
Find ${{\left[ \begin{matrix} 1 & 3 \\ 3 & 10 \\ \end{matrix} \right]}^{-1}}= $ [EAMCET 1994; DCE 1999]
A. $\left[ \begin{matrix} 10 & 3 \\ 3 & 1 \\ \end{matrix} \right]$
B. $\left[ \begin{matrix} 10 & -3 \\ -3 & 1 \\ \end{matrix} \right]$
C. $\left[ \begin{matrix} 1 & 3 \\ 3 & 10 \\ \end{matrix} \right]$
D. $\left[ \begin{matrix} -1 & -3 \\ -3 & -10 \\ \end{matrix} \right]$
Answer
164.1k+ views
Hint: Here you can find the inverse either by using elementary row operation or by inverse formula. Using the determinant and adjoint of the given matrix you can determine the inverse. Divide each component of an adjoint matrix obtained by the determinant of that matrix.
Formula Used:
Determinant of $2 \times 2$ matrix A (say) is given by $|A| or det A=(a_{11} a_{22}-a_{12} a_{21})$
Adjoint of $2 \times 2$ matrix A(say) is given by $adjA=\left[\begin{matrix} a_{22} & -a_{12} \\ -a_{21} & a_{11} \\ \end{matrix}\right]$
Inverse matrix formula $A^{-1}=\dfrac{adjA}{|A|}$
Complete step-by-step Solution:
Let $A= \left[ \begin{matrix} 1 & 3 \\ 3 & 10 \\ \end{matrix} \right]$
Determinant;
$|A|=(1\times10)-(3\times3)\\
|A|=10-9\\
|A|=1$
Now, the Adjoint of A can be determined by alternating the main diagonal's components. Simply swap the signs of the components in the other diagonal, taking care not to interchange them. So,
$A=Adj A=\left[ \begin{matrix} 10 & -3 \\ -3 & 1 \\ \end{matrix} \right]$
Therefore,
$A^{-1}=\dfrac{1}{1}.\left[ \begin{matrix} 10 & -3 \\ -3 & 1 \\ \end{matrix} \right]\\
A^{-1}=\left[ \begin{matrix} 10 & -3 \\ -3 & 1 \\ \end{matrix} \right]$
So, option is B correct.
Additional Information: Inverse using Row Operations for $2\times 2$ Matrix
To determine the inverse of a $2\times 2$ matrix, $A$, we can use the fundamental row operations.
1. As the identity matrix of order $2\times 2$, $A$ and $I$ are first written as an augmented matrix with a line separating them so that $A$ is on the left and $I$ is on the right.
2. Apply row operations so that identity matrix $I$ is created from the left side matrix.
3. Following that, $A^{-1}$ is the right-side matrix.
Note: The alternative way to determine the inverse of the $2 \times 2$ matrix is using elementary operations. If $A$ is a matrix such that $A^{-1}$ exists, then write $A = IA$ and do a series of row operations on $A = IA$ until we get $I = BA$. This will find the inverse of A, or $A^{-1}$, using simple row operations. The inverse of matrix $A$ will be matrix $B$. In a similar manner, if finding $A^{-1}$ via column operations is what you're after, then write $A = AI$ and use a series of column operations on it until we get $AB = I$.
Formula Used:
Determinant of $2 \times 2$ matrix A (say) is given by $|A| or det A=(a_{11} a_{22}-a_{12} a_{21})$
Adjoint of $2 \times 2$ matrix A(say) is given by $adjA=\left[\begin{matrix} a_{22} & -a_{12} \\ -a_{21} & a_{11} \\ \end{matrix}\right]$
Inverse matrix formula $A^{-1}=\dfrac{adjA}{|A|}$
Complete step-by-step Solution:
Let $A= \left[ \begin{matrix} 1 & 3 \\ 3 & 10 \\ \end{matrix} \right]$
Determinant;
$|A|=(1\times10)-(3\times3)\\
|A|=10-9\\
|A|=1$
Now, the Adjoint of A can be determined by alternating the main diagonal's components. Simply swap the signs of the components in the other diagonal, taking care not to interchange them. So,
$A=Adj A=\left[ \begin{matrix} 10 & -3 \\ -3 & 1 \\ \end{matrix} \right]$
Therefore,
$A^{-1}=\dfrac{1}{1}.\left[ \begin{matrix} 10 & -3 \\ -3 & 1 \\ \end{matrix} \right]\\
A^{-1}=\left[ \begin{matrix} 10 & -3 \\ -3 & 1 \\ \end{matrix} \right]$
So, option is B correct.
Additional Information: Inverse using Row Operations for $2\times 2$ Matrix
To determine the inverse of a $2\times 2$ matrix, $A$, we can use the fundamental row operations.
1. As the identity matrix of order $2\times 2$, $A$ and $I$ are first written as an augmented matrix with a line separating them so that $A$ is on the left and $I$ is on the right.
2. Apply row operations so that identity matrix $I$ is created from the left side matrix.
3. Following that, $A^{-1}$ is the right-side matrix.
Note: The alternative way to determine the inverse of the $2 \times 2$ matrix is using elementary operations. If $A$ is a matrix such that $A^{-1}$ exists, then write $A = IA$ and do a series of row operations on $A = IA$ until we get $I = BA$. This will find the inverse of A, or $A^{-1}$, using simple row operations. The inverse of matrix $A$ will be matrix $B$. In a similar manner, if finding $A^{-1}$ via column operations is what you're after, then write $A = AI$ and use a series of column operations on it until we get $AB = I$.
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