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Compute (AB)¹ where:
$straight A space equals space open square brackets table row 1 cell space space space space 1 end cell cell space space space 2 end cell row 0 cell space space space 2 end cell cell space minus 3 end cell row 3 cell space minus 2 end cell cell space space space space 4 end cell end table close square brackets space space space space space space space straight B to the power of negative 1 end exponent space equals space open square brackets table row 1 cell space space space space space 2 end cell cell space space space space space 0 end cell row 0 cell space space space 3 end cell cell space space minus 1 end cell row 1 cell space space 0 end cell cell space space space space 2 end cell end table close square brackets$

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222.

If $straight A space equals space open square brackets table row 2 cell space space space 2 end cell cell space space space 1 end cell row cell negative 2 end cell cell space space space 1 end cell cell space space 2 end cell row 1 cell space minus 2 end cell cell space space 2 end cell end table close square brackets space space space and space straight B space equals space open square brackets table row 1 cell space space 3 end cell cell space space 2 end cell row 1 cell space space space 1 end cell cell space 1 end cell row 2 cell negative 3 end cell cell space 1 end cell end table close square brackets comma$ verify that (AB)^–1 = B"^–1 A^–1.

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223.

If $straight A to the power of negative 1 end exponent space equals space open square brackets table row 3 cell space space minus 1 end cell cell space space space space space 1 end cell row cell negative 15 end cell cell space space space space space 6 end cell cell space minus 5 end cell row 5 cell space minus 2 end cell cell space space space space 2 end cell end table close square brackets space space space and space straight B space equals space open square brackets table row cell space space space 1 end cell cell space space space space 2 end cell cell space space space minus 2 end cell row cell negative 1 end cell cell space space space space space 3 end cell cell space space space space space 0 end cell row cell space space 0 end cell cell space space minus 2 end cell cell space space space space space 1 end cell end table close square brackets comma$ find (AB)^–1.

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Multiple Choice Questions

224. Let A be a non-singular square matrix of order 3 × 3. Then | adj A| j is equal to

| A |
| A |²
| A |³
| A |³

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225.

If A is an invertible matrix of order 2, then det (A^–1) is equal to

det (A)
$fraction numerator 1 over denominator det space left parenthesis straight A right parenthesis end fraction$
1
1

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226. If x, y, z are non-zero real numbers, then the inverse of matrix

straight A space equals open square brackets table row straight x cell space space space space 0 end cell cell space space space 0 end cell row 0 cell space space space straight y end cell cell space space space 0 end cell row 0 cell space space space 0 end cell cell space space space straight z end cell end table close square brackets space is

$open square brackets table row cell straight x to the power of negative 1 end exponent end cell cell space space 0 end cell cell space space 0 end cell row 0 cell space space straight y to the power of negative 1 space end exponent end cell cell space space 0 end cell row 0 0 cell space straight z to the power of negative 1 end exponent end cell end table close square brackets$
$xyz open square brackets table row cell straight x to the power of negative 1 end exponent end cell cell space space 0 end cell cell space space space 0 end cell row 0 cell space space straight y to the power of negative 1 end exponent space end cell cell space space 0 end cell row 0 0 cell space straight z to the power of negative 1 end exponent end cell end table close square brackets$
$1 over xyz open square brackets table row straight x cell space space 0 end cell cell space space 0 end cell row 0 cell space straight y end cell cell space space 0 end cell row 0 cell space 0 end cell cell space space straight z end cell end table close square brackets$
$1 over xyz open square brackets table row straight x cell space space 0 end cell cell space space 0 end cell row 0 cell space straight y end cell cell space space 0 end cell row 0 cell space 0 end cell cell space space straight z end cell end table close square brackets$

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Short Answer Type

227.
Examine the consistency of the system of equations:
x + 2y = 2
2x + 3y = 3

The given equations are
x + 2y = 2
2x + 3y = 3
These equations can be written as
$open square brackets table row 1 cell space space space 2 end cell row 2 cell space space space 3 end cell end table close square brackets space open square brackets table row straight x row straight y end table close square brackets space equals space open square brackets table row 2 row 3 end table close square brackets$
$or space AX space equals space straight B space where space straight A space equals space open square brackets table row 1 cell space space 2 end cell row 2 cell space space 3 end cell end table close square brackets comma space space straight X space equals space open square brackets table row straight x row straight y end table close square brackets comma space space straight B space equals space open square brackets table row 2 row 3 end table close square brackets space space space space space space space space space space space space space space space space space space space space open vertical bar straight A close vertical bar space equals space open vertical bar table row 1 cell space space space 2 end cell row 2 cell space space space 3 end cell end table close vertical bar space equals space 3 minus 4 space equals space minus 1 space not equal to space 0 space$
∴ A^–1 exists
∴ system of equations has a unique solution
∴ system of equations is consistent.