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

1.	Without expanding the determinant, prove that \(\begin{bmatrix} 41& 1 & 5 \\[0.3em] 79 & 7 & 9 \\[0.3em] 29 &5 & 3 \end{bmatrix}\) = 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 \(\begin{bmatrix} 41& 1 & 5 \\[0.3em] 79 & 7 & 9 \\[0.3em] 29 &5 & 3 \end{bmatrix}\) Now it can clearly be seen that C₁ = 8 x C₃ Applying above equation we get, \(\begin{bmatrix} 0& 1 & 5 \\[0.3em] 0 & 7 & 9 \\[0.3em] 0 & 3 & 3 \end{bmatrix}\) We know that if a row or column of a determinant is 0. Then it is singular determinant.

Without expanding the determinant, prove that \(\begin{bmatrix} 41& 1 & 5 \\[0.3em] 79 & 7 & 9 \\[0.3em] 29 &5 & 3 \end{bmatrix}\) = 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

\(\begin{bmatrix} 41& 1 & 5 \\[0.3em] 79 & 7 & 9 \\[0.3em] 29 &5 & 3 \end{bmatrix}\) Now it can clearly be seen that C₁ = 8 x C₃

Applying above equation we get,

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

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