
Inverse of the matrix $\begin{bmatrix} 3 & -2 & -1 \\ -4 & 1 & -1 \\ 2& 0 & 1 \\ \end{bmatrix}$ is
A. $\begin{bmatrix} 1 & 2 & 3 \\ 3 & 3 & 7 \\ -2 & -4 & -5 \\ \end{bmatrix}$
B. $\begin{bmatrix} 1 & -3 & 5 \\ 7 & 4 & 6 \\ 4 & 2 & 7 \\ \end{bmatrix}$
C. $\begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \\ \end{bmatrix}$
D. $\begin{bmatrix} 1 & -3 & 5 \\ 7 & 4 & 6 \\ 4 & 2 & -7 \\ \end{bmatrix}$
Answer
161.1k+ views
Hint: You can use two techniques to determine a matrix's inverse. By utilising an adjoint of a matrix and simple operations, it is possible to calculate the inverse of a matrix. Row or column transformations can carry out the basic functions on a matrix. Additionally, the adjoint and determinant of the matrix can be used to apply the inverse of the matrix formula to calculate the inverse of a matrix.
Formula Used:
Inverse Matrix Formula$=A^{-1}=\frac{1}{|A|}.AdjA$
Complete step by step Solution:
Let the given matrix as $A=\begin{bmatrix} 3 & -2 & -1 \\ -4 & 1 & -1 \\ 2 & 0 & 1 \\ \end{bmatrix}$
Then, the determinant of A is given by;
$|A|=\begin{vmatrix} 3 & -2 & -1 \\ -4 & 1 & -1 \\ 2 & 0 & 1 \\ \end{vmatrix}\\
|A|=1$
we can now determine the adjoint of the matrix A by calculating the cofactors for each element and then transposing the cofactor matrix,
The matrix of cofactors of $A = \begin{bmatrix} {{c}_{11}} & {{c}_{12}} & {{c}_{13}} \\ {{c}_{21}} & {{c}_{22}} & {{c}_{23}} \\ {{c}_{31}} & {{c}_{32}} & {{c}_{33}} \\ \end{bmatrix}=\begin{bmatrix} 1 & 2 & -2 \\ 2 & 5 & -4 \\ 3 & 7 & -5 \\ \end{bmatrix}$
Therefore, the transpose of the cofactor matrix is an Adjoint matrix.
$Adj(A)= \begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \\ \end{bmatrix}$
The inverse of a matrix A is:
${{A}^{-1}}=\frac{1}{|A|}\,.\,adjA\\
{{A}^{-1}}=1.\begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \\ \end{bmatrix}\\
{{A}^{-1}}= \begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \\ \end{bmatrix}$
So, option C is correct.
Note: The inverse of a matrix can only exist if the matrix's determinant has a non-zero value |A| i.e., $|A|\neq 0$. The provided matrix must be square. In making the cofactor matrix reverse the sign of the alternating terms to create the adjoint or adjugate matrix.
Formula Used:
Inverse Matrix Formula$=A^{-1}=\frac{1}{|A|}.AdjA$
Complete step by step Solution:
Let the given matrix as $A=\begin{bmatrix} 3 & -2 & -1 \\ -4 & 1 & -1 \\ 2 & 0 & 1 \\ \end{bmatrix}$
Then, the determinant of A is given by;
$|A|=\begin{vmatrix} 3 & -2 & -1 \\ -4 & 1 & -1 \\ 2 & 0 & 1 \\ \end{vmatrix}\\
|A|=1$
we can now determine the adjoint of the matrix A by calculating the cofactors for each element and then transposing the cofactor matrix,
The matrix of cofactors of $A = \begin{bmatrix} {{c}_{11}} & {{c}_{12}} & {{c}_{13}} \\ {{c}_{21}} & {{c}_{22}} & {{c}_{23}} \\ {{c}_{31}} & {{c}_{32}} & {{c}_{33}} \\ \end{bmatrix}=\begin{bmatrix} 1 & 2 & -2 \\ 2 & 5 & -4 \\ 3 & 7 & -5 \\ \end{bmatrix}$
Therefore, the transpose of the cofactor matrix is an Adjoint matrix.
$Adj(A)= \begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \\ \end{bmatrix}$
The inverse of a matrix A is:
${{A}^{-1}}=\frac{1}{|A|}\,.\,adjA\\
{{A}^{-1}}=1.\begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \\ \end{bmatrix}\\
{{A}^{-1}}= \begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \\ \end{bmatrix}$
So, option C is correct.
Note: The inverse of a matrix can only exist if the matrix's determinant has a non-zero value |A| i.e., $|A|\neq 0$. The provided matrix must be square. In making the cofactor matrix reverse the sign of the alternating terms to create the adjoint or adjugate matrix.
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