031 Review Part 2, Problem 3

Let  $B={\begin{bmatrix}1&-2&3&4\\0&3&0&0\\0&5&1&2\\0&-1&3&6\end{bmatrix}}.$

(a) Is  $B$  invertible? Explain.

(b) Define a linear transformation  $T$  by the formula  $T({\vec {x}})=B{\vec {x}}.$  Is  $T$  onto? Explain.

Foundations:
1. A matrix  $A$  is invertible if and only if  ${\text{det }}A\neq 0.$
2. A linear transformation  $T$  given by  $T({\vec {x}})=A{\vec {x}},$  where  $A$  is a  $m\times n$  matrix, is onto
if and only if the columns of  $A$  span  $\mathbb {R} ^{m}.$

Solution:

(a)

Step 1:
We begin by calculating  ${\text{det }}B.$
To do this, we use cofactor expansion along the second row first and then the first column.
So, we have
${\begin{array}{rcl}\displaystyle {{\text{det }}B}&=&\displaystyle {3(-1)^{2+2}\left|{\begin{array}{ccc}1&3&4\\0&1&2\\0&3&6\end{array}}\right|}\\&&\\&=&\displaystyle {3\cdot 1\cdot (-1)^{1+1}\left|{\begin{array}{cc}1&2\\3&6\end{array}}\right|}\\&&\\&=&\displaystyle {3(6-6)}\\&&\\&=&\displaystyle {0.}\end{array}}$

Step 2:
Since  ${\text{det }}B=0,$  we know that  $B$  is not invertible.

(b)

Step 1:
If  $T$  was onto, then  $B$  spans  $\mathbb {R} ^{4}.$
This would mean that  $B$  contains 4 pivots.

Step 2:
But, if  $B$  has 4 pivots, then  $B$  would be invertible, which is not true.
Hence,  $T$  is not onto.

Final Answer:
(a)     Since  ${\text{det }}B=0,$  we have that  $B$  is not invertible.
(b)     No, see explaination above.

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