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let A = [ aij] be a 3x3 invertible matrix with real entries and B = [bij] be a matrix which is formed such that bij is the sum of all the elements except aij in the ith row. Answer the following:
If there exist a matrix X with Constant elements such that A X = B, then X is
If there exist a matrix X with Constant elements such that A X = B, then X is
- a)Skew symmetric
- b)Null Matrix
- c)Diagonal matrix
- d)Symmetric matrix
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
let A = [ aij]be a 3x3 invertible matrix with real entries and B = [bi...
Now ginen A is invertiable ⇒ |A| ≠ 0 and A-1 exit and AX = B
and
Most Upvoted Answer
let A = [ aij]be a 3x3 invertible matrix with real entries and B = [bi...
Explanation:
To find the matrix X, we need to solve the equation AX = B, where A is a 3x3 invertible matrix and B is a matrix formed by taking the sum of all elements in each row of A, except the corresponding diagonal element.
Let's consider the given matrix A:
A = [aij]
We can write the equation AX = B as:
[ a11 a12 a13 ] [ x11 x12 x13 ] [ b11 b12 b13 ]
[ a21 a22 a23 ] * [ x21 x22 x23 ] = [ b21 b22 b23 ]
[ a31 a32 a33 ] [ x31 x32 x33 ] [ b31 b32 b33 ]
Expanding the product, we get:
a11*x11 + a12*x21 + a13*x31 = b11
a11*x12 + a12*x22 + a13*x32 = b12
a11*x13 + a12*x23 + a13*x33 = b13
a21*x11 + a22*x21 + a23*x31 = b21
a21*x12 + a22*x22 + a23*x32 = b22
a21*x13 + a22*x23 + a23*x33 = b23
a31*x11 + a32*x21 + a33*x31 = b31
a31*x12 + a32*x22 + a33*x32 = b32
a31*x13 + a32*x23 + a33*x33 = b33
We can rearrange these equations in matrix form as:
[ a11 a12 a13 ] [ x11 x12 x13 ] [ b11 b12 b13 ]
[ a21 a22 a23 ] * [ x21 x22 x23 ] = [ b21 b22 b23 ]
[ a31 a32 a33 ] [ x31 x32 x33 ] [ b31 b32 b33 ]
This equation can be solved using the inverse of matrix A:
X = A^(-1) * B
Since A is an invertible matrix, its inverse exists. Therefore, we can find the matrix X by multiplying the inverse of A with B.
Answer:
The matrix X obtained by solving the equation AX = B is a symmetric matrix.
To find the matrix X, we need to solve the equation AX = B, where A is a 3x3 invertible matrix and B is a matrix formed by taking the sum of all elements in each row of A, except the corresponding diagonal element.
Let's consider the given matrix A:
A = [aij]
We can write the equation AX = B as:
[ a11 a12 a13 ] [ x11 x12 x13 ] [ b11 b12 b13 ]
[ a21 a22 a23 ] * [ x21 x22 x23 ] = [ b21 b22 b23 ]
[ a31 a32 a33 ] [ x31 x32 x33 ] [ b31 b32 b33 ]
Expanding the product, we get:
a11*x11 + a12*x21 + a13*x31 = b11
a11*x12 + a12*x22 + a13*x32 = b12
a11*x13 + a12*x23 + a13*x33 = b13
a21*x11 + a22*x21 + a23*x31 = b21
a21*x12 + a22*x22 + a23*x32 = b22
a21*x13 + a22*x23 + a23*x33 = b23
a31*x11 + a32*x21 + a33*x31 = b31
a31*x12 + a32*x22 + a33*x32 = b32
a31*x13 + a32*x23 + a33*x33 = b33
We can rearrange these equations in matrix form as:
[ a11 a12 a13 ] [ x11 x12 x13 ] [ b11 b12 b13 ]
[ a21 a22 a23 ] * [ x21 x22 x23 ] = [ b21 b22 b23 ]
[ a31 a32 a33 ] [ x31 x32 x33 ] [ b31 b32 b33 ]
This equation can be solved using the inverse of matrix A:
X = A^(-1) * B
Since A is an invertible matrix, its inverse exists. Therefore, we can find the matrix X by multiplying the inverse of A with B.
Answer:
The matrix X obtained by solving the equation AX = B is a symmetric matrix.
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