Determinant of Inverse Matrix — GATE CE 2014

Step-by-step solution using determinant properties.

Engineering Mathematics

Determinant of \(A^{-1}\)

Question

If

\[ A= \begin{bmatrix} 1 & 2 & 3\\ 0 & 1 & 2\\ 0 & 0 & 4 \end{bmatrix} \]

then \(\det(A^{-1})\) is:

A. \(0.25\)

B. \(0.50\)

C. \(0.75\)

D. \(1.25\)

Given

The matrix is

\[ A= \begin{bmatrix} 1 & 2 & 3\\ 0 & 1 & 2\\ 0 & 0 & 4 \end{bmatrix} \]

We need to find

\[ \det(A^{-1}) \]

Concept Used

1. Determinant of a triangular matrix

For an upper triangular or lower triangular matrix, the determinant is equal to the product of its diagonal elements.

\[ \boxed{ \det(A) = a_{11}a_{22}\cdots a_{nn} } \]

2. Determinant of an inverse matrix

For an invertible square matrix \(A\),

\[ \boxed{ \det(A^{-1}) = \frac{1}{\det(A)} } \]

This follows from the property

\[ AA^{-1}=I \]

Taking determinants on both sides,

\[ \det(AA^{-1}) = \det(I) \]

Using

\[ \det(AB) = \det(A)\det(B) \]

we get

\[ \det(A)\det(A^{-1}) = 1 \]

Therefore,

\[ \boxed{ \det(A^{-1}) = \frac{1}{\det(A)} } \]

Solution

Step 1: Identify the matrix type

We are given

\[ A= \begin{bmatrix} 1 & 2 & 3\\ 0 & 1 & 2\\ 0 & 0 & 4 \end{bmatrix} \]

Observe that all the elements below the main diagonal are zero:

\[ a_{21}=0, \qquad a_{31}=0, \qquad a_{32}=0 \]

Therefore, \(A\) is an upper triangular matrix.

Step 2: Calculate \(\det(A)\)

For a triangular matrix, the determinant is the product of its diagonal elements.

\[ \det(A) = (1)(1)(4) \]

Therefore,

\[ \boxed{ \det(A)=4 } \]

Step 3: Find \(\det(A^{-1})\)

We use the determinant inverse formula:

\[ \boxed{ \det(A^{-1}) = \frac{1}{\det(A)} } \]

Since

\[ \det(A)=4 \]

substitute this value:

\[ \det(A^{-1}) = \frac{1}{4} \]

Hence,

\[ \boxed{ \det(A^{-1})=0.25 } \]

Step 4: Match with the options

We obtained

\[ \det(A^{-1})=0.25 \]

Comparing with the given options:

\[ \boxed{ \text{Option A}=0.25 } \]

Therefore, the correct answer is Option A.

Quick GATE Method

This question does not require calculating \(A^{-1}\).

The fastest approach is:

\[ \det(A) = 1\times1\times4 = 4 \]

Then immediately use:

\[ \det(A^{-1}) = \frac{1}{\det(A)} = \frac14 = 0.25 \]

Therefore:

\[ \boxed{\text{Option A}} \]

Final Answer

\[ \boxed{ \det(A^{-1})=0.25 } \]

Correct option: A

Formulas & Concepts to Remember

1. Determinant of a triangular matrix

\[ \boxed{ \det(A) = \prod_{i=1}^{n}a_{ii} } \]

2. Determinant of product

\[ \boxed{ \det(AB) = \det(A)\det(B) } \]

3. Determinant of inverse

\[ \boxed{ \det(A^{-1}) = \frac{1}{\det(A)} } \]

4. Identity matrix

\[ \boxed{ \det(I)=1 } \]

5. Condition for inverse

\[ \boxed{ A^{-1}\text{ exists} \iff \det(A)\neq0 } \]

GATE Tip

The biggest trap in this question is trying to calculate the complete inverse matrix \(A^{-1}\).

That is completely unnecessary.

The question asks only for its determinant:

\[ \det(A^{-1}) \]

So use the property directly:

\[ \boxed{ \det(A^{-1}) = \frac{1}{\det(A)} } \]

Since \(A\) is triangular:

\[ \det(A) = 1\times1\times4 = 4 \]

Therefore:

\[ \boxed{ \det(A^{-1})=\frac14=0.25 } \]

No inverse matrix calculation is required.

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