Which of the following matrices is invertible | vedclass

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(C) square matrix $A$ is invertible if and only if its determinant $|A| 
eq 0$.Let us calculate the determinant for each matrix:$1$. For $A_{1} = \be

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$A_{4}$

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(C) square matrix $A$ is invertible if and only if its determinant $|A| 
eq 0$.Let us calculate the determinant for each matrix:$1$. For $A_{1} = \begin{bmatrix} 4 & 2 \ 2 & 1 \end{bmatrix}$,$|A_{1}| = (4 	imes 1) - (2 	imes 2) = 4 - 4 = 0$. Thus,$A_{1}$ is not invertible.$2$. For $A_{2} = \begin{bmatrix} -1 & -2 & 3 \ 4 & 5 & 7 \ 2 & 4 & -6 \end{bmatrix}$,notice that row $3$ is $-2$ times row $1$ $(R_{3} = -2R_{1})$. Since two rows are proportional,$|A_{2}| = 0$. Thus,$A_{2}$ is not invertible.$3$. For $A_{3} = \begin{bmatrix} 1 & 0 & 0 \ 5 & 2 & 1 \ 7 & 2 & 1 \end{bmatrix}$,notice that row $2$ and row $3$ are not proportional,but let's calculate: $|A_{3}| = 1(2-2) - 0 + 0 = 0$. Thus,$A_{3}$ is not invertible.$4$. For $A_{4} = \begin{bmatrix} 1 & 0 & 1 \ 0 & 2 & 3 \ 1 & 2 & 1 \end{bmatrix}$,$|A_{4}| = 1(2-6) - 0(0-3) + 1(0-2) = 1(-4) + 1(-2) = -4 - 2 = -6$.Since $|A_{4}| = -6 
eq 0$,the matrix $A_{4}$ is invertible.

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