Algebra

Matrices & Determinants

Easy

applied

If A is an invertible matrix of order 3 and the determinant of A is 9, then the determinant of $A^{-1}$ is:

1/9

✅ Correct Option: 4

For any invertible matrix $A$ :

$\text{det}(A) \times \text{det}(A^{-1}) = 1$

Given that $A$ is an invertible matrix of order 3 with $\text{det}(A) = 9$ .

Using the property:

$\text{det}(A) \times \text{det}(A^{-1}) = 1$

$9 \times \text{det}(A^{-1}) = 1$

$\text{det}(A^{-1}) = \dfrac{1}{9}$

This property holds because when a matrix is multiplied by its inverse, the result is the identity matrix:

$A \times A^{-1} = I$

Taking determinant on both sides:

$\text{det}(A \times A^{-1}) = \text{det}(I)$

$\text{det}(A) \times \text{det}(A^{-1}) = 1$

The determinant of an inverse matrix is always the reciprocal of the original determinant.

Therefore, $\text{det}(A^{-1}) = \dfrac{1}{9}$

2026: 31 May Shift 2

The number of all possible matrices of order $2 \times 2$ with each entry 0 or 1 having their determinants as positive value is :

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If A and B are symmetric matrices of same order, then which of the following statements are true?

(A) $AB^T+B^TA$ is a skew-symmetric matrix.

(B) $AB^T-BA^T$ is a skew-symmetric matrix

(D) $(AB)^{-1}(BA)^T=A$

Choose the correct answer from the options given below:

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If $\begin{vmatrix} 2x & 2 \\ 4 & x \end{vmatrix} = 10$ , then $x$ is:

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If A is a square matrix such that $A^2 = I$ , then $(A+I)^3 + (A-I)^3 - 7A$ is equal to

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For matrix $A = \begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}$ the value of $A^2 - 6A$ is equal to:(where I is an identity matrix of order 2)

If A is an invertible matrix of order 3 and the determinant of A is 9, then the determinant of A−1A^{-1}A−1 is: