Question:

Find 1​/2(A+$\begin{array}{}{A}^{\mathrm{T}}\end{array}$) and 1​/2(A−$\begin{array}{}{A}^{\mathrm{T}}\end{array}$), when A=$\left[\begin{array}{ccc}0& \mathrm{a}& \mathrm{b}\\ \mathrm{-a}& 0& \mathrm{c}\\ \mathrm{-b}& \mathrm{-c}& 0\end{array}\right]$

Answer:

1. Find A^T:

A^T = $\left[\begin{array}{ccc}0& \mathrm{-a}& \mathrm{-b}\\ \mathrm{a}& 0& \mathrm{c}\\ \mathrm{b}& \mathrm{-c}& 0\end{array}\right]$

2. Find 1/2(A + A^T):

1/2(A + A^T) = $\left[\begin{array}{ccc}0& 0& 0\\ 0& \mathrm{a}& \mathrm{c}\\ 0& \mathrm{c}& 0\end{array}\right]$

3. Find 1/2(A - A^T):

1/2(A - A^T) = $\left[\begin{array}{ccc}0& \mathrm{a}& \mathrm{b}\\ \mathrm{-a}& 0& \mathrm{-c}\\ \mathrm{-b}& \mathrm{c}& 0\end{array}\right]$

Question:

If A=$\left[\begin{array}{ccc}-1& 2& 3\\ 5& 7& 9\\ -2& 1& 1\end{array}\right]$ and B=$\left[\begin{array}{ccc}-4& 1& -5\\ 1& 2& 0\\ 1& 3& 1\end{array}\right]$, then verify that (i)$\begin{array}{}{\mathrm{(A+B)}}^{\mathrm{T}}\end{array}$=$\begin{array}{}{A}^{\mathrm{T}}\end{array}$+$\begin{array}{}{B}^{\mathrm{T}}\end{array}$ (ii)$\begin{array}{}{\mathrm{(A-B)}}^{\mathrm{T}}\end{array}$=$\begin{array}{}{A}^{\mathrm{T}}\end{array}$−$\begin{array}{}{B}^{\mathrm{T}}\end{array}$

Answer:

(i) A+B = $\left[\begin{array}{ccc}-5& 3& -2\\ 6& 9& 9\\ -1& 4& 2\end{array}\right]$

(A+B)^T = $\left[\begin{array}{ccc}-5& 6& -1\\ 3& 9& 4\\ -2& 9& 2\end{array}\right]$

A^T = $\left[\begin{array}{ccc}-1& 5& -2\\ 2& 7& 1\\ 3& 9& 1\end{array}\right]$

B^T = $\left[\begin{array}{ccc}-4& 1& 1\\ 1& 2& 3\\ -5& 0& 1\end{array}\right]$

A^T + B^T = $\left[\begin{array}{ccc}-5& 6& -1\\ 3& 9& 4\\ -2& 9& 2\end{array}\right]$

Therefore, (A+B)^T = A^T + B^T

(ii) A-B = $\left[\begin{array}{ccc}3& 1& 8\\ 4& 5& 9\\ -3& -2& 0\end{array}\right]$

(A-B)^T = $\left[\begin{array}{ccc}3& 4& -3\\ 1& 5& -2\\ 8& 9& 0\end{array}\right]$

A^T = $\left[\begin{array}{ccc}-1& 5& -2\\ 2& 7& 1\\ 3& 9& 1\end{array}\right]$

B^T = $\left[\begin{array}{ccc}-4& 1& 1\\ 1& 2& 3\\ -5& 0& 1\end{array}\right]$

A^T - B^T = $\left[\begin{array}{ccc}3& 4& -3\\ 1& 5& -2\\ 8& 9& 0\end{array}\right]$

Therefore, (A-B)^T = A^T - B^T

Question:

If (i) A=$\left[\begin{array}{cc}\mathrm{cos\alpha }& \mathrm{sin\alpha }\\ \mathrm{-sin\alpha }& \mathrm{cos\alpha }\end{array}\right]$, then verify that A′A=I (ii) A=$\left[\begin{array}{cc}\mathrm{sin\alpha }& \mathrm{cos\alpha }\\ \mathrm{-cos\alpha }& \mathrm{sin\alpha }\end{array}\right]$, then verify that A′A=I

Answer:

(i) A’A = $\left[\begin{array}{cc}\mathrm{cos²\alpha }& \mathrm{sin\alpha cos\alpha }\\ \mathrm{sin\alpha cos\alpha }& \mathrm{sin²\alpha }\end{array}\right]$

Since cos2α + sin2α = 1, A’A = $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

which is the identity matrix I.

(ii) A’A = $\left[\begin{array}{cc}\mathrm{sin²\alpha }& \mathrm{sin\alpha cos\alpha }\\ \mathrm{sin\alpha cos\alpha }& \mathrm{cos²\alpha }\end{array}\right]$

Since sin2α + cos2α = 1, A’A = $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

which is the identity matrix I.

Question:

Find the transpose of each of the following matrices: (i) $\left[\begin{array}{c}5\\ \mathrm{1/2}\\ -1\end{array}\right]$ (ii) $\left[\begin{array}{cc}1& -1\\ 2& 3\end{array}\right]$ (iii) $\left[\begin{array}{ccc}-1& 5& 6\\ \mathrm{\surd 3}& 5& 6\\ 2& 3& -1\end{array}\right]$

Answer:

(i) $\left[\begin{array}{ccc}5& \mathrm{1/2}& -1\end{array}\right]$

(ii) $\left[\begin{array}{cc}1& 2\\ -1& 3\end{array}\right]$

(iii) $\left[\begin{array}{ccc}-1& \mathrm{\surd 3}& 2\\ 5& 5& 3\\ 6& 6& -1\end{array}\right]$

Question:

If A=$\left[\begin{array}{cc}\mathrm{cos\alpha }& \mathrm{-sin\alpha }\\ \mathrm{sin\alpha }& \mathrm{cos\alpha }\end{array}\right]$, then A+A′=I, if the value of α is A π​/6 B π​/3 C n D 3π/2

Answer:

A π​/6

A=$\left[\begin{array}{cc}\cos \pi /6& \mathrm{-sin}\pi /6\\ \sin \pi /6& \cos \pi /6\end{array}\right]$

A′=$\left[\begin{array}{cc}\cos \pi /6& \sin \pi /6\\ \mathrm{-sin}\pi /6& \cos \pi /6\end{array}\right]$

A+A′=$\left[\begin{array}{cc}2\cos \pi /6& 0\\ 0& 2\cos \pi /6\end{array}\right]$

A+A′=$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

Therefore, A+A′=I and the value of α is A π​/6.

Question:

For the matrices A and B, verify that (AB)′=B′A′ where (i) A=$\left[\begin{array}{c}1\\ -4\\ 3\end{array}\right]$,B=$\left[\begin{array}{ccc}-1& 2& 1\end{array}\right]$ (ii) A=$\left[\begin{array}{c}0\\ 1\\ 2\end{array}\right]$,B=$\left[\begin{array}{ccc}1& 5& 7\end{array}\right]$

Answer:

(i) A=$\left[\begin{array}{c}1\\ -4\\ 3\end{array}\right]$, B=$\left[\begin{array}{ccc}-1& 2& 1\end{array}\right]$

Step 1: Calculate AB

$\left[\begin{array}{c}-1\\ -8\\ 5\end{array}\right]$

Step 2: Calculate (AB)′

$\left[\begin{array}{ccc}-1& -8& 5\end{array}\right]$

Step 3: Calculate B′

$\left[\begin{array}{ccc}-1& 2& 1\end{array}\right]$

Step 4: Calculate B′A′

$\left[\begin{array}{ccc}1& -4& 3\end{array}\right]$

Therefore, (AB)′=B′A′.

(ii) A=$\left[\begin{array}{c}0\\ 1\\ 2\end{array}\right]$, B=$\left[\begin{array}{ccc}1& 5& 7\end{array}\right]$

Step 1: Calculate AB

$\left[\begin{array}{c}5\\ 7\\ 17\end{array}\right]$

Step 2: Calculate (AB)′

Question:

For the matrix A=$\left[\begin{array}{cc}1& 5\\ 6& 7\end{array}\right]$, verify that (i) (A+A′) is a symmetric matrix (ii) (A−A′) is a skew symmetric matrix.

Answer:

(i) To verify that (A+A′) is a symmetric matrix, we need to show that it is equal to its transpose.

A+A′ = $\left[\begin{array}{cc}1& 5\\ 6& 7\end{array}\right]+\left[\begin{array}{cc}1& 6\\ 5& 7\end{array}\right]$

= $\left[\begin{array}{cc}2& 11\\ 11& 14\end{array}\right]$

A′+A = $\left[\begin{array}{cc}1& 6\\ 5& 7\end{array}\right]+\left[\begin{array}{cc}1& 5\\ 6& 7\end{array}\right]$

= $\left[\begin{array}{cc}2& 11\\ 11& 14\end{array}\right]$

Since A+A′ = A′+A, it is a symmetric matrix.

(ii) To verify that (A−A′) is a skew symmetric matrix, we need to show that it is equal to its negative transpose.

A−A′ = $\left[\begin{array}{cc}1& 5\\ 6& 7\end{array}\right]-\left[\begin{array}{cc}1& 6\\ 5& 7\end{array}\right]$

= $\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]$

A′−A = <math xmlns = “http://www.w3.org/1998

Question:

If A,B are symmetric matrices of same order, then AB−BA is a , A Skew symmetric matrix B Symmetric matrix C Zero matrix D Identity matrix

Answer:

Answer: C Zero matrix

Question:

Express the following matrices as the sum of a symmetric and a skew symmetric matrix: (i) $\left[\begin{array}{cc}3& 5\\ 1& -1\end{array}\right]$ (ii) $\left[\begin{array}{ccc}6& -2& 2\\ -2& 3& -1\\ 2& -1& 3\end{array}\right]$ (iii) $\left[\begin{array}{ccc}3& 3& -1\\ -2& -2& 1\\ -4& -5& 2\end{array}\right]$ (iv) $\left[\begin{array}{cc}1& 5\\ -1& 2\end{array}\right]$

Answer:

(i) $\left[\begin{array}{cc}3& 5\\ 1& -1\end{array}\right]$

Symmetric Matrix: $\left[\begin{array}{cc}2& 4\\ 4& -2\end{array}\right]$

Skew Symmetric Matrix: $\left[\begin{array}{cc}1& 1\\ -1& 1\end{array}\right]$

(ii) $\left[\begin{array}{ccc}6& -2& 2\\ -2& 3& -1\\ 2& -1& 3\end{array}\right]$

Symmetric Matrix: $\left[\begin{array}{ccc}4& 0& 2\\ 0& 2& 0\\ 2& 0& 4\end{array}\right]$

Skew Symmetric Matrix: $\left[\begin{array}{ccc}2& 2& 0\\ -2& 1& -1\\ 0& 1& -2\end{array}\right]$

(iii)

Question:

(i) Show that the matrix A=$\left[\begin{array}{ccc}1& -1& 5\\ -1& 2& 1\\ 5& 1& 3\end{array}\right]$ is a symmetric matrix (ii) Show that the matrix A=$\left[\begin{array}{ccc}0& 1& -1\\ -1& 0& 1\\ 1& -1& 0\end{array}\right]$ is a skew symmetric matrix.

Answer:

(i) A matrix A is symmetric if A = AT, where AT is the transpose of A.

Let A = $\left[\begin{array}{ccc}1& -1& 5\\ -1& 2& 1\\ 5& 1& 3\end{array}\right]$

Then, AT = $\left[\begin{array}{ccc}1& -1& 5\\ -1& 2& 1\\ 5& 1& 3\end{array}\right]$

Since A = AT, the matrix A is symmetric.

(ii) A matrix A is skew symmetric if A = -AT, where AT is the transpose of A.

Let A = $\left[\begin{array}{ccc}0& 1& -1\\ -1& 0& 1\\ 1& -1& 0\end{array}\right]$

Then, AT = $\left[\begin{array}{ccc}0& -1& 1\\ 1& 0& -1\\ -1& 1& 0\end{array}\right]$

Since A = -AT, the matrix A is skew symmetric.

Question:

If A′=$\left[\begin{array}{cc}3& 4\\ -1& 2\\ 0& 1\end{array}\right]$ and B=$\left[\begin{array}{ccc}-1& 2& 1\\ 1& 2& 3\end{array}\right]$, then verify that: (i)(A+B)′=A′+B′ (ii)(A−B)′=A′−B′

Answer:

(i) To verify that (A+B)′=A′+B′, we need to calculate (A+B)′ and A′+B′ separately and then compare them.

(A+B)′ = $\left[\begin{array}{ccc}2& 6& 1\\ 0& 4& 4\\ 1& 3& 4\end{array}\right]$

A′+B′ = $\left[\begin{array}{ccc}2& 6& 1\\ 0& 4& 4\\ 1& 3& 4\end{array}\right]$

Since (A+B)′=A′+B′, we can conclude that (i) is true.

(ii) To verify that (A−B)′=A′−B′, we need to calculate (A−B)′ and A′−B′ separately and then compare them.

(A−B)′ = $\left[\begin{array}{ccc}4& 2& -1\\ -1& 0& -2\\ 1& -1& 2\end{array}\right]$

A′−B′ = $\left[\begin{array}{ccc}4& 2& -1\\ -1& 0& -2\\ 1& -1& 2\end{array}\right]$

Since (A−B)′=A′−B′, we can conclude that (ii) is true.

Question:

If A′=$\left[\begin{array}{cc}-2& 3\\ 1& 2\end{array}\right]$ and B=$\left[\begin{array}{cc}-1& 0\\ 1& 2\end{array}\right]$, then find (A+2B)′.

Answer:

Step 1: Add A and 2B to get (A + 2B) A + 2B = $\left[\begin{array}{cc}-3& 6\\ 3& 6\end{array}\right]$

Step 2: Find the transpose of (A + 2B) (A + 2B)′ = $\left[\begin{array}{cc}-3& 3\\ 6& 6\end{array}\right]$