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Question

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

The correct answer is \({\rm{x}} = \frac{{11 \pm \sqrt 5 }}{2}\)

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 \([A | I]\).

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Important Questions from Adjoint and Inverse of a Square Matrix

  1. Let A be a matrix of order 3 × 3 and |A| = 4. If |2adj(3A)| = 2α 3β, then what is the value of (α + β)? 

  2. If A is a square matrix, then what is adj (A -1 ) – (adj A) -1 equal to?

  3. 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?

  4. Find the value of $|adj (2 \cdot adj A)|$ if matrix $A$ is of the order of $3$ and $|A| = 15$.

  5. If $A$ is a singular matrix of order $n \ge 2$, then which of the following statements is true?
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