The matrix \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} 1&3&2\\ 1&{{\rm{x}} - 1}&1\\ 2&7&{{\rm{x}} - 3} \end{array}} \right]\)
Understanding Matrix Inverses
A square matrix has an inverse if and only if its determinant is non-zero. If the determinant of a matrix is zero, the matrix is called singular, and it does not have an inverse. The question asks for the values of the real number x for which the given matrix \({\rm{A}}\) will not have an inverse. This means we need to find the values of x for which the determinant of matrix A is equal to zero.
Condition for Matrix Inverse
For a square matrix A to have an inverse, the following condition must be met:
\(\det({\rm{A}}) \neq 0\)
Conversely, for a matrix A to not have an inverse (i.e., be a singular matrix), the condition is:
\(\det({\rm{A}}) = 0\)
Calculating the Determinant of Matrix A
The given matrix is:
\({\rm{A}} = \left[ {\begin{array}{*{20}{c}} 1&3&2\\ 1&{{\rm{x}} - 1}&1\\ 2&7&{{\rm{x}} - 3} \end{array}} \right]\)
We will calculate the determinant of this 3x3 matrix. The determinant can be calculated by expanding along the first row:
\(\det({\rm{A}}) = 1 \cdot \left| {\begin{array}{*{20}{c}} {{\rm{x}} - 1}&1\\ 7&{{\rm{x}} - 3} \end{array}} \right| - 3 \cdot \left| {\begin{array}{*{20}{c}} 1&1\\ 2&{{\rm{x}} - 3} \end{array}} \right| + 2 \cdot \left| {\begin{array}{*{20}{c}} 1&{{\rm{x}} - 1}\\ 2&7 \end{array}} \right|\)
Now, we calculate the 2x2 determinants:
- \( \left| {\begin{array}{*{20}{c}} {{\rm{x}} - 1}&1\\ 7&{{\rm{x}} - 3} \end{array}} \right| = ({{\rm{x}} - 1})({\rm{x}} - 3) - (1)(7) = {{\rm{x}}^2} - 3{\rm{x}} - {\rm{x}} + 3 - 7 = {{\rm{x}}^2} - 4{\rm{x}} - 4 \)
- \( \left| {\begin{array}{*{20}{c}} 1&1\\ 2&{{\rm{x}} - 3} \end{array}} \right| = (1)({{\rm{x}} - 3}) - (1)(2) = {\rm{x}} - 3 - 2 = {\rm{x}} - 5 \)
- \( \left| {\begin{array}{*{20}{c}} 1&{{\rm{x}} - 1}\\ 2&7 \end{array}} \right| = (1)(7) - ({{\rm{x}} - 1})(2) = 7 - (2{\rm{x}} - 2) = 7 - 2{\rm{x}} + 2 = 9 - 2{\rm{x}} \)
Substitute these back into the determinant formula:
\(\det({\rm{A}}) = 1 \cdot ({{\rm{x}}^2} - 4{\rm{x}} - 4) - 3 \cdot ({\rm{x}} - 5) + 2 \cdot (9 - 2{\rm{x}})\)
\(\det({\rm{A}}) = {{\rm{x}}^2} - 4{\rm{x}} - 4 - 3{\rm{x}} + 15 + 18 - 4{\rm{x}}\)
Combine like terms:
\(\det({\rm{A}}) = {{\rm{x}}^2} + (-4{\rm{x}} - 3{\rm{x}} - 4{\rm{x}}) + (-4 + 15 + 18)\)
\(\det({\rm{A}}) = {{\rm{x}}^2} - 11{\rm{x}} + 29\)
Finding Values Where Inverse Doesn't Exist
The matrix A will not have an inverse when its determinant is zero. So, we set the determinant equal to zero and solve for x:
\({{\rm{x}}^2} - 11{\rm{x}} + 29 = 0\)
This is a quadratic equation in the form \(a{\rm{x}}^2 + b{\rm{x}} + c = 0\), where \(a=1\), \(b=-11\), and \(c=29\). We can use the quadratic formula to find the values of x:
\({\rm{x}} = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\)
Substitute the values of a, b, and c:
\({\rm{x}} = \frac{{ - ( - 11) \pm \sqrt {{{( - 11)}^2} - 4(1)(29)} }}{{2(1)}}\)
\({\rm{x}} = \frac{{11 \pm \sqrt {121 - 116} }}{2}\)
\({\rm{x}} = \frac{{11 \pm \sqrt 5 }}{2}\)
These are the two values of x for which the determinant of matrix A is zero. For any other real value of x, the determinant will be non-zero, and the matrix will have an inverse.
Conclusion on Matrix Inverse Existence
The matrix \({\rm{A}}\) will have an inverse for every real number x except for the values where its determinant is zero. We found that the determinant is zero when \({\rm{x}} = \frac{{11 \pm \sqrt 5 }}{2}\). Therefore, the matrix will have an inverse for every real number x except for these specific values.
Revision Table: Key Concepts
| Concept | Condition for Inverse | Determinant Calculation (3x3) | Quadratic Formula |
|---|---|---|---|
| Matrix Inverse | Exists if \(\det({\rm{A}}) \neq 0\). Does not exist if \(\det({\rm{A}}) = 0\) (Singular Matrix). | For matrix \(\left[ {\begin{array}{*{20}{c}} a&b&c\\ d&e&f\\ g&h&i \end{array}} \right]\), det is \(a(ei-fh) - b(di-fg) + c(dh-eg)\). | For \(ax^2 + bx + c = 0\), \(x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\). |
Additional Information: Related Matrix Properties
When a matrix is singular (\(\det(A) = 0\)), it has several implications in linear algebra:
- The system of linear equations \(Ax = b\) might have no solutions or infinitely many solutions.
- The columns (or rows) of the matrix are linearly dependent.
- The rank of the matrix is less than its dimension (for an n x n matrix, rank < n).
- The matrix transforms space in a way that collapses dimensions (e.g., maps a plane to a line or a point).
Finding the inverse of a non-singular matrix involves methods like using the adjugate matrix (\(A^{-1} = \frac{1}{\det(A)} \text{adj}(A)\)) or using Gaussian elimination (row operations) on the augmented matrix \([A | I]\).
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?
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?
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?
\(A = \left( {\begin{array}{*{20}{c}} {\cos \theta }&{\sin \theta }&0\\ { - \sin \theta }&{\cos \theta }&0\\ 0&0&1 \end{array}} \right)\)
