GATE 2018 | MATHS | Q-42 DA Practice Questions

Question

Consider the following two statements:

\(P\): The matrix \(\begin{bmatrix} 0 & 5 \\ 0 & 7 \end{bmatrix}\) has infinitely many LU factorizations, where \(L\) is lower triangular with each diagonal entry 1 and \(U\) is upper triangular.

\(Q\): The matrix \(\begin{bmatrix} 0 & 0 \\ 2 & 5 \end{bmatrix}\) has no LU factorization, where \(L\) is lower triangular with each diagonal entry 1 and \(U\) is upper triangular.

Then which one of the following options is correct?

(A) \(P\) is TRUE and \(Q\) is FALSE

(B) Both \(P\) and \(Q\) are TRUE

(C) \(P\) is FALSE and \(Q\) is TRUE

(D) Both \(P\) and \(Q\) are FALSE

Answer 1

(B) Both P and Q are TRUE.

Explanation:

Statement P:

• The matrix (\begin{bmatrix} 0 & 5 \\ 0 & 7 \end{bmatrix}) has infinitely many LU factorizations. This statement is true.

• To see why, consider the following LU factorization:

L = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, 

U = \begin{bmatrix} 0 & 5 \\ 0 & 7 \end{bmatrix}

• This factorization works because the product LU gives the original matrix. However, any value of $l_{21}$ (the element in the second row, first column of L) would also produce a valid LU factorization, since it gets multiplied by 0 in the product. Therefore, there are infinitely many possible choices for $l_{21}$, leading to infinitely many LU factorizations.

Statement Q:

• The matrix (\begin{bmatrix} 0 & 0 \\ 2 & 5 \end{bmatrix}) has no LU factorization. This statement is true.

• To see why, consider the process of Gaussian elimination used to find an LU factorization. The first step would be to eliminate the leading 0 in the first row. However, this is impossible without introducing a non-zero element below the leading 1 in the first column of L, violating the requirement that L be lower triangular with 1s on the diagonal.

• Therefore, no matter how we try to arrange L and U, we cannot achieve an LU factorization for this matrix.