & 0 & -4\\
\end{bmatrix}\) 6R₃-4R₂
\(? ?(A)= 3.\)

Hint Answer

6. Rank of following matrix is A = \(\begin{bmatrix}
3 & 2 & -1\\
4 & 2 & 6\\
7 & 4 & 5\\
\end{bmatrix} \) is?

1 2 3 0

2

\(|A| = 3(10-24) + 2(42-20) – (16-14) = 10\)
\(? |A| = 0\)
\( ~ \begin{bmatrix}
3 & 2\\
4 & 2\\
\end{bmatrix} ?0 = -2 \)
\(? ? (A) = 2.\)

Hint Answer

7. If every minor of order ‘r’ of a matrix is zero n ρ (A) =?

>r =r ≤r

By definition of ?Rank of a matrix?
A matrix is said to have rank ?r? if
(i) At least one minor of order r is non-zero
(ii) All minors of order r+1 is zero
? The given matrix (ii) condition seems to be applied
Hence, rank of matrix ? (A) = < r.

Hint Answer

8. Find inverse of matrix by using Cayley Hamilton Theorem.
A = \( ~ \begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 1\\
0 & -2 & 4\\
\end{bmatrix} \)

\(\frac{1}{6} \begin{bmatrix}
1 & 0 & 0\\
0 & -1 & -1\\
0 & 2 & -4\\
\end{bmatrix} \) \(\frac{1}{6} \begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 1\\
0 & 2 & 4\\
\end{bmatrix} \) \(\frac{1}{6} \begin{bmatrix}
1 & 0 & 0\\
0 & -1 & -1\\
0 & -2 & -4\\
\end{bmatrix} \) \(\frac{1}{6} \begin{bmatrix}

\(\frac{1}{6} \begin{bmatrix}
1 & 0 & 0\\
0 & -1 & -1\\
0 & 2 & -4\\
\end{bmatrix} \)

By Cayley Hamliton Theorem
We have ?³ – S₁ ?² + S₂ ? – |A| = 0 characteristics equation of matrix A
Also A satisfies above equation according to orem.
? A³– S₁A²+ S₂A – 6 = 0……… (i) Where S₁=6, S₂=6 & |A|=6
? A³ – 6A² + 6A – 6 = 0
Multiplying equation (i) by A^-1
A^-1 (A³) – 6 A^-1 (A²) + A^-1 (A) – 6 A^-1 = 0
=> A² – 6A + 6I – 6 A^-1 = \(\frac{1}{6}\) A^-1
=> A^-1 = \(\frac{1}{6} \Bigg\{\begin{bmatrix}
1 & 0 & 0\\
0 & -1 & 5\\
0 & -10 & 14\\
\end{bmatrix} – \begin{bmatrix}
6 & 0 & 0\\
0 & 6 & 6\\
0 & -12 & -24\\
\end{bmatrix} + \begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1\\
\end{bmatrix} \Bigg\} \)
\(? A^{-1} = \frac{1}{6} \begin{bmatrix}
1 & 0 & 0\\
0 & -1 & -1\\
0 & 2 & -4\\
\end{bmatrix} \).

Hint Answer

9. Find value of p for which, rank of given matrix is 1.
\(\begin{bmatrix}3&p&p\\p&3&p\\p&p&3\end{bmatrix}\)

4 2 3 1

3

In this Question we have,
A=\(\begin{bmatrix}3&p&p\\p&3&p\\p&p&3\end{bmatrix}\)
If we put 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, rank of given matrix is 1.
Therefore, value of p must be 3.

Hint Answer

10. Which of following Linear Transformations is not correct for given matrix?
\(\begin{bmatrix}1&2&-3\\-1&0&1\\2&1&0\end{bmatrix}\)

x₁=1y₁-3y₂-3y₃ x₁=1y₁-2y₂-3y₃ x₂=-1y₁+1y₃ x₃=2y₁+y₂

x₁=1y₁-3y₂-3y₃

In given question,
X=\(\begin{bmatrix}1&2&-3\\-1&0&1\\2&1&0\end{bmatrix}\)Y
Thus,
x₁=1y₁-2y₂-3y₃
x₂=-1y₁+1y₃
x₃=2y₁+y₂.

Hint Answer

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Topics

This Chapter Laplace-Transform consists of the following topics

Matrices

Which of following matrix is Skew Symmetric?

Which of following matrix are singular?

If \(A = \begin{bmatrix}
a+ib & c+id\\
-c+id & a-ib\\
\end{bmatrix} and \, a^2+b^2+c^2+d^2=1\) n find A^-1.

The inverse of a symmetric matrix (if it exists) is?

Find rank of matrix \(\begin{bmatrix}
4 & 2 & -1 & 2\\
1 & -1 & 2 & 1\\
2 & 2 & -2 & 0\\
\end{bmatrix} \)?

Rank of following matrix is A = \(\begin{bmatrix}
3 & 2 & -1\\
4 & 2 & 6\\
7 & 4 & 5\\
\end{bmatrix} \) is?

If every minor of order ‘r’ of a matrix is zero n ρ (A) =?

Find inverse of matrix by using Cayley Hamilton Theorem.
A = \( ~ \begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 1\\
0 & -2 & 4\\
\end{bmatrix} \)

Find value of p for which, rank of given matrix is 1.
\(\begin{bmatrix}3&p&p\\p&3&p\\p&p&3\end{bmatrix}\)

Which of following Linear Transformations is not correct for given matrix?
\(\begin{bmatrix}1&2&-3\\-1&0&1\\2&1&0\end{bmatrix}\)

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