Which of the following matrices is not invertible?

[\begin{matrix} 1 & 1 0 & 1 \end{matrix}]

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[\begin{matrix} - 1 & - 1 - 1 & 2 \end{matrix}]

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[\begin{matrix} 2 & 3 4 & 6 \end{matrix}]

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[\begin{matrix} 2 & - 2 1 & 1 \end{matrix}]

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Solution

The correct option is C

[\begin{matrix} 2 & 3 4 & 6 \end{matrix}]

A square matrix that is not invertible is called singular matrix in which its determinant is 0.

[\begin{matrix} 2 & 3 4 & 6 \end{matrix}]

is a non invertible matrix.

∵

its determinant is

12 - 12 = 0

Option C is correct.

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Similar questions

Express the following matrix as the sum of a symmetric and a skew-symmetric matrices;

$⎡ ⎢ ⎣ \begin{matrix} 6 & - 2 & 2 - 2 & 3 & - 1 2 & - 1 & 3 \end{matrix} ⎤ ⎥ ⎦$

Q. If A and B are invertible matrices, which of the following statement is not correct.
(a)

adj A = |A| A^{- 1}

(b)

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

(c)

{(A + B)}^{- 1} = A^{- 1} + B^{- 1}

(d)

{(A B)}^{- 1} = B^{- 1} A^{- 1}

Q. If A and B are invertible matrices, then which one of the following is not correct?
(a) adj A =

|A|

A^-1 (b) det (A^-1) = [det(A)]^-1
(c) (AB)^-1 = B^-1A^-1 (d) (A+B)^-1 = B^-1 + A^-1

If A and B are invertible matrices, then which of the following is not correct?

(a) $a d j A = | A | . A^{- 1}$
(b) $d e t (A)^{- 1} = [d e t (A)]^{- 1}$
(c) $(A B)^{- 1} = B^{- 1} A^{- 1}$
(d) $(A + B)^{- 1} = B^{- 1} + A^{- 1}$

If possible, using elementary row transformations, find the inverse of the following matrices.
$⎡ ⎢ ⎣ \begin{matrix} 2 & 3 & - 3 - 1 & - 2 & 2 1 & 1 & - 1 \end{matrix} ⎤ ⎥ ⎦$

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Which of the following matrices is not invertible?

The correct option is C [2346] A square matrix that is not invertible is called singular matrix in which its determinant is 0.[2346] is a non invertible matrix. ∵ its determinant is 12−12=0Option C is correct.

The correct option is C $[\begin{matrix} 2 & 3 4 & 6 \end{matrix}]$
A square matrix that is not invertible is called singular matrix in which its determinant is 0.
$[\begin{matrix} 2 & 3 4 & 6 \end{matrix}]$ is a non invertible matrix.
$∵$ its determinant is $12 - 12 = 0$
Option C is correct.