Without expanding the determinant, prove that \(\begin{vmatrix} 41 &1

1.	Without expanding the determinant, prove that \(\begin{vmatrix} 41 &1 & 5 \\[0.3em] 79 & 7 &9\\[0.3em] 29 & 5 & 3 \end{vmatrix}\) = 0SINGULAR MATRIX A square matrix A is said to be singular if \|A\| = 0. Also, A is called non singular if \|A\| ≠ 0.
Answer» We know that C₁⇒ C₁-C₂, would not change anything for the determinant. Applying the same in above determinant, we get \(\begin{bmatrix} 40 &1 & 5 \\[0.3em] 72 & 7 &9\\[0.3em] 24 & 5 & 3 \end{bmatrix}\) Now it can clearly be seen that C₁=8 × C₃ Applying above equation we get, \(\begin{bmatrix} 0 &1 & 5 \\[0.3em] 0 & 7 &9\\[0.3em] 0 & 5 & 3 \end{bmatrix}\) We know that if a row or column of a determinant is 0. Then it is singular determinant.

Without expanding the determinant, prove that \(\begin{vmatrix} 41 &1 & 5 \\[0.3em] 79 & 7 &9\\[0.3em] 29 & 5 & 3 \end{vmatrix}\) = 0SINGULAR MATRIX A square matrix A is said to be singular if |A| = 0. Also, A is called non singular if |A| ≠ 0.

Answer»

We know that C₁⇒ C₁-C₂, would not change anything for the determinant.

Applying the same in above determinant, we get

\(\begin{bmatrix} 40 &1 & 5 \\[0.3em] 72 & 7 &9\\[0.3em] 24 & 5 & 3 \end{bmatrix}\) Now it can clearly be seen that C₁=8 × C₃

Applying above equation we get,

\(\begin{bmatrix} 0 &1 & 5 \\[0.3em] 0 & 7 &9\\[0.3em] 0 & 5 & 3 \end{bmatrix}\)

We know that if a row or column of a determinant is 0. Then it is singular determinant.