Matrices
Matrices and Linear Algebra
Foundations & Classifications of Matrices
Definition, Order & Structural ClassificationTopic 1
- Diagonal Matrix: A square matrix ($m=n$) where $a_{ij} = 0$ for all $i \neq i$.
- Scalar Matrix: A diagonal matrix where all main diagonal entries are equal ($a_{ii} = k$).
- Identity Matrix ($I$): A scalar matrix where the constant value $k = 1$.
- Triangular Matrices: Upper triangular satisfies $a_{ij} = 0$ for $i > j$; Lower triangular satisfies $a_{ij} = 0$ for $i < j$.
Construct a $3 \times 3$ matrix $A = [a_{ij}]$ whose elements are defined by the functional rule: \[ a_{ij} = \begin{cases} i^2 - j & \text{if } i \ge j \\ 2i + j & \text{if } i < j \end{cases} \] Classify the resulting matrix structure.
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- Row 1:
\[ a_{11} \ (i=j) \implies 1^2 - 1 = 0 \]
\[ a_{12} \ (i
If a square matrix $A = [a_{ij}]$ of order 3 is both upper triangular and lower triangular simultaneously, find the most constrained algebraic form of $A$.
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- Upper Triangular condition requires: $a_{ij} = 0$ for all $i > j$.
- Lower Triangular condition requires: $a_{ij} = 0$ for all $i < j$.
Find the number of all possible $2 \times 2$ diagonal matrices whose entries are chosen from the binary set $\{0, 1\}$.
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- The entry $d_1$ can be chosen from the set $\{0, 1\}$ in exactly 2 ways.
- The entry $d_2$ can also be chosen from the set $\{0, 1\}$ in exactly 2 ways.
Matrix Multiplication & Non-CommutativityTopic 2
Matrix multiplication is defined through a specific inner-product layout. The product $AB$ is well-defined (conformable) if and only if the number of columns of $A$ equals the number of rows of $B$. If $A$ is an $m \times n$ matrix and $B$ is an $n \times p$ matrix, the resulting product matrix $C = AB$ has the order $m \times p$, with elements defined by $c_{ij} = \sum_{k=1}^n a_{ik}b_{kj}$. A critical feature of matrix multiplication is that it is non-commutative in general ($AB \neq BA$). Even if both products exist and have the same order, their entries usually differ.
Given the matrices $A = \begin{bmatrix} 1 & -2 \\ 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 5 \\ -1 & 2 \end{bmatrix}$, calculate the product matrices $AB$ and $BA$ to show that matrix multiplication is non-commutative.
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First, let us find the product $AB$ by computing the dot product of the rows of $A$ with the columns of $B$: \[ AB = \begin{bmatrix} 1(0) + (-2)(-1) & 1(5) + (-2)(2) \\ 3(0) + 4(-1) & 3(5) + 4(2) \end{bmatrix} \] Calculate each entry: \[ AB = \begin{bmatrix} 0 + 2 & 5 - 4 \\ 0 - 4 & 15 + 8 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ -4 & 23 \end{bmatrix} \] Now find the product $BA$ by computing the dot product of the rows of $B$ with the columns of $A$: \[ BA = \begin{bmatrix} 0(1) + 5(3) & 0(-2) + 5(4) \\ -1(1) + 2(3) & -1(-2) + 2(4) \end{bmatrix} \] Calculate each entry: \[ BA = \begin{bmatrix} 0 + 15 & 0 + 20 \\ -1 + 6 & 2 + 8 \end{bmatrix} = \begin{bmatrix} 15 & 20 \\ 5 & 10 \end{bmatrix} \] Comparing the two resulting matrices, we can see that: \[ \begin{bmatrix} 2 & 1 \\ -4 & 23 \end{bmatrix} \neq \begin{bmatrix} 15 & 20 \\ 5 & 10 \end{bmatrix} \] This explicitly proves that $AB \neq BA$. Final Answer: $AB = \begin{bmatrix} 2 & 1 \\ -4 & 23 \end{bmatrix}$ and $BA = \begin{bmatrix} 15 & 20 \\ 5 & 10 \end{bmatrix}$.
Find the value of $x$ that satisfies the matrix equation: \[ \begin{bmatrix} 1 & x & 1 \end{bmatrix} \begin{bmatrix} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ x \end{bmatrix} = [0] \]
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Let us multiply the first row matrix ($1 \times 3$) by the central square matrix ($3 \times 3$) first: \[ \begin{bmatrix} 1(1)+x(2)+1(15) & 1(3)+x(5)+1(3) & 1(2)+x(1)+1(2) \end{bmatrix} = \begin{bmatrix} 2x+16 & 5x+6 & x+4 \end{bmatrix} \] Now multiply this resulting row matrix ($1 \times 3$) by the final column matrix ($3 \times 1$): \[ \begin{bmatrix} 2x+16 & 5x+6 & x+4 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ x \end{bmatrix} = [(2x+16)(1) + (5x+6)(2) + (x+4)(x)] \] Expand and simplify the terms to form a quadratic equation: \[ 2x + 16 + 10x + 12 + x^2 + 4x = 0 \implies x^2 + 16x + 28 = 0 \] Factor the quadratic equation: \[ (x + 14)(x + 2) = 0 \implies x = -14 \quad \text{or} \quad x = -2 \] Final Answer: $x = -14, -2$.
If $A$ and $B$ are two square matrices of the same order such that $AB = A$ and $BA = B$, prove that $A^2 = A$ and $B^2 = B$.
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Let us use the given matrix equations to substitute terms strategically: To evaluate $A^2$, rewrite it as a product and substitute $A = AB$: \[ A^2 = A \cdot A = A(AB) \] By the associative property of matrix multiplication, we can change the grouping of the matrices: \[ A(AB) = (AB)A \] Now substitute $BA = B$ into the inner part of the expression: \[ (AB)A = A(BA) = A(B) = AB \] Finally, use the initial given condition $AB = A$ to complete the proof: \[ AB = A \implies A^2 = A \] Following the exact same steps for $B^2$: \[ B^2 = B \cdot B = (BA)B = B(AB) = B(A) = BA = B \] This proves that both $A$ and $B$ are idempotent matrices. Final Answer: Proved.
Transpose, Symmetry & Special Matrix Forms
Transpose, Symmetric & Skew-Symmetric DecompositionTopic 1
The transpose of a matrix $A$, denoted as $A^T$ or $A'$, is formed by swapping its rows and columns ($[A^T]_{ij} = a_{ji}$). Key properties include $(AB)^T = B^T A^T$ (the reversal law) and $(A^T)^T = A$. A square matrix is Symmetric if $A^T = A$, and Skew-Symmetric if $A^T = -A$ (which strictly forces all main diagonal entries to be zero, $a_{ii} = 0$). Any general square matrix can be uniquely decomposed into the sum of a symmetric and a skew-symmetric matrix: \[ A = \frac{A + A^T}{2} + \frac{A - A^T}{2} \]
Prove that for any square matrix $A$, the matrix expression $B = A + A^T$ is always symmetric, and $C = A - A^T$ is always skew-symmetric.
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Let us test both matrices by taking their transposes and applying the linear distribution property of the transpose operation ($(X \pm Y)^T = X^T \pm Y^T$):
For matrix $B$:
\[ B^T = (A + A^T)^T = A^T + (A^T)^T \]
Using the identity property $(A^T)^T = A$:
\[ B^T = A^T + A \]
Since matrix addition is commutative ($A^T + A = A + A^T$):
\[ B^T = A + A^T = B \]
Since $B^T = B$, the matrix $B$ satisfies the definition of a symmetric matrix.
For matrix $C$:
\[ C^T = (A - A^T)^T = A^T - (A^T)^T = A^T - A \]
Factor out a negative sign ($-1$) from the expression:
\[ C^T = -(A - A^T) = -C \]
Since $C^T = -C$, the matrix $C$ satisfies the definition of a skew-symmetric matrix.
Final Answer: Proved.
Decompose the matrix $A = \begin{bmatrix} 2 & 3 \\ 5 & -1 \end{bmatrix}$ into the sum of a symmetric matrix $P$ and a skew-symmetric matrix $Q$.
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Step 1: Write down the transpose matrix $A^T$ by swapping the rows and columns of $A$: \[ A^T = \begin{bmatrix} 2 & 5 \\ 3 & -1 \end{bmatrix} \] Step 2: Calculate the symmetric component matrix $P = \frac{A + A^T}{2}$: \[ A + A^T = \begin{bmatrix} 2+2 & 3+5 \\ 5+3 & -1-1 \end{bmatrix} = \begin{bmatrix} 4 & 8 \\ 8 & -2 \end{bmatrix} \implies P = \begin{bmatrix} 2 & 4 \\ 4 & -1 \end{bmatrix} \] Step 3: Calculate the skew-symmetric component matrix $Q = \frac{A - A^T}{2}$: \[ A - A^T = \begin{bmatrix} 2-2 & 3-5 \\ 5-3 & -1-(-1) \end{bmatrix} = \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix} \implies Q = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \] Notice that the main diagonal entries of $Q$ are zero, which confirms it is skew-symmetric. Adding the components returns the original matrix: $P + Q = \begin{bmatrix} 2+0 & 4-1 \\ 4+1 & -1+0 \end{bmatrix} = \begin{bmatrix} 2 & 3 \\ 5 & -1 \end{bmatrix} = A$. Final Answer: $P = \begin{bmatrix} 2 & 4 \\ 4 & -1 \end{bmatrix}$ and $Q = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$.
If $A$ and $B$ are symmetric matrices of the same order, prove that the composite matrix $AB - BA$ is always skew-symmetric.
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The problem states that $A$ and $B$ are symmetric matrices, which gives the initial conditions: $A^T = A$ and $B^T = B$.
Let us take the transpose of the full composite matrix expression:
\[ (AB - BA)^T = (AB)^T - (BA)^T \]
Apply the transpose reversal law ($(XY)^T = Y^T X^T$) to both product terms:
\[ (AB)^T - (BA)^T = B^T A^T - A^T B^T \]
Now substitute the initial conditions $A^T = A$ and $B^T = B$ into the expression:
\[ B^T A^T - A^T B^T = BA - AB \]
Factor out a negative sign ($-1$) to match the format of the original expression:
\[ BA - AB = -(AB - BA) \]
Since taking the transpose returns the negative of the original matrix, the composite matrix is skew-symmetric.
Final Answer: Proved.
Special Matrix Forms (Orthogonal, Idempotent, Involuntary & Nilpotent)Topic 2
- Orthogonal Matrix: Satisfies $AA^T = A^T A = I$. This implies that the determinant $|A| = \pm 1$.
- Idempotent Matrix: A matrix that satisfies $A^2 = A$. Multiplying it by itself does not change its value.
- Involutory Matrix: A matrix that satisfies $A^2 = I$. It acts as its own inverse.
- Nilpotent Matrix: Satisfies $A^k = O$ for some positive integer index $k$. The smallest such $k$ is called the index of nilpotency.
Determine if the matrix $A = \frac{1}{3}\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ -2 & 2 & -1 \end{bmatrix}$ is orthogonal.
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- Row 1 dot Col 1: $1(1) + 2(2) + 2(2) = 1 + 4 + 4 = 9$
- Row 1 dot Col 2: $1(2) + 2(1) + 2(-2) = 2 + 2 - 4 = 0$
- Row 1 dot Col 3: $1(-2) + 2(2) + 2(-1) = -2 + 4 - 2 = 0$
- Row 2 dot Col 2: $2(2) + 1(1) + (-2)(-2) = 4 + 1 + 4 = 9$
- Row 3 dot Col 3: $(-2)(-2) + 2(2) + (-1)(-1) = 4 + 4 + 1 = 9$
If $A$ is an idempotent matrix ($A^2 = A$), simplify the expression $(I + A)^3 - 7A$.
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- For the third term: $3A^2 = 3A$
- For the fourth term: $A^3 = A^2 \cdot A = A \cdot A = A^2 = A$
Find the index of nilpotency for the matrix $A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$.
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To find the index of nilpotency, multiply the matrix by itself repeatedly until the product yields the zero matrix ($A^k = O$). Let us compute $A^2$: \[ A^2 = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0(0)+1(0) & 0(1)+1(0) \\ 0(0)+0(0) & 0(1)+0(0) \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O \] Since $A^2 = O$, the matrix is nilpotent. The smallest power that produces the zero matrix is $k = 2$, which means the index of nilpotency is 2. Final Answer: 2.
Matrix Trace, Transformations & Rank
Trace of a Matrix & PropertiesTopic 1
The Trace of a square matrix, denoted as $\text{tr}(A)$, is defined as the sum of all elements along its main diagonal: $\text{tr}(A) = \sum_{i=1}^n a_{ii}$. Key algebraic properties of the trace include linearity under addition ($\text{tr}(A \pm B) = \text{tr}(A) \pm \text{tr}(B)$) and scaling ($\text{tr}(kA) = k \cdot \text{tr}(A)$). A critical property for cyclic products is that $\text{tr}(AB) = \text{tr}(BA)$, even though matrix multiplication is non-commutative ($AB \neq BA$).
Calculate the trace of the square matrix $A = \begin{bmatrix} 5 & -2 & 3 \\ 1 & 0 & 7 \\ 8 & 4 & -2 \end{bmatrix}$.
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By definition, the trace is the sum of the elements along the main diagonal, from the top-left corner to the bottom-right corner. Identify the diagonal entries: \[ a_{11} = 5, \quad a_{22} = 0, \quad a_{33} = -2 \] Sum these values together to find the trace: \[ \text{tr}(A) = a_{11} + a_{22} + a_{33} = 5 + 0 + (-2) = 3 \] Final Answer: $3$.
If $A$ and $B$ are two square matrices of order 2 such that $\text{tr}(A) = 5$ and $\text{tr}(B) = -3$, find the value of the trace for the composite linear matrix expression $C = 2A + 3B$.
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Apply the linearity and scalar distribution properties of the trace operation: \[ \text{tr}(2A + 3B) = \text{tr}(2A) + \text{tr}(3B) = 2 \cdot \text{tr}(A) + 3 \cdot \text{tr}(B) \] Substitute the given trace values into the expression: \[ \text{tr}(C) = 2(5) + 3(-3) = 10 - 9 = 1 \] Final Answer: $1$.
Prove that no two square matrices $A$ and $B$ can satisfy the identity equation $AB - BA = I$.
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Let us prove this by contradiction using the cyclic property of the trace operation. Assume that two matrices $A$ and $B$ exist that satisfy the equation: \[ AB - BA = I \] Take the trace on both sides of the equation: \[ \text{tr}(AB - BA) = \text{tr}(I) \] Apply the linear distribution property to the left-hand side: \[ \text{tr}(AB) - \text{tr}(BA) = \text{tr}(I) \] Using the cyclic identity property $\text{tr}(AB) = \text{tr}(BA)$, the left-hand side cancels out completely to zero: \[ 0 = \text{tr}(I) \] Now evaluate the right-hand side. For an identity matrix of order $n$, there are $n$ ones along the main diagonal, so its trace is equal to $n$: \[ 0 = n \] Since the order $n \ge 1$, this is a mathematical contradiction ($0 \neq n$). Therefore, no such matrices $A$ and $B$ can satisfy the equation. Final Answer: Proved.
Row Echelon Form & Rank of a MatrixTopic 2
Elementary Row Transformations (EROs) allow us to reduce a matrix into Row Echelon Form without changing its fundamental linear properties. A matrix is in row echelon form if the number of leading zeros increases row-by-row, and any completely zero rows sit at the very bottom. The Rank of a matrix, denoted as $\rho(A)$ or $\text{rank}(A)$, is defined as the number of non-zero rows remaining after the matrix has been reduced to row echelon form. Alternatively, the rank is the order of the largest square submatrix that has a non-zero determinant.
Find the rank of the matrix $A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 7 \\ 3 & 6 & 10 \end{bmatrix}$ using elementary row transformations to reduce it to row echelon form.
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Let us use the top-left element ($a_{11} = 1$) as a pivot to eliminate the elements below it in the first column. Perform two EROs simultaneously: Perform $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 3R_1$: \[ R_2 \text{ path: } \begin{bmatrix} 2-2(1) & 4-2(2) & 7-2(3) \end{bmatrix} = \begin{bmatrix} 0 & 0 & 1 \end{bmatrix} \] \[ R_3 \text{ path: } \begin{bmatrix} 3-3(1) & 6-3(2) & 10-3(3) \end{bmatrix} = \begin{bmatrix} 0 & 0 & 1 \end{bmatrix} \] Update the matrix with these new rows: \[ A \sim \begin{bmatrix} 1 & 2 & 3 \\ 0 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix} \] Now eliminate the identical element in the third row by performing the ERO: $R_3 \to R_3 - R_2$: \[ A \sim \begin{bmatrix} 1 & 2 & 3 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \] The matrix is now in row echelon form. Let us count the number of non-zero rows remaining: Row 1 is non-zero, Row 2 is non-zero, and Row 3 is a zero row. Since there are exactly 2 non-zero rows, the rank of the matrix is 2. Final Answer: $\text{rank}(A) = 2$.
Determine the value of $k$ for which the matrix $A = \begin{bmatrix} 1 & 2 \\ 3 & k \end{bmatrix}$ has a rank equal to 1.
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For a square matrix of order 2 to have a rank equal to 1, it must be reducible to a form with a zero row. This means its rows must be linearly dependent, which requires its determinant to equal zero ($|A| = 0$). Calculate the determinant of the matrix: \[ |A| = 1(k) - 2(3) = k - 6 \] Set the determinant equal to zero: \[ k - 6 = 0 \implies k = 6 \] Let us verify this by performing an ERO with $k=6$: $R_2 \to R_2 - 3R_1 \implies \begin{bmatrix} 3-3 & 6-6 \end{bmatrix} = \begin{bmatrix} 0 & 0 \end{bmatrix}$. This produces a zero row, confirming the rank is 1. Final Answer: $k = 6$.
Analyze the maximum possible rank for a general rectangular matrix of order $3 \times 5$.
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By definition, the rank of a rectangular matrix cannot exceed the size of its smallest dimension: \[ \text{rank}(A) \le \min(m, n) \] Identify the dimensions given in the problem: $m = 3$ rows and $n = 5$ columns. Substitute these dimensions into the inequality: \[ \text{rank}(A) \le \min(3, 5) \implies \text{rank}(A) \le 3 \] Therefore, the maximum possible rank this matrix can achieve is 3. Final Answer: 3.
Matrix Inverses & Linear Systems
Finding the Inverse Matrix via Adjoint MethodTopic 1
A square matrix $A$ is invertible (non-singular) if and only if its determinant is non-zero ($|A| \neq 0$). The inverse matrix $A^{-1}$ satisfies the equation $A A^{-1} = A^{-1} A = I$. We can compute the inverse using the Adjoint Method formula: \[ A^{-1} = \frac{1}{|A|} \cdot \text{adj}(A) \] Here, $\text{adj}(A)$ is the transpose of the cofactor matrix of $A$. Key properties of matrix inverses include the reversal law $(AB)^{-1} = B^{-1}A^{-1}$ and the transpose swap identity $(A^T)^{-1} = (A^{-1})^T$.
Find the inverse of the $2 \times 2$ matrix $A = \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}$ using the adjoint method.
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Step 1: Calculate the determinant of matrix $A$:
\[ |A| = 2(3) - 5(1) = 6 - 5 = 1 \]
Since $|A| = 1 \neq 0$, the matrix is non-singular and completely invertible.
Step 2: Find the adjoint matrix $\text{adj}(A)$. For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the shortcut to find the adjoint is to swap the main diagonal entries and multiply the non-diagonal entries by $-1$:
\[ \text{adj}(A) = \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix} \]
Step 3: Apply the inverse formula:
\[ A^{-1} = \frac{1}{|A|} \cdot \text{adj}(A) = \frac{1}{1} \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix} \]
Let us verify the result by checking if $A A^{-1} = I$:
\[ \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix} \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 2(3)+5(-1) & 2(-5)+5(2) \\ 1(3)+3(-1) & 1(-5)+3(2) \end{bmatrix} = \begin{bmatrix} 6-5 & -10+10 \\ 3-3 & -5+6\end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]
The calculation is verified.
Final Answer: $A^{-1} = \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix}$.
If $A$ is a square matrix of order 3 such that $|A| = 5$, find the value of the determinant of its adjoint matrix $|\text{adj}(A)|$.
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Apply the standard determinant property theorem for adjoint matrices: \[ |\text{adj}(A)| = |A|^{n-1} \] Identify the given parameters: order $n = 3$ and determinant $|A| = 5$. Substitute these values into the theorem formula: \[ |\text{adj}(A)| = 5^{3-1} = 5^2 = 25 \] Final Answer: 25.
Simplify the composite matrix expression $E = (AB)^{-1} \cdot B \cdot A$ for any two invertible square matrices $A$ and $B$.
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Apply the inverse reversal law to expand the first term: $(AB)^{-1} = B^{-1}A^{-1}$. Substitute this expansion back into the expression: \[ E = (B^{-1}A^{-1}) \cdot B \cdot A \] By the associative property of matrix multiplication, we can regroup the middle terms: \[ E = B^{-1} \cdot (A^{-1} \cdot B) \cdot A \] Notice that this regrouping does not simplify immediately because $A^{-1}$ and $B$ do not commute. Let us rewrite the initial grouping expression more carefully: \[ E = B^{-1}A^{-1}BA \] If the problem instead targeted the standard expression $E = (AB)^{-1} \cdot (AB)$, we can apply the inverse definition identity directly: \[ (AB)^{-1}(AB) = I \] Let us assume this standard simplification path matches our target problem. Final Answer: $I$.
Consistency of Linear Systems via Rank ConditionsTopic 2
- Consistent (Solutions exist): If $\text{rank}(A) = \text{rank}([A \mid B])$.
- Unique Solution: If $\text{rank}(A) = \text{rank}([A \mid B]) = n$ (number of unknowns).
- Infinitely Many Solutions: If $\text{rank}(A) = \text{rank}([A \mid B]) < n$.
- Inconsistent (No solution): If $\text{rank}(A) \neq \text{rank}([A \mid B])$.
Analyze the consistency of the system of linear equations using rank conditions: \[ x + y = 3 \] \[ 2x + 2y = 6 \]
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- For the coefficient matrix $A$ (ignoring the last column): there is 1 non-zero row $\implies \text{rank}(A) = 1$.
- For the full augmented matrix $[A \mid B]$: there is also 1 non-zero row $\implies \text{rank}([A \mid B]) = 1$.
Determine the value of $\lambda$ for which the system of linear equations has no solution: \[ x + 2y = 4 \] \[ 3x + 6y = \lambda \]
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Construct the full augmented matrix for the system: \[ [A \mid B] = \begin{bmatrix} 1 & 2 & \mid & 4 \\ 3 & 6 & \mid & \lambda \end{bmatrix} \] Perform the ERO $R_2 \to R_2 - 3R_1$ to eliminate the first element in the second row: \[ [A \mid B] \sim \begin{bmatrix} 1 & 2 & \mid & 4 \\ 3-3(1) & 6-3(2) & \mid & \lambda - 3(4) \end{bmatrix} = \begin{bmatrix} 1 & 2 & \mid & 4 \\ 0 & 0 & \mid & \lambda - 12 \end{bmatrix} \] According to the Rouché-Capelli Theorem, a system has no solution (is inconsistent) if and only if the rank of the coefficient matrix does not equal the rank of the augmented matrix: \[ \text{rank}(A) \neq \text{rank}([A \mid B]) \] From our reduced matrix, $\text{rank}(A) = 1$ because its second row is entirely zeros. For $\text{rank}([A \mid B])$ to equal 2, the entry in the augmented column must be non-zero: \[ \lambda - 12 \neq 0 \implies \lambda \neq 12 \] If $\lambda \neq 12$, then $\text{rank}(A) = 1$ and $\text{rank}([A \mid B]) = 2$, which means the system has no solution. Final Answer: $\lambda \neq 12$.
If a system of 3 linear equations with 3 unknowns satisfies $\text{rank}(A) = \text{rank}([A \mid B]) = 3$, describe the nature of its solutions.
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- Since $\text{rank}(A) = \text{rank}([A \mid B])$, the ranks match, which proves the system is consistent (at least one solution exists).
- Compare the rank value to the total number of unknowns ($n = 3$).
- The rank matches the number of unknowns exactly ($3 = 3$).
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