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Linear Algebra

Applications and Exercises

The Inverse of a Block Matrix

Block-Diagonal Matrices

The Inverse of a Block-Diagonal Matrix

If A=(A100A2)A = \begin{pmatrix} A_1 & 0 \\ 0 & A_2 \end{pmatrix}, and both A1A_1 and A2A_2 are invertible, then: A−1=(A1−100A2−1)A^{-1} = \begin{pmatrix} A_1^{-1} & 0 \\ 0 & A_2^{-1} \end{pmatrix}

General Block Matrices

The Inverse of a General Block Matrix

For the block matrix (ABCD)\begin{pmatrix} A & B \\ C & D \end{pmatrix}, if AA is invertible then: (ABCD)−1=(A−1+A−1B(D−CA−1B)−1CA−1−A−1B(D−CA−1B)−1−(D−CA−1B)−1CA−1(D−CA−1B)−1)\begin{pmatrix} A & B \\ C & D \end{pmatrix}^{-1} = \begin{pmatrix} A^{-1} + A^{-1}B(D-CA^{-1}B)^{-1}CA^{-1} & -A^{-1}B(D-CA^{-1}B)^{-1} \\ -(D-CA^{-1}B)^{-1}CA^{-1} & (D-CA^{-1}B)^{-1} \end{pmatrix}

Applications of the Inverse Matrix

Solving Systems of Linear Equations

Method: for the system of linear equations AX=BAX = B, if AA is invertible then:

X=A−1BX = A^{-1}B

Example: Solve the system {x+2y=53x+4y=7\begin{cases} x + 2y = 5 \\ 3x + 4y = 7 \end{cases}

Solution:

  1. A=(1234)A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}, b=(57)b = \begin{pmatrix} 5 \\ 7 \end{pmatrix}
  2. A−1=(−211.5−0.5)A^{-1} = \begin{pmatrix} -2 & 1 \\ 1.5 & -0.5 \end{pmatrix}
  3. x=A−1b=(−211.5−0.5)(57)=(−34)x = A^{-1}b = \begin{pmatrix} -2 & 1 \\ 1.5 & -0.5 \end{pmatrix} \begin{pmatrix} 5 \\ 7 \end{pmatrix} = \begin{pmatrix} -3 \\ 4 \end{pmatrix}
  4. So x=−3x = -3, y=4y = 4

Matrix Equations

Method: for the matrix equation AX=BAX = B, if AA is invertible then:

X=A−1BX = A^{-1}B

Example: Solve the matrix equation AX=BAX = B, where: A=(1234)A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}, B=(5678)B = \begin{pmatrix} 5 & 6 \\ 7 & 8 \end{pmatrix}

Solution:

  1. A−1=(−211.5−0.5)A^{-1} = \begin{pmatrix} -2 & 1 \\ 1.5 & -0.5 \end{pmatrix}
  2. X=A−1B=(−211.5−0.5)(5678)X = A^{-1}B = \begin{pmatrix} -2 & 1 \\ 1.5 & -0.5 \end{pmatrix} \begin{pmatrix} 5 & 6 \\ 7 & 8 \end{pmatrix}
  3. =(−3−445)= \begin{pmatrix} -3 & -4 \\ 4 & 5 \end{pmatrix}

Exercises

Exercise 1

Decide whether the matrix A=(1234)A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} is invertible, and if so find its inverse.

block matrixinverse matrixmatrix equationdeterminant

Approach: First compute the determinant, then find the inverse matrix.

Steps:

  1. ∣A∣=1×4−2×3=4−6=−2≠0|A| = 1 \times 4 - 2 \times 3 = 4 - 6 = -2 \neq 0, so AA is invertible

  2. A11=4A_{11} = 4, A12=−3A_{12} = -3, A21=−2A_{21} = -2, A22=1A_{22} = 1

  3. A∗=(4−2−31)A^* = \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix}

  4. A−1=1−2(4−2−31)=(−211.5−0.5)A^{-1} = \frac{1}{-2} \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} -2 & 1 \\ 1.5 & -0.5 \end{pmatrix}

Answer: AA is invertible, and A−1=(−211.5−0.5)A^{-1} = \begin{pmatrix} -2 & 1 \\ 1.5 & -0.5 \end{pmatrix}.

Exercise 2

Find the inverse of the matrix A=(2113)A = \begin{pmatrix} 2 & 1 \\ 1 & 3 \end{pmatrix} by the elementary transformation method.

block matrixinverse matrixmatrix equationelementary transformation

Approach: Use the elementary transformation method.

Steps:

  1. Form the augmented matrix: (21∣1013∣01)\begin{pmatrix} 2 & 1 & | & 1 & 0 \\ 1 & 3 & | & 0 & 1 \end{pmatrix}

  2. Interchange the two rows: (13∣0121∣10)\begin{pmatrix} 1 & 3 & | & 0 & 1 \\ 2 & 1 & | & 1 & 0 \end{pmatrix}

  3. Add −2-2 times the first row to the second row: (13∣010−5∣1−2)\begin{pmatrix} 1 & 3 & | & 0 & 1 \\ 0 & -5 & | & 1 & -2 \end{pmatrix}

  4. Multiply the second row by −15-\frac{1}{5}: (13∣0101∣−1525)\begin{pmatrix} 1 & 3 & | & 0 & 1 \\ 0 & 1 & | & -\frac{1}{5} & \frac{2}{5} \end{pmatrix}

  5. Add −3-3 times the second row to the first row: (10∣35−1501∣−1525)\begin{pmatrix} 1 & 0 & | & \frac{3}{5} & -\frac{1}{5} \\ 0 & 1 & | & -\frac{1}{5} & \frac{2}{5} \end{pmatrix}

Answer: A−1=(35−15−1525)A^{-1} = \begin{pmatrix} \frac{3}{5} & -\frac{1}{5} \\ -\frac{1}{5} & \frac{2}{5} \end{pmatrix}.

Exercise 3

Find the adjugate matrix of the matrix A=(123014560)A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix}.

block matrixinverse matrixmatrix equationalgebraic cofactor

Approach: Compute the algebraic cofactor of each entry.

Steps:

  1. Compute the algebraic cofactor of each entry:

    • A11=∣1460∣=−24A_{11} = \begin{vmatrix} 1 & 4 \\ 6 & 0 \end{vmatrix} = -24
    • A12=−∣0450∣=20A_{12} = -\begin{vmatrix} 0 & 4 \\ 5 & 0 \end{vmatrix} = 20
    • A13=∣0156∣=−5A_{13} = \begin{vmatrix} 0 & 1 \\ 5 & 6 \end{vmatrix} = -5
    • A21=−∣2360∣=18A_{21} = -\begin{vmatrix} 2 & 3 \\ 6 & 0 \end{vmatrix} = 18
    • A22=∣1350∣=−15A_{22} = \begin{vmatrix} 1 & 3 \\ 5 & 0 \end{vmatrix} = -15
    • A23=−∣1256∣=4A_{23} = -\begin{vmatrix} 1 & 2 \\ 5 & 6 \end{vmatrix} = 4
    • A31=∣2314∣=5A_{31} = \begin{vmatrix} 2 & 3 \\ 1 & 4 \end{vmatrix} = 5
    • A32=−∣1304∣=−4A_{32} = -\begin{vmatrix} 1 & 3 \\ 0 & 4 \end{vmatrix} = -4
    • A33=∣1201∣=1A_{33} = \begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} = 1

  2. Arrange them in the transposed positions: adj(A)=(−2418520−15−4−541)\text{adj}(A) = \begin{pmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{pmatrix}

Answer: adj(A)=(−2418520−15−4−541)\text{adj}(A) = \begin{pmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{pmatrix}.

Exercise 4

Solve the matrix equation AX=BAX = B, where: A=(1234)A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}, B=(57)B = \begin{pmatrix} 5 \\ 7 \end{pmatrix}

block matrixinverse matrixmatrix equation

Approach: First find A−1A^{-1}, then compute X=A−1BX = A^{-1}B.

Steps:

  1. A−1=(−211.5−0.5)A^{-1} = \begin{pmatrix} -2 & 1 \\ 1.5 & -0.5 \end{pmatrix}

  2. X=A−1B=(−211.5−0.5)(57)X = A^{-1}B = \begin{pmatrix} -2 & 1 \\ 1.5 & -0.5 \end{pmatrix} \begin{pmatrix} 5 \\ 7 \end{pmatrix}

  3. =(−34)= \begin{pmatrix} -3 \\ 4 \end{pmatrix}

Answer: X=(−34)X = \begin{pmatrix} -3 \\ 4 \end{pmatrix}.

Exercise 5

Prove that: if AA is an invertible matrix, then (AT)−1=(A−1)T(A^T)^{-1} = (A^{-1})^T.

block matrixinverse matrixmatrix equationtranspose

Approach: Use the definition of the inverse matrix and the property of transposition.

Steps:

  1. Since AA is invertible, AA−1=A−1A=IAA^{-1} = A^{-1}A = I

  2. Taking transposes: (AA−1)T=IT=I(AA^{-1})^T = I^T = I

  3. Using the property of transposition: (AA−1)T=(A−1)TAT=I(AA^{-1})^T = (A^{-1})^T A^T = I

  4. Likewise: (A−1A)T=AT(A−1)T=I(A^{-1}A)^T = A^T(A^{-1})^T = I

  5. Therefore (A−1)T(A^{-1})^T is the inverse matrix of ATA^T

Answer: the proof is complete.

Tip: inverse matrices and adjugate matrices are important concepts in matrix theory, so you should master their definitions, properties and methods of computation.

Summary

Symbols Used in This Article

SymbolTypePronunciation / NoteMeaning in This Article
A−1A^{-1}Mathematical symbolA inverseThe inverse matrix of the matrix AA
AX=BAX = BMathematical symbolA X equals BThe matrix equation, whose solution is X=A−1BX = A^{-1}B when AA is invertible
X=A−1BX = A^{-1}BMathematical symbolX equals A inverse times BThe solution of the matrix equation AX=BAX=B
(A100A2)\begin{pmatrix} A_1 & 0 \\ 0 & A_2 \end{pmatrix}Mathematical symbolblock diagonal matrixA block-diagonal matrix, whose inverse is obtained by inverting each block separately
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