find which of the given matrices are nonsingular. a) begin{bmatrix}1 & 2 &-3 -1 & 2&3 0 &8&0 end{bmatrix} b)begin{bmatrix}1 & 2 &-3 -1 & 2&3 0 &1&1 end{bmatrix} c) begin{bmatrix}1 & 1 &2 -1 & 3&4 -5 &7&8 end{bmatrix} d) begin{bmatrix}1 & 1 &4&-1 1 & 2&3&2 -1 &3&2&1-2&6&12&-4 end{bmatrix}

Question
Matrices
asked 2021-02-09
find which of the given matrices are nonsingular.
a) \(\begin{bmatrix}1 & 2 &-3 \\-1 & 2&3 \\ 0 &8&0 \end{bmatrix}\)
b)\(\begin{bmatrix}1 & 2 &-3 \\-1 & 2&3 \\ 0 &1&1 \end{bmatrix}\)
c) \(\begin{bmatrix}1 & 1 &2 \\-1 & 3&4 \\ -5 &7&8 \end{bmatrix}\)
d) \(\begin{bmatrix}1 & 1 &4&-1 \\1 & 2&3&2 \\ -1 &3&2&1\\-2&6&12&-4 \end{bmatrix}\)

Answers (1)

2021-02-10
Step 1
Since you have posted a multiple sub-parts problems ,we solve first 3 sub-parts for you .To get the remaining sub-parts solved please repost the complete question ,and mention sub-parts to be solved.
We have given some matrices and we have to determine which matrices are non-singular among them.
Step 2 First note that a matrix A is said to be nonsingular if \(det(A) \neq 0\).We will use this to determine the non singularity of the matrices .
a) we have \(A=\begin{bmatrix}1 & 2 &-3 \\-1 & 2&3 \\ 0 &8&0 \end{bmatrix}\)
Now \(det(A)=1 \times (0-24)-2 \times (0-0)-3 \times (-8-0)=-24+24=0\)
Since determinant is 0 so the matrix is singular ,not non-singular.
b) \(B=\begin{bmatrix}1 & 2 &-3 \\-1 & 2&3 \\ 0 &1&1 \end{bmatrix}\)
Now \(det(B)=1 \times (2-3)- 2 \times (-1-0)-3 \times (-1-0)=-1+2+3=4 \neq 0\)
Sine the determinant is non-zero so the matrix is non-singular .
c) \(C=\begin{bmatrix}1 & 1&2 \\-1 & 3&4 \\ -5 &7&8 \end{bmatrix}\)
Now, \(det(C)=1 \times (24-28)-1 \times (-8+20)+2 \times (-7+15)=0\)
Since determinant is 0 so the matrix is singular ,not non-singular.
0

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