Linear Algebra MCQ - 1

As it's upper traingular matrix ... So determinant will be product of main diagonal element.
det(A) = 6*2*4* -1 = -48.
Similar concept can be appliead , if Matrix is lower triangular or Diagonal Matrix

Hi,
For this kind of matrices Determinant is zero.
A will be a 3x3 matrix where the first row will be 2 [1 9 5], second row will be -4 [1 9 5] and third will be 7 [1 9 5]. That is, all the rows of A are linearly dependent which means A is singular.
When matrix is singular |A| = 0

The determinant of a real matrix can never be imaginary. So, if one eigen value is complex, the other eigen value has to be its conjugate.
 
So, the eigen values of the matrix will be 2+i, 2-i and 3.
 
Also, determinant is the product of all eigen values.
So, the required answer is (2+i)*(2-i)*(3) = (4-i²)*(3) = (5)*(3) = 15.

now consider each of the triplets as the value of x, y, z and put in these equations the one which satisfies is the answer.
why so because an eigen vector represents a vector which passes through all the points which can solve these equations.
so we can observe that only option B is satisfying the equations.

This is because eigen vectors corresponding to DIFFERENT eigen values of a REAL symmetric matrix are ORTHOGONAL to each other.
However, same eigen values they may not be.
And Dot -product of orthogonal vectors(perpendicular vectors ) is 0 (ZERO)

Eigen values of   Therefore second statement is false.
Since the rank of matrix  is 2, therefore atleast one eigen value would be zero for n>3.
For n= 2,  It can be proven that

Both  λ₁ and λ₂  would be real because  is a real symmetric matrix. Which implies that atleast one eigen value would be in
Now, to prove  matrix, let us consider the matrix is  is the eigen value
of this matrix.


(For real  symmetric matrix, b=c and < would be replaced by equal sign)

x = [x¹ , x²] be a unit eigen vector
i.e. x¹ 2 + x² 2 = 1
∴ { x is a unit Eigen vector}
x'Ax = x'Lx=  Lx'x  = L [x1,x2]' [x¹,x²] = L [x1² + x2²] = L(1) = L .

So, for our given matrix, we have

This is a
 block matrix where the first and last and the second and third elements are the same. So, applying the formulafor determinant of a block matrix as given here (second last case)

Each of the eigen value of A is the solution of the equation  . So, we can equate   to any of the eigen value of A, and that will get our value of  and that is one of the choice. For no other choice, this equation holds. So, (c) 2 is the answer.

det (P + Q), So Rank cannot be 3, but there exists a  submatrix such that determinant of submatrix is not 0.
 

The given matrix is a upper triangular matrix and the eigenvalues of upper or lower traingular matrix are the diagonal values itself. (Property)

Trace is the sum of all diagonal elements of a square matrix.
Determinant of a matrix = Product of eigen values.
A) Is the right answer. To have the determinant negative ,atleast one eigen value has to be negative(but reverse may not be true). {you can take simple example with upper or lower triangular matrices. In the case option (b) , (c) and (d) reverse is always true .}

We can see that the rank of the given matrix is 2 (since 3 rows are same, and other 2 rows are also same). Sum of eigen values = sum of diagonals. So, we have two eigen values which sum to 5. This information can be used to get answer in between the following solution. Let Eigen value be X. Now, equating the determinant of the following to 0 gives us the values for X. To find X in the following matrix, we can equate the determinant to 0. For finding the determinant we can use row and column additions and make the matrix a triangular one. Then determinant will just be the product of the diagonals which should equate to 0. 

Taking X out from R4, 2-X from R1, (so, X = 2 is one eigen value)










Now, we got a triangular matrix and determinant of a triangular matrix is product of the diagonal.
So (3-X) (-X) = 0 => X = 3 or X = 0. So, X = 3 is another eigen value and product of non-zero eigen values = 2 * 3 = 6. 

Given that λ = 2 is an eigen value. So, it must satisfy characterstic equation.

For finding the Eigen Values of a Matrix we need to build the Characteristic equation which is of the form,

A - λI
Where A is the given Matrix.
λ is a constant

I is the identity matrix.
We'll have a Linear equation after solving A - λI. Which will give us 2 roots for λ.

6 is larger and hence is the Answer.