GO Classes CS/DA 2025 | Weekly Quiz 4 | Linear Algebra | Question: 7

Answer 1

Since
$$
\left[\begin{array}{l}
1 \\
0 \\
1
\end{array}\right]+\left[\begin{array}{l}
0 \\
1 \\
0
\end{array}\right]=\left[\begin{array}{l}
1 \\
1 \\
1
\end{array}\right],
$$
we have
$$
A \mathbf{x}+\mathrm{Ay}=\mathrm{Az}
$$
It follows that we have
$$
\mathrm{A}(\mathbf{x}+\mathbf{y}-\mathbf{z})=\mathbf{0}
$$
Since the vectors $\mathbf{x}, \mathbf{y}, \mathbf{z}$ are linearly independent, the linear combination $\mathbf{x}+\mathbf{y}-\mathbf{z} \neq \mathbf{0}$. Hence the matrix $\mathrm{A}$ is singular, and the determinant of $\mathrm{A}$ is zero.
(Recall that a matrix $\mathrm{A}$ is singular if and only if there exist nonzero vector $\mathbf{v}$ such that $A \mathbf{u}=\mathbf{0}$.)

Comments

Nonameishere just check this is in the Form of $AX=B$

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But same can be written as x + y = z. But that is a contradiction to given fact that x, y, z are linearly independent. What about that?

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since it is a contradiction then only we say that det A is 0 since there linear combination can't be 0.

Answer 2

Here you can easily solve this without any difficult technical or mathematical terms.

Just combine Ax, Ay, and Az together to create a Single Matrix B and find out its Determinant. 

This Determinant of B is the Determinant of A (the answer). this will always work if Linear Dependency is th

Here Matrix B = | Ax  Ay  Az |  or 

1	0	1
0	1	1
1	0	1

This Determinant will come out to be Zero. This trick will surely work if there is a linear dependency. Of course remember the actual process too