If $A = \begin{bmatrix} 2 & 7 \\ 1 & 5 \end{bmatrix}$, then what is $A + 3A^{-1}$ equal to, where $A$ is a matrix of order 2?
7I
Given the matrix:
$$A = \begin{bmatrix} 2 & 7 \\ 1 & 5 \end{bmatrix}$$
Step 1: Find the determinant of \(A\)
$$|A| = (2 \times 5) - (7 \times 1) = 10 - 7 = 3$$
Step 2: Find the inverse of \(A\) (\(A^{-1}\))
For a \(2 \times 2\) matrix, \(A^{-1} = \frac{1}{|A|} \text{adj}(A)\):
$$A^{-1} = \frac{1}{3} \begin{bmatrix} 5 & -7 \\ -1 & 2 \end{bmatrix}$$
Step 3: Calculate \(3A^{-1}\)
$$3A^{-1} = 3 \times \frac{1}{3} \begin{bmatrix} 5 & -7 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & -7 \\ -1 & 2 \end{bmatrix}$$
Step 4: Calculate \(A + 3A^{-1}\)
$$A + 3A^{-1} = \begin{bmatrix} 2 & 7 \\ 1 & 5 \end{bmatrix} + \begin{bmatrix} 5 & -7 \\ -1 & 2 \end{bmatrix}$$
$$A + 3A^{-1} = \begin{bmatrix} 2+5 & 7-7 \\ 1-1 & 5+2 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$$
Final Answer:
$$A + 3A^{-1} = 7I$$
Similar Questions
What should be the value of x so that the matrix \(\left( {\begin{array}{*{20}{c}} 2&4\\ { - 8}&{\rm{x}} \end{array}} \right)\) does not have an inverse?
What is the adjoint of the matrix \(\left( {\begin{array}{*{20}{c}} {\cos \left( { - \theta } \right)}&{ - \sin \left( { - \theta } \right)}\\ { - \sin \left( { - \theta } \right)}&{\cos \left( { - \theta } \right)} \end{array}} \right)\) ?
If A is an identity matrix of order 3, then its inverse (A -1 )
What is the inverse of the matrix?
\(A = \left( {\begin{array}{*{20}{c}} {\cos \theta }&{\sin \theta }&0\\ { - \sin \theta }&{\cos \theta }&0\\ 0&0&1 \end{array}} \right)\)
If A and B are two invertible square matrices of same order, then what is (AB) -1 equal to?
If \(B = \left[ {\begin{array}{*{20}{c}} 3&2&0\\ 2&4&0\\ 1&1&0 \end{array}} \right]\) , then what is adjoint of B equal to?
For a square matrix A, which of the following properties hold?
1) (A -1 )-1 = A
2) \(\det \left( {{A^{ - 1}}} \right) = \frac{1}{{detA}}\)
3) (λA) -1 = λA -1 where λ is a scalar
Select the correct answer using the code given below:The adjoint of the matrix \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} 1&0&2\\ 2&1&0\\ 0&3&1 \end{array}} \right]\) is
If A is a square matrix, then what is adj (A -1 ) – (adj A) -1 equal to?
The matrix \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} 1&3&2\\ 1&{{\rm{x}} - 1}&1\\ 2&7&{{\rm{x}} - 3} \end{array}} \right]\)
Will have inverse for every real number x except for
Important Questions from Adjoint and Inverse of a Square Matrix
What should be the value of x so that the matrix \(\left( {\begin{array}{*{20}{c}} 2&4\\ { - 8}&{\rm{x}} \end{array}} \right)\) does not have an inverse?
What is the adjoint of the matrix \(\left( {\begin{array}{*{20}{c}} {\cos \left( { - \theta } \right)}&{ - \sin \left( { - \theta } \right)}\\ { - \sin \left( { - \theta } \right)}&{\cos \left( { - \theta } \right)} \end{array}} \right)\) ?
The inverse of the matrix A = \(\left( {\begin{array}{} 1&1&3\\ 1&3&{ - 3}\\ { - 2}&{ - 4}&{ - 4} \end{array}} \right)\) is:
If A is an identity matrix of order 3, then its inverse (A -1 )
What is the inverse of the matrix?
\(A = \left( {\begin{array}{*{20}{c}} {\cos \theta }&{\sin \theta }&0\\ { - \sin \theta }&{\cos \theta }&0\\ 0&0&1 \end{array}} \right)\)
