Subjective Type

Find inverse, by elementary row operations (if possible), of the following matrices $$\begin{bmatrix} 1 & -3 \\ -2 & 6 \end{bmatrix}$$

Solution

To check if the inverse exist we find the determinant:
We have:
A=[ 1 −3 −2 6 ]

So, |A|=1×6−(−2×−3)

⇒|A|=6−6=0

Since, |A|=0, hence the inverse does not exist

SIMILAR QUESTIONS

Matrices

Which of the following is the new row that results when you add rows $$1$$ and $$3$$? $$\begin{bmatrix}3&4&2&11\\9&1&0&0\\0&1&0&2\\0&0&6&1\end{bmatrix}$$

Matrices

If $$I=I=\left[ \begin{matrix} 1 \\ 0 \end{matrix}\begin{matrix} 0 \\ 1 \end{matrix} \right] ,j=\left[ \begin{matrix} 0 \\ -1 \end{matrix}\begin{matrix} 1 \\ 0 \end{matrix} \right] and B=\left[ \begin{matrix} cos\theta \\ -sin\theta \end{matrix}\begin{matrix} sin\theta \\ cos\theta \end{matrix} \right] ,$$ then B =

Matrices

Apply the elementary transformation of the following matrix. $$ B = \begin{bmatrix} 1 & -1 & 3 \\ 2 & 5 & 4 \end{bmatrix}, R_{1} \rightarrow R_{1} - R_{2} $$

Matrices

Find the inverse of the matrix $$A=\begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix}$$ by using column transformations.

Matrices

If $$A = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 5 & 3 \\ 0 & 2 & 1 \end{bmatrix} $$ find $$A^{-1}$$, using elementary row transformations.

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