EXERCISE 3.3 (Matrix)

class 12 maths exercise 3.3 solution

Question1: Find the transpose of each of the following matrices:

(i) $\begin{bmatrix}5\\\frac{1}{2}\\-1\end{bmatrix}$

(ii) $\begin{bmatrix}1&-1\\2&3\end{bmatrix}$

(iii) $\begin{bmatrix}-1&5&6\\\sqrt3&5&6\\2&3&1\end{bmatrix}$

Solution: (i) Let $A=\begin{bmatrix}5\\\frac{1}{2}\\-1\end{bmatrix}$

Then $A^T=\begin{bmatrix}5&\frac{1}{2}&-1\end{bmatrix}$

(ii) Let $A=\begin{bmatrix}1&-1\\2&3\end{bmatrix}$
Then
$A^T=\begin{bmatrix}1&2\\-1&3\end{bmatrix}$

(iii) Let $A=\begin{bmatrix}-1&5&6\\\sqrt3&5&6\\2&3&1\end{bmatrix}$
Then
$A^T=\begin{bmatrix}-1&\sqrt3&2\\5&5&3\\6&6&1\end{bmatrix}$

Question2: $A=\begin{bmatrix}-1&2&3\\5&7&9\\-2&1&1\end{bmatrix}$ and $B=\begin{bmatrix}-4&1&-5\\1&2&0\\1&3&1\end{bmatrix}$ Then verify that

(i) $(A+B)^T=A^T+B^T$

(ii) $(A-B)^T=A^T-B^T$

Solution: (i) $A=\begin{bmatrix}-1&2&3\\5&7&9\\-2&1&1\end{bmatrix}$ and $B=\begin{bmatrix}-4&1&-5\\1&2&0\\1&3&1\end{bmatrix}$
$A^T=\begin{bmatrix}-1&5&-2\\2&7&1\\3&9&1\end{bmatrix}$
$B^T=\begin{bmatrix}-4&1&1\\1&2&3\\-5&0&1\end{bmatrix}$

$(A+B)=\begin{bmatrix}-1&2&3\\5&7&9\\-2&1&1\end{bmatrix}+\begin{bmatrix}-4&1&-5\\1&2&0\\1&3&1\end{bmatrix}$

$=\begin{bmatrix}-1-4&2+1&3-5\\5+1&7+2&9+0\\-2+1&1+3&1+1\end{bmatrix}$
$=\begin{bmatrix}-5&3&-2\\6&9&9\\-1&4&2\end{bmatrix}$
$(A+B)^T=\begin{bmatrix}-5&6&-1\\3&9&4\\-2&9&2\end{bmatrix}$
Now
$A^T+B^T=\begin{bmatrix}-1&5&-2\\2&7&1\\3&9&1\end{bmatrix}+\begin{bmatrix}-4&1&1\\1&2&3\\-5&0&1\end{bmatrix}$

$=\begin{bmatrix}-1-4&5+1&-2+1\\2+1&7+2&1+3\\3-5&9+0&1+1\end{bmatrix}$
$=\begin{bmatrix}-5&6&-1\\3&9&4\\-2&9&2\end{bmatrix}$

Hence $(A+B)^T=A^T+B^T$

(ii) $(A-B)=\begin{bmatrix}-1&2&3\\5&7&9\\-2&1&1\end{bmatrix}-\begin{bmatrix}-4&1&-5\\1&2&0\\1&3&1\end{bmatrix}$
$=\begin{bmatrix}3&1&8\\4&5&9\\-3&-2&0\end{bmatrix}$
Then
$(A-B)^T=\begin{bmatrix}3&4&-3\\1&5&-2\\8&9&0\end{bmatrix}$

$A^T-B^T=\begin{bmatrix}-1&5&-2\\2&7&1\\3&9&1\end{bmatrix}-\begin{bmatrix}-4&1&1\\1&2&3\\-5&0&1\end{bmatrix}$

$=\begin{bmatrix}3&4&-3\\1&5&-2\\8&9&0\end{bmatrix}$
Thus
$(A-B)^T=A^T-B^T$

Question 5: For the matrices A and B, verify that $(AB)^T=B^TA^T$

(i) $A=\begin{bmatrix}1\\-4\\3\end{bmatrix},B=\begin{bmatrix}-1&2&1\end{bmatrix}$

(ii) $A=\begin{bmatrix}0\\1\\2\end{bmatrix},B=\begin{bmatrix}1&5&7\end{bmatrix}$

Solution: (i) It is given that $A=\begin{bmatrix}1\\-4\\3\end{bmatrix}$ and $B=\begin{bmatrix}-1&2&1\end{bmatrix}$

$AB=\begin{bmatrix}1\\-4\\3\end{bmatrix}\begin{bmatrix}-1&2&1\end{bmatrix}$

$=\begin{bmatrix}-1&2&1\\4&-8&-4\\-3&6&3\end{bmatrix}$

Therefore,

$(AB)^T=\begin{bmatrix}-1&4&-3\\2&-8&6\\1&-4&3\end{bmatrix}$
Now,

$A^T=\begin{bmatrix}1&-4&3\end{bmatrix}$

$B^T=\begin{bmatrix}-1\\2\\1\end{bmatrix}$

$B^TA^T=\begin{bmatrix}-1\\2\\1\end{bmatrix}\begin{bmatrix}1&-4&3\end{bmatrix}$

$\Rightarrow B^TA^T=\begin{bmatrix}-1&4&-3\\2&-8&6\\1&-4&3\end{bmatrix}$
Thus,

$(AB)^T=B^TA^T$

(ii) It is given that $A=\begin{bmatrix}0\\1\\2\end{bmatrix}$ and $B=\begin{bmatrix}1&5&7\end{bmatrix}$

Hence

$AB=\begin{bmatrix}0\\1\\2\end{bmatrix}\begin{bmatrix}1&5&7\end{bmatrix}$

$AB=\begin{bmatrix}0&0&0\\1&5&7\\2&10&14\end{bmatrix}$
Therefore,

$(AB)^T=\begin{bmatrix}0&1&2\\0&5&10\\0&7&14\end{bmatrix}$
Now,

$A^T=\begin{bmatrix}0&1&2\end{bmatrix}$ and
$B^T=\begin{bmatrix}1\\5\\7\end{bmatrix}$

Therefore,

$B^TA^T=\begin{bmatrix}1\\5\\7\end{bmatrix}\begin{bmatrix}0&1&2\end{bmatrix}$

$B^TA^T=\begin{bmatrix}0&1&2\\0&5&10\\0&7&14\end{bmatrix}$
Thus,

$(AB)^T=B^TA^T$

Question 6: If (i) $\begin{bmatrix}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\end{bmatrix}$ Then verify $A'A=I$

(ii) $\begin{bmatrix}\sin\alpha&\cos\alpha\\-\cos\alpha&\sin\alpha\end{bmatrix}$ Then verify $A'A=I$

Solution: (i) Since $A=\begin{bmatrix}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\end{bmatrix}$
Therefore,

$A'=\begin{bmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{bmatrix}$
Now,

$A'A=\begin{bmatrix}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\end{bmatrix}\begin{bmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{bmatrix}$

$=\begin{bmatrix}\cos^2\alpha+\sin^2\alpha&-\sin\alpha\cos\alpha+\sin\alpha\cos\alpha\\-\sin\alpha\cos\alpha+\sin\alpha\cos\alpha&\cos^2\alpha+\sin^2\alpha\end{bmatrix}$

$=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
Thus

$A'A=I$

(ii) It is given that $A=\begin{bmatrix}\sin\alpha&\cos\alpha\\-\cos\alpha&\sin\alpha\end{bmatrix}$
Therefore

$A'=\begin{bmatrix}\sin\alpha&-\cos\alpha\\\cos\alpha&\sin\alpha\end{bmatrix}$

Now

$A'A=\begin{bmatrix}\sin\alpha&\cos\alpha\\-\cos\alpha&\sin\alpha\end{bmatrix}\begin{bmatrix}\sin\alpha&-\cos\alpha\\\cos\alpha&\sin\alpha\end{bmatrix}$

$=\begin{bmatrix}\sin^2\alpha+\cos^2\alpha&-\sin\alpha\cos\alpha+\sin\alpha\cos\alpha\\-\sin\alpha\cos\alpha+\sin\alpha\cos\alpha&\cos^2\alpha+\sin^2\alpha\end{bmatrix}$

$=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
Thus
$A'A=I$

Question 7:(i) Show that the matrix $A=\begin{bmatrix}1&-1&5\\-1&2&1\\5&1&3\end{bmatrix}$ is a symmetric matrix.

(ii) Show that the matrix $A=\begin{bmatrix}0&1&-1\\-1&0&1\\1&-1&0\end{bmatrix}$ is a skew symmetric matrix.

Solution: (i) Since $A=\begin{bmatrix}1&-1&5\\-1&2&1\\5&1&3\end{bmatrix}$
Now

$A'=\begin{bmatrix}1&-1&5\\-1&2&1\\5&1&3\end{bmatrix}$

$=A$
Hence, A is a symmetric matrix.

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

$A'=\begin{bmatrix}0&-1&1\\1&0&-1\\-1&1&0\end{bmatrix}$

$\Rightarrow A'=-\begin{bmatrix}0&1&-1\\-1&0&1\\1&-1&0\end{bmatrix}$

$\Rightarrow A'=-A$
Hence, A is a skew symmetric matrix.

Question 8: For the matrix $A=\begin{bmatrix}1&0\\0&1\end{bmatrix}$ , verify that

(i) $</strong>(A+A')$ is a symmetric matrix.

(ii) $(A-A')$ is a skew symmetric matrix.

Solution: It is given that $A=\begin{bmatrix}1&5\\6&7\end{bmatrix}$

$A'=\begin{bmatrix}1&6\\5&7\end{bmatrix}$
Hence,

(i) $(A+A')=\begin{bmatrix}1&5\\6&7\end{bmatrix}+\begin{bmatrix}1&6\\5&7\end{bmatrix}$

$=\begin{bmatrix}2&11\\11&14\end{bmatrix}$
Therefore,

$(A+A')'=\begin{bmatrix}2&11\\11&14\end{bmatrix}$

$=(A+A')$

Thus (A+A’) is a symmetric matrix.

(ii) $(A-A')=\begin{bmatrix}1&5\\6&7\end{bmatrix}-\begin{bmatrix}1&6\\5&7\end{bmatrix}$

$=\begin{bmatrix}0&1\\-1&0\end{bmatrix}$
Therefore,

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

$(A-A')'=-(A-A')$

Thus, $(A-A')$ is a skew symmetric matrix.

Question 9: Find $\frac{1}{2}(A+A')$ and $\frac{1}{2}(A-A')$ ,When

$A=\begin{bmatrix}0&a&b\\-a&0&c\\-b&-c&0\end{bmatrix}$

Solution: It is given that $A=\begin{bmatrix}0&a&b\\-a&0&c\\-b&-c&0\end{bmatrix}$

Hence,

$A'=\begin{bmatrix}0&-a&-b\\a&0&-c\\b&c&0\end{bmatrix}$
Now,

$(A+A')=\begin{bmatrix}0&a&b\\-a&0&c\\-b&-c&0\end{bmatrix}+\begin{bmatrix}0&-a&-b\\a&0&-c\\b&c&0\end{bmatrix}$

$=\begin{bmatrix}0&a&b\\-a&0&c\\-b&-c&0\end{bmatrix}$

$=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$
Therefore,

$\frac{1}{2}(A+A')=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$

Now,

$(A-A')=\begin{bmatrix}0&a&b\\-a&0&c\\-b&-c&0\end{bmatrix}-\begin{bmatrix}0&-a&-b\\a&0&-c\\b&c&0\end{bmatrix}$

$=\begin{bmatrix}0&2a&2b\\-2a&0&2c\\-2b&-2c&0\end{bmatrix}$
Thus,

$\frac{1}{2}(A-A')=\begin{bmatrix}0&a&b\\-a&0&c\\-b&-c&0\end{bmatrix}$

Question 10: Express the following as the sum of a symmetric and skew symmetric matrix:

(i) $\begin{bmatrix}3&5\\1&-1\end{bmatrix}$

(ii) $\begin{bmatrix}6&-2&2\\-2&3&-1\\2&-1&3\end{bmatrix}$

(iii) $\begin{bmatrix}3&3&-1\\-2&-2&1\\-4&-5&2\end{bmatrix}$

(iv) $\begin{bmatrix}1&5\\-1&2\end{bmatrix}$

Solution:(i) Let $A=\begin{bmatrix}3&5\\1&-1\end{bmatrix}$
Hence,

$A'=\begin{bmatrix}3&1\\5&-1\end{bmatrix}$
Now,

$(A+A')=\begin{bmatrix}3&5\\1&-1\end{bmatrix}+\begin{bmatrix}3&1\\5&-1\end{bmatrix}$ +

$=\begin{bmatrix}6&6\\6&-2\end{bmatrix}$
Let,

P= $\frac{1}{2}(A+A')$

$=\frac{1}{2}\begin{bmatrix}6&6\\6&-2\end{bmatrix}$

$=\begin{bmatrix}3&3\\3&-1\end{bmatrix}$
Now,

P’= $\begin{bmatrix}3&3\\3&-1\end{bmatrix}$

$=P$

Thus,

P= $\frac{1}{2}(A+A')$ is a symmetric matrix.
Now,

$(A-A')=\begin{bmatrix}3&5\\1&-1\end{bmatrix}-\begin{bmatrix}3&1\\5&-1\end{bmatrix}$

$=\begin{bmatrix}0&4\\-4&0\end{bmatrix}$
Let,

Q= $\frac{1}{2}(A-A')$

$=\frac{1}{2}\begin{bmatrix}0&4\\-4&0\end{bmatrix}$

$Q=\begin{bmatrix}0&2\\-2&0\end{bmatrix}$
Now,

$Q'=\begin{bmatrix}0&-2\\2&0\end{bmatrix}$

$=-\begin{bmatrix}0&2\\-2&0\end{bmatrix}$

$Q'=-Q$

$Q=\frac{1}{2}(A-A')$ is a skew symmetric matrix.
Representing $A$ as the sum of P and Q:

$P+Q=\begin{bmatrix}3&3\\3&-1\end{bmatrix}+\begin{bmatrix}0&2\\-2&0\end{bmatrix}$

$=\begin{bmatrix}3&5\\1&-1\end{bmatrix}$

$=A$

(ii) Let $A=\begin{bmatrix}6&-2&2\\-2&3&-1\\2&-1&3\end{bmatrix}$
Hence,

$A'=\begin{bmatrix}6&-2&2\\-2&3&-1\\2&-1&3\end{bmatrix}$

Now,
.
$(A+A')=\begin{bmatrix}6&-2&2\\-2&3&-1\\2&-1&3\end{bmatrix}+\begin{bmatrix}6&-2&2\\-2&3&-1\\2&-1&3\end{bmatrix}$

$=\begin{bmatrix}12&-4&4\\-4&6&-2\\4&-2&6\end{bmatrix}$
Let,

$P=\frac{1}{2}(A+A')$

$=\frac{1}{2}\begin{bmatrix}12&-4&4\\-4&6&-2\\4&-2&6\end{bmatrix}$

$=\begin{bmatrix}6&-2&2\\-2&3&-1\\2&-1&3\end{bmatrix}$
Now,

$P'=\begin{bmatrix}6&-2&2\\-2&3&-1\\2&-1&3\end{bmatrix}$

$=P$

Thus $P=\frac{1}{2}(A+A')$ is a symmetric matrix.
Now,

$(A-A')=\begin{bmatrix}6&-2&2\\-2&3&-1\\2&-1&3\end{bmatrix}-\begin{bmatrix}6&-2&2\\-2&3&-1\\2&-1&3\end{bmatrix}$

$=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$
Let,

$Q=\frac{1}{2}(A-A')$

$=\frac{1}{2}\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$

$Q'=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$

$=-Q$

Thus, $Q=\frac{1}{2}(A-A')$ is a skew symmetric matrix.
Representing $A$ as the sum of $P$ and $Q$ :

$P+Q=\begin{bmatrix}6&-2&2\\-2&3&-1\\2&-1&3\end{bmatrix}+\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$

$=\begin{bmatrix}6&-2&2\\-2&3&-1\\2&-1&3\end{bmatrix}$

$=A$

(iii) Let $A=\begin{bmatrix}3&3&-1\\-2&-2&1\\-4&-5&2\end{bmatrix}$
Hence,

$A'=\begin{bmatrix}3&-2&-4\\3&-2&5\\-1&1&2\end{bmatrix}$

Now,

. $(A+A')=\begin{bmatrix}3&3&-1\\-2&-2&1\\-4&-5&2\end{bmatrix}+\begin{bmatrix}3&-2&-4\\3&-2&5\\-1&1&2\end{bmatrix}$

$=\begin{bmatrix}6&1&-5\\1&-4&-4\\-5&-4&4\end{bmatrix}$
Let,

$P=\frac{1}{2}(A+A')$

$=\frac{1}{2}\begin{bmatrix}6&1&-5\\1&-4&-4\\-5&-4&4\end{bmatrix}$

$=\begin{bmatrix}3&\frac{1}{2}&-\frac{5}{2}\\\frac{1}{2}&-2&-2\\-\frac{5}{2}&-2&2\end{bmatrix}$
Now,

$P'=\begin{bmatrix}3&\frac{1}{2}&-\frac{5}{2}\\\frac{1}{2}&-2&-2\\-\frac{5}{2}&-2&2\end{bmatrix}$

$=P$

Thus $P=\frac{1}{2}(A+A')$ is a symmetric matrix.

Now,

$(A-A')=\begin{bmatrix}3&3&-1\\-2&-2&1\\-4&-5&2\end{bmatrix}-\begin{bmatrix}3&-2&-4\\3&-2&5\\-1&1&2\end{bmatrix}$

$=\begin{bmatrix}0&5&3\\-5&0&6\\-3&-6&0\end{bmatrix}$
Let,

$Q=\frac{1}{2}(A-A')$

$=\frac{1}{2}\begin{bmatrix}0&5&3\\-5&0&6\\-3&-6&0\end{bmatrix}$

$=\begin{bmatrix}0&\frac{5}{2}&\frac{3}{2}\\-\frac{5}{2}&0&3\\-\frac{3}{2}&-3&0\end{bmatrix}$

$Q'=\begin{bmatrix}0&-\frac{5}{2}&-\frac{3}{2}\\\frac{5}{2}&0&-3\\\frac{3}{2}&3&0\end{bmatrix}$

$=-\begin{bmatrix}0&\frac{5}{2}&\frac{3}{2}\\-\frac{5}{2}&0&3\\-\frac{3}{2}&-3&0\end{bmatrix}$

$=-Q$
Thus, $Q=\frac{1}{2}(A-A')$ is a skew symmetric matrix.
Representing $A$ as the sum of $P$ and $Q$ :

$P+Q=\begin{bmatrix}3&\frac{1}{2}&-\frac{5}{2}\\\frac{1}{2}&-2&-2\\-\frac{5}{2}&-2&2\end{bmatrix}+\begin{bmatrix}0&\frac{5}{2}&\frac{3}{2}\\-\frac{5}{2}&0&3\\-\frac{3}{2}&-3&0\end{bmatrix}$

$=\begin{bmatrix}3&3&-1\\-2&-2&1\\-4&-5&2\end{bmatrix}$

$=A$

(iv) Let $A=\begin{bmatrix}1&5\\-1&2\end{bmatrix}$
Hence,

$A'=\begin{bmatrix}1&-1\\5&2\end{bmatrix}$
Now,

$(A+A')=\begin{bmatrix}1&5\\-1&2\end{bmatrix}+\begin{bmatrix}1&-1\\5&2\end{bmatrix}$

$=\begin{bmatrix}2&4\\4&4\end{bmatrix}$
Let,

P= $\frac{1}{2}(A+A')$

$=\frac{1}{2}\begin{bmatrix}2&4\\4&4\end{bmatrix}$

$=\begin{bmatrix}1&2\\2&2\end{bmatrix}$
Now,

P’= $\begin{bmatrix}1&2\\2&2\end{bmatrix}$

$=P$
Thus,

P= $\frac{1}{2}(A+A')$ is a symmetric matrix.

Now,

$(A-A')=\begin{bmatrix}1&5\\-1&2\end{bmatrix}-\begin{bmatrix}1&-1\\5&2\end{bmatrix}$

$=\begin{bmatrix}0&6\\-6&0\end{bmatrix}$
Let,

Q= $\frac{1}{2}(A-A')$

$=\frac{1}{2}\begin{bmatrix}0&6\\-6&0\end{bmatrix}$

$Q=\begin{bmatrix}0&3\\-3&0\end{bmatrix}$
Now,

$Q'=\begin{bmatrix}0&-3\\3&0\end{bmatrix}$

$=-\begin{bmatrix}0&3\\-3&0\end{bmatrix}$

$Q'=-Q$

$Q=\frac{1}{2}(A-A')$ is a skew symmetric matrix.
Representing $A$ as the sum of P and Q:

$P+Q=\begin{bmatrix}1&2\\2&2\end{bmatrix}+\begin{bmatrix}0&3\\-3&0\end{bmatrix}$

$=\begin{bmatrix}1&5\\-1&2\end{bmatrix}$

$=A$

Question 11: If $A,B$ are symmetric matrices of the same order, then $AB=BA$ is a

(A) Skew symmetric matrix

(B) Symmetric matrix

(C) Zero matrix

(D) Identity matrix

Solution:The Correct option is $A$ .

If and are symmetric matrices of the same order, then

Solution: If $A$ and $B$ are symmetric matrices of the same order, then

$A'=A$ and $B'=B$ —(1)

Now consider,

$(AB-BA)'=(AB)'-(BA)'$

$=B'A'-A'B'$

$=BA-AB$

$=-(AB-BA)$

Therefore,

$(AB-BA)'=-(AB-BA)$

Thus, $AB-BA$ is a skew symmetric matrix.

Question 12: If $A=\begin{bmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{bmatrix}$ then $A+A''=I$ ,if the value of $\alpha$ is:

(A) $\frac{\pi}{6}$ (B) $\frac{\pi}{3}$

(C) $\pi$ (D) $\frac{3\pi}{2}$

Solution:Thus, the correct option is $B$ .

It is given that

$A=\begin{bmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{bmatrix}$
Hence,

$A'=\begin{bmatrix}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\end{bmatrix}$
NOW,

$A+A'=I$

Therefore,

$\begin{bmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{bmatrix}+\begin{bmatrix}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$

$\begin{bmatrix}2\cos\alpha&0\\0&2\cos\alpha\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
Comparing the corresponding elements of the two matrices, we have:

$\Rightarrow2\cos\alpha=1$

$\Rightarrow\cos\alpha=\frac{1}{2}$

$\alpha=\frac{\pi}{3}$

Amit, Biraj and chirag were giventhe task of creating a square matrix of order 2

Three schools DPS, CVC and KVS decided to organize a fair for collecting money

EXERCISE 3.3 (Matrix)

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