Answer to Question #200609 in Linear Algebra for Mpopo

Question

(5.1)Find det(−2A) and compare it to det(A) for

A= 1 1

3 -1

(5.2)Find det(−2A) and compare it to det(A) for

-2 1 3

A = 1 4 5

2 3 1

Answer 1

Solution:

(5.1):

$A=\begin{bmatrix}1&1\\ \:\:3&-1\end{bmatrix} \\|A|=\begin{vmatrix}1&1\\ \:\:3&-1\end{vmatrix}=1(-1)-3(1)=-4\ ...(i)$

$-2A=-2\begin{bmatrix}1&1\\ \:\:3&-1\end{bmatrix}=\begin{bmatrix}-2&-2\\ \:\:-6&2\end{bmatrix} \\|-2A|=\begin{vmatrix}-2&-2\\ \:\:-6&2\end{vmatrix}=-2(2)-(-2)(-6)=-4-12=-16$

Now, $|-2A|=-16=-4(4)=4A$ [Using (i)]

(5.2):

$A=\begin{bmatrix}-2&1&3\\ \:\:1&4&5\\ \:\:2&3&1\end{bmatrix} \\|A|=\begin{vmatrix}-2&1&3\\ \:\:1&4&5\\ \:\:2&3&1\end{vmatrix} \\=-2\cdot \det \begin{pmatrix}4&5\\ 3&1\end{pmatrix}-1\cdot \det \begin{pmatrix}1&5\\ 2&1\end{pmatrix}+3\cdot \det \begin{pmatrix}1&4\\ 2&3\end{pmatrix}$

$=-2\left(-11\right)-1\cdot \left(-9\right)+3\left(-5\right)=16$ ...(i)

Now, $-2A=-2\begin{bmatrix}-2&1&3\\ \:\:1&4&5\\ \:\:2&3&1\end{bmatrix}$

$=\begin{bmatrix}4&-2&-6\\ -2&-8&-10\\ -4&-6&-2\end{bmatrix}$

$|-2A|=4\cdot \det \begin{pmatrix}-8&-10\\ -6&-2\end{pmatrix}-\left(-2\right)\det \begin{pmatrix}-2&-10\\ -4&-2\end{pmatrix}-6\cdot \det \begin{pmatrix}-2&-8\\ -4&-6\end{pmatrix}$

$=4\left(-44\right)-\left(-2\right)\left(-36\right)-6\left(-20\right)=-128 \\=16(-8)=-8|A| \ \ [\text{using (i)}]$