A lower triangular matrix A=(aiij)n ×n is singular if and only if (a) aii=0 for all i=1,2, …n (b

step 1 So in the given question we are told that there is a lower triangular matrix A, a lower triangul

A lower triangular matrix A=(aiij)n ×n is singular if and...

A lower triangular matrix $\mathrm{A}=\left(\mathrm{a}_{\mathrm{iij}}\right)_{\mathrm{n} \times \mathrm{n}}$ is singular if and only if (a) $\mathrm{a}_{\mathrm{ii}}=0$ for all $\mathrm{i}=1,2, \ldots \mathrm{n}$ (b) $\mathrm{a}_{\mathrm{ii}}=0$ for at least one $\mathrm{i}=1,2, \ldots \mathrm{n}$ (c) $\mathrm{a}_{\mathrm{ii}} \neq 0$ for all $\mathrm{i}=1,2, \ldots \mathrm{n}$ (d) $\mathrm{a}_{\mathrm{ii}} \neq 0$ for at least one $\mathrm{i}, \mathrm{i}=1,2, \ldots \mathrm{n}$

A lower triangular matrix $\mathrm{A}=\left(\mathrm{a}_{\mathrm{iij}}\right)_{\mathrm{n} \times \mathrm{n}}$ is singular if and only if
(a) $\mathrm{a}_{\mathrm{ii}}=0$ for all $\mathrm{i}=1,2, \ldots \mathrm{n}$
(b) $\mathrm{a}_{\mathrm{ii}}=0$ for at least one $\mathrm{i}=1,2, \ldots \mathrm{n}$
(c) $\mathrm{a}_{\mathrm{ii}} \neq 0$ for all $\mathrm{i}=1,2, \ldots \mathrm{n}$
(d) $\mathrm{a}_{\mathrm{ii}} \neq 0$ for at least one $\mathrm{i}, \mathrm{i}=1,2, \ldots \mathrm{n}$

Key Concepts

Singular Matrix

A matrix is singular if its determinant is zero. In the context of triangular matrices, since the determinant is the product of the diagonal elements, the matrix is singular if and only if at least one diagonal entry is zero.

Lower Triangular Matrix

A lower triangular matrix is a square matrix where all entries above the main diagonal are zero. This structure simplifies many computations, especially the calculation of the determinant and solving linear systems.

Determinant of a Triangular Matrix

For any triangular matrix, whether lower or upper, the determinant is the product of its diagonal elements. This property is key in quickly determining whether a matrix is singular or nonsingular.

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