Która Macierz jest Odwracalna? Test Macierzy Osobliwej i Wyznaczanie k dla det = 0

Direct Answer / AI Overview: A square matrix $A$ is invertible (non-singular) if and only if its determinant is not equal to zero: $\det(A) \neq 0$. If $\det(A) = 0$, the matrix is singular, meaning it has no multiplicative inverse ($A^{-1}$ does not exist), its rows and columns are linearly dependent, and its geometric transformation collapses $n$-dimensional space into a lower dimension (zero area or volume). In algebra exams, questions asking to “find the value of $k$ for which the matrix is singular” require you to set up the determinant expression as an algebraic equation $\det(A(k)) = 0$ and solve for the unknown parameter $k$.

The Invertibility Test: Singular vs. Invertible Matrices

When analyzing a matrix, the fastest test of invertibility is evaluating its determinant:

To test any matrix instantly, input its coefficients into our free Determinant Solver.

How to Determine Which Matrix is Invertible (Step-by-Step)

Suppose an exam or assignment asks: “Which of the following matrices is invertible?”

$$\mathbf{M_1} = \begin{bmatrix} 3 & 6 \ 2 & 4 \end{bmatrix}, \quad \mathbf{M_2} = \begin{bmatrix} 5 & -2 \ 3 & 4 \end{bmatrix}$$

Testing Matrix $M_1$:

Using the 2x2 determinant formula ($\det = ad - bc$): $$\det(M_1) = (3)(4) - (6)(2) = 12 - 12 = 0$$ Since $\det(M_1) = 0$, $M_1$ is singular (NOT invertible). Notice that Row 1 is a scalar multiple of Row 2 ($[3, 6] = 1.5 \times [2, 4]$).

Testing Matrix $M_2$:

$$\det(M_2) = (5)(4) - (-2)(3) = 20 - (-6) = 20 + 6 = 26$$ Since $\det(M_2) = 26 \neq 0$, $M_2$ is invertible.

Exam Problem 1: Finding $k$ Such That a 2x2 Matrix is Singular

Problem: Find all values of $k$ for which the matrix $A$ is singular:

$$A = \begin{bmatrix} k - 2 & 3 \ 4 & k + 2 \end{bmatrix}$$

Step-by-Step Solution (Show Work Format):

1. State the Invertibility Criterion:
   A matrix is singular if and only if its determinant is zero: $\det(A) = 0$.

2. Set up the Determinant Formula:
   $$\det(A) = (k - 2)(k + 2) - (3)(4)$$

3. Expand and Simplify:
   Using the difference of squares:
   $$(k^2 - 4) - 12 = k^2 - 16$$

4. Set Equal to Zero and Solve for $k$:
   $$k^2 - 16 = 0 \implies k^2 = 16 \implies k = \pm 4$$

Conclusion: The matrix is singular when $k = 4$ or $k = -4$. For any other real number $k \neq \pm 4$, the matrix is invertible.

Exam Problem 2: Finding $k$ for a 3x3 Singular Matrix

Problem: For what value(s) of $k$ does the following $3 \times 3$ matrix have NO inverse?

$$B = \begin{bmatrix} 1 & 2 & k \ 0 & k & 4 \ 2 & 1 & 0 \end{bmatrix}$$

Step-by-Step Solution (Show Work Format):

1. Condition for No Inverse:
   The matrix has no inverse if $\det(B) = 0$.

2. Evaluate the 3x3 Determinant via First Column Expansion:
   Expanding down Column 1 (which contains a zero) using our How to Find 3x3 Determinant method:

   $$\det(B) = 1 \cdot \det\begin{bmatrix} k & 4 \ 1 & 0 \end{bmatrix} - 0 \cdot \det\begin{bmatrix} 2 & k \ 1 & 0 \end{bmatrix} + 2 \cdot \det\begin{bmatrix} 2 & k \ k & 4 \end{bmatrix}$$

3. Calculate the 2x2 Minors:

   • Minor 1: $(k)(0) - (4)(1) = -4$
   • Minor 3: $(2)(4) - (k)(k) = 8 - k^2$

4. Combine the Terms:
   $$\det(B) = 1(-4) + 2(8 - k^2) = -4 + 16 - 2k^2 = 12 - 2k^2$$

5. Solve $\det(B) = 0$:
   $$12 - 2k^2 = 0 \implies 2k^2 = 12 \implies k^2 = 6 \implies k = \pm \sqrt{6}$$

Final Answer: Matrix $B$ has no inverse if and only if $k = \sqrt{6}$ or $k = -\sqrt{6}$.

Geometric Meaning: Why Does $\det(A) = 0$ Mean Non-Invertible?

In linear algebra, the absolute value of the determinant measures the factor by which a linear transformation scales $n$-dimensional volume:

• In 2D, $|\det(A)|$ represents the area of the parallelogram formed by the column vectors.
• In 3D, $|\det(A)|$ represents the volume of the parallelepiped formed by the column vectors.

When $\det(A) = 0$, the transformation crushes the entire space into a lower dimension:

• In 2D, a 2D plane collapses onto a 1D line or a single 0D point (Area $= 0$).
• In 3D, a 3D space flattens onto a 2D plane, line, or point (Volume $= 0$).

Because information is lost when a higher dimension is collapsed into a lower one, no inverse function can reconstruct the original vectors. For visual examples, explore our guide on Calculating Area and Volume Using Determinants.