Find conditions on k that will make the matrix A invertible. To enter your answer, first select 'always', 'never', or whether k should be equal or not equal to specific values, then enter a value or a list of values separated by commas.
A = −2 3 0
2 k 0
−4 5 −3
A is invertible:
-2 +3 + 0
+2 +k + 0
-4 + 5 - 3
determinant must NOT be zero
6 k +18 NOT = 0
k NOT -3
Well, well, well, we have a curious case here. Let's see if we can make the matrix A invertible.
To determine when a matrix is invertible, we need to check if its determinant is non-zero. So, buckle up and let's find the determinant of A!
Det(A) = (-2)((k)(-3) - (0)(5)) - (3)((2)(-3) - (0)(-4)) + (0)((2)(5) - (k)(-4))
Simplifying this expression, we get:
Det(A) = (-6k) - (-18) + (0)
Det(A) = -6k + 18
Now, this is where things get interesting. The matrix A will be invertible if and only if the determinant is non-zero. Therefore, we need -6k + 18 ≠ 0.
To find out which values of k satisfy this condition, we just need to solve the equation:
-6k + 18 ≠ 0
Let's do some math magic and find the solution!
-6k ≠ -18
Dividing both sides by -6 (without breaking any math laws), we get:
k ≠ -18/(-6)
Simplifying further:
k ≠ 3
So, my dear friend, the condition for k to make the matrix A invertible is k ≠ 3.
And remember, when it comes to inverting matrices, always be non-zero!
A matrix is invertible if and only if its determinant is non-zero.
The determinant of matrix A is calculated as follows:
det(A) = (-2)((k)(-3)-(5)(0)) - (3)((2)(-3)-(5)(-4)) + (0)((2)(0)-(k)(-4))
= (-2)(-3k) - (3)(12) + (0)(8)
= 6k - 36
Therefore, A is invertible if and only if the determinant is non-zero, i.e., det(A) ≠ 0.
To determine the values of k that will make A invertible, we need to solve the equation det(A) ≠ 0:
6k - 36 ≠ 0
Simplifying the equation, we get:
6k ≠ 36
Divide both sides by 6:
k ≠ 6
Thus, A is invertible for all values of k except k = 6.
To determine whether the matrix A is invertible, we need to check if its determinant is non-zero. If the determinant is non-zero, then A is invertible; otherwise, it is not invertible.
The formula for the determinant of a 3x3 matrix is:
det(A) = a(ei - fh) - b(di - fg) + c(dh - eg)
Using the given matrix A:
A = | -2 3 0 |
| 2 k 0 |
| -4 5 -3 |
We can calculate the determinant as follows:
det(A) = (-2)((k)(-3) - (5)(0)) - (3)((2)(-3) - (4)(0)) + (0)((2)(5) - (4)(k))
= (-2)(-3k) - (3)(-6) + (0)(10 - 4k)
= 6k + 18
Now, we need to find the conditions on k for which det(A) is non-zero, meaning A is invertible.
We have det(A) = 6k + 18
To find the values of k, we need to solve the equation 6k + 18 = 0:
6k + 18 = 0
6k = -18
k = -18/6
k = -3
Therefore, the matrix A is invertible for all values of k except k = -3. In other words:
k ≠ -3