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Without expanding the determinant, prove that $\left|\begin{array}{ccc}41 & 1 & 5 \\ 79 & 7 & 9 \\ 29 & 5 & 3\end{array}\right|=0$. SINGULAR MATRIX A square matrix $A$ is said to be singular if $|A|=0$. Also, $A$ is called non singular if $|A| \neq 0$.
Without expanding the determinant, prove that $\left|\begin{array}{ccc}41 & 1 & 5 \\ 79 & 7 & 9 \\ 29 & 5 & 3\end{array}\right|=0$. SINGULAR MATRIX A square matrix $A$ is said to be singular if $|A|=0$. Also, $A$ is called non singular if $|A| \neq 0$.
CBSE | Class 12 | Determinants | Exercise 6A | Maths | RS Aggarwal
Without expanding the determinant, prove that $\left|\begin{array}{ccc}41 & 1 & 5 \\ 79 & 7 & 9 \\ 29 & 5 & 3\end{array}\right|=0$. SINGULAR MATRIX A square matrix $A$ is said to be singular if $|A|=0$. Also, $A$ is called non singular if $|A| \neq 0$.

Solution:

We know that $C_{1} \Rightarrow C_{1}-C_{2}$, would not change anything for the determinant.
Applying the same in above determinant, we get
$\left[\begin{array}{lll}40 & 1 & 5 \\ 72 & 7 & 9 \\ 24 & 5 & 3\end{array}\right]$ Now it can clearly be seen that $C_{1}=8 \times C_{3}$
Applying above equation we get,
$\left[\begin{array}{lll}
0 & 1 & 5 \\
0 & 7 & 9 \\
0 & 3 & 3
\end{array}\right]$
We know that if a row or column of a determinant is 0 . Then it is singular determinant.

i

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